Difference between revisions of "The Definite Integral"

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Suppose we are given a function and would like to determine the area underneath its graph over an interval. We could guess, but how could we figure out the exact area?  Below, using a few clever ideas, we actually ''define'' such an area and show that by using what is called the '''definite integral''' we can indeed determine the exact area underneath a curve.
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==Introduction==
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[[Image:Riemann_Integration_3.png|thumb|right|300px|Figure 1: Approximation of the area under the curve <math>f(x)</math> from <math>x=x_0</math> to <math>x=x_4</math>.]]
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[[Image:Riemann_Integration_2.png|thumb|right|300px|Figure 2: Rectangle approximating the area under the curve from <math>x_2</math> to <math>x_3</math> with sample point <math>x_3^*</math> .]]
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The rough idea of defining the area under the graph of <math>f</math> is to approximate this area with a finite number of rectangles. Since we can easily work out the area of the rectangles, we get an estimate of the area under the graph. If we use a larger number of smaller-sized rectangles we expect greater accuracy with respect to the area under the curve and hence a better approximation. Somehow, it seems that we could use our old friend from differentiation, the limit, and "approach" an infinite number of rectangles to get the exact area. Let's look at such an idea more closely.
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Suppose we have a function <math>f</math> that is positive on the interval <math>[a,b]</math> and we want to find the area <math>S</math> under <math>f</math> between <math>a</math> and <math>b</math> . Let's pick an integer <math>n</math> and divide the interval into <math>n</math> subintervals of equal width (see [[:File:Riemann_Integration_3.png|Figure 1]]). As the interval <math>[a,b]</math> has width <math>b-a</math>, each subinterval has width <math>\Delta x=\frac{b-a}{n}</math> . We denote the endpoints of the subintervals by <math>x_0,x_1,\ldots,x_n</math> . This gives us
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:<math>x_i=a+i\Delta x\text{ for }i=0,1,\dots,n</math>
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[[Image:Riemann2.gif|thumb|right|300px|Figure 3: Riemann sums with an increasing number of subdivisions yielding better approximations.]]
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Now for each <math>i=1,\ldots,n</math> pick a ''sample point'' <math>x_i^*</math> in the interval <math>[x_{i-1},x_i]</math> and consider the rectangle of height <math>f(x_i^*)</math> and width <math>\Delta x</math> (see [[:File:Riemann_Integration_2.png|Figure 2]]). The area of this rectangle is <math>f(x_i^*)\Delta x</math> . By adding up the area of all the rectangles for <math>i=1,\ldots,n</math> we get that the area <math>S</math> is approximated by
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:<math>A_n=f(x_1^*)\Delta x+\dots+f(x_n^*)\Delta x</math>
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A more convenient way to write this is with summation notation.
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:<math>A_n=\sum_{i=1}^n f(x_i^*)\Delta x</math>
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For each number <math>n</math> we get a different approximation. As <math>n</math> gets larger the width of the rectangles gets smaller which yields a better approximation (see [[:File:Riemann2.gif|Figure 3]]). In the limit of <math>A_n</math> as <math>n</math> tends to infinity we get the area <math>S</math> .
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==Definition of the Definite Integral==
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Suppose <math>f</math> is a continuous function on <math>[a,b]</math> and <math>\Delta x=\frac{b-a}{n}</math> . Then the ''definite integral'' of <math>f</math> between <math>a</math> and <math>b</math> is
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:<math>\int\limits_a^b f(x)dx=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x</math>
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where <math>x_i^*</math> are any sample points in the interval <math>[x_{i-1},x_i]</math> and <math>x_k=a+k\cdot\Delta x</math> for <math>k=0,\dots,n</math> .
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It is a fact that if <math>f</math> is continuous on <math>[a,b]</math> then this limit always exists and does not depend on the choice of the points <math>x_i^*\in[x_{i-1},x_i]</math> . For instance they may be evenly spaced, or distributed ambiguously throughout the interval. The proof of this is technical and is beyond the scope of this section.
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===Notation===
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When considering the expression, <math>\int\limits_a^b f(x)dx</math> (read "the integral from <math>a</math> to <math>b</math> of the <math>f</math> of <math>x</math> <math>dx</math>"), the function <math>f</math> is called the ''integrand'' and the interval <math>[a,b]</math> is the interval of integration. Also <math>a</math> is called the ''lower limit'' and <math>b</math> the ''upper limit'' of integration.
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[[Image:signed-area.png|thumb|right|300px|Figure 4: The integral gives the signed area under the graph.]]
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One important feature of this definition is that we also allow functions which take negative values. If <math>f(x)<0</math> for all <math>x</math> then  <math>f(x_i^*)<0</math> so <math>f(x_i^*)\Delta x<0</math> . So the definite integral of <math>f</math> will be strictly negative. More generally if <math>f</math> takes on both positive and negative values then <math>\int\limits_a^b f(x)dx</math> will be the area under the positive part of the graph of <math>f</math> '''minus''' the area above the graph of the negative part of the graph (see [[:File:Signed-area.png|Figure 4]]). For this reason we say that <math>\int\limits_a^b f(x)dx</math> is the '''signed area''' under the graph.
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===Independence of Variable===
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It is important to notice that the variable <math>x</math> did not play an important role in the definition of the integral. In fact we can replace it with any other letter, so the following are all equal:
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:<math>\int\limits_a^b f(x)dx=\int\limits_a^b f(t)dt=\int\limits_a^b f(u)du=\int\limits_a^b f(w)dw</math>
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Each of these is the signed area under the graph of <math>f</math> between <math>a</math> and <math>b</math> . Such a variable is often referred to as a '''dummy variable''' or a '''bound variable'''.
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===Left and Right Handed Riemann Sums===
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[[Image:Riemann-Sum-right-hand.png|thumb|350px|Figure 5: Right-handed Riemann sum]]
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[[Image:Riemann_Sum_Left_Hand.png|thumb|350px|Figure 6: Left-handed Riemann sum]]
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The following methods are sometimes referred to as L-RAM and R-RAM, RAM standing for "Rectangular Approximation Method."
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We could have decided to choose all our sample points <math>x_i^*</math> to be on the right hand side of the interval <math>[x_{i-1},x_i]</math> (see [[:File:Riemann-Sum-right-hand.png|Figure 5]]). Then <math>x_i^*=x_i</math> for all <math>i</math> and the approximation that we called <math>A_n</math> for the area becomes
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:<math>A_n=\sum_{i=1}^n f(x_i)\Delta x</math>
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This is called the ''right-handed Riemann sum'', and the integral is the limit
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:<math>\int\limits_a^b f(x)dx=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x</math>
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Alternatively we could have taken each sample point on the left hand side of the interval. In this case <math>x_i^*=x_{i-1}</math> (see [[:File:Riemann_Sum_Left_Hand.png|Figure 6]]) and the approximation becomes
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:<math>A_n=\sum_{i=1}^n f(x_{i-1})\Delta x</math>
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Then the integral of <math>f</math> is
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:<math>\int\limits_a^b f(x)dx=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\sum_{i=1}^n f(x_{i-1})\Delta x</math>
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The key point is that, as long as <math>f</math> is continuous, these two definitions give the same answer for the integral.
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===Examples===
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'''Example 1'''<br/>
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In this example we will calculate the area under the curve given by the graph of <math>f(x)=x</math> for <math>x</math> between 0 and 1. First we fix an integer <math>n</math> and divide the interval <math>[0,1]</math> into <math>n</math> subintervals of equal width. So each subinterval has width
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:<math>\Delta x=\frac{1}{n}</math>
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To calculate the integral we will use the right-handed Riemann sum. (We could have used the left-handed sum instead, and this would give the same answer in the end). For the right-handed sum the sample points are
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:<math>x_i^*=0+i\Delta x=\frac{i}{n}\quad i=1,\ldots,n</math>
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Notice that <math>f(x_i^*)=x_i^*=\frac{i}{n}</math> . Putting this into the formula for the approximation,
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:<math>A_n=\sum_{i=1}^n f(x_{i}^*)\Delta x=\sum_{i=1}^n f\left(\frac{i}{n}\right)\Delta x=\sum_{i=1}^n \frac{i}{n}\cdot\frac{1}{n}=\frac{1}{n^2}\sum_{i=1}^n i</math>
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Now we use the formula
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:<math>\sum_{i=1}^n i=\frac{n(n+1)}{2}</math>
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to get
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:<math>A_n=\frac{1}{n^2}\cdot\frac{n(n+1)}{2}=\frac{n+1}{2n}</math>
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To calculate the integral of <math>f</math> between <math>0</math> and <math>1</math> we take the limit as <math>n</math> tends to infinity,
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:<math>\int\limits_0^1 f(x)dx=\lim_{n\to\infty}\frac{n+1}{2n}=\frac{1}{2}</math>
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'''Example 2'''<br/>
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Next we show how to find the integral of the function <math>f(x)=x^2</math> between <math>x=a</math> and <math>x=b</math> . This time the interval <math>[a,b]</math> has width <math>b-a</math> so
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:<math>\Delta x=\frac{b-a}{n}</math>
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Once again we will use the right-handed Riemann sum. So the sample points we choose are
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:<math>x_i^*=a+i\Delta x=a+\frac{i(b-a)}{n}</math>
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Thus
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:{|
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|<math>A_n</math>
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|<math>=\sum_{i=1}^n f(x_{i}^*)\Delta x</math>
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|-
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|
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|<math>=\sum_{i=1}^n f\left(a+\frac{(b-a)i}{n}\right)\Delta x</math>
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|-
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|
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|<math>=\frac{b-a}{n}\sum_{i=1}^n \left(a+\frac{(b-a)i}{n}\right)^2</math>
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|-
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|
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|<math>=\frac{b-a}{n}\sum_{i=1}^n \left(a^2+\frac{2a(b-a)i}{n}+\frac{(b-a)^2i^2}{n^2}\right)</math>
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|}
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We have to calculate each piece on the right hand side of this equation. For the first two,
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:<math>\sum_{i=1}^n a^2=a^2\sum_{i=1}^n 1=na^2</math>
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:<math>\sum_{i=1}^n \frac{2a(b-a)i}{n}=\frac{2a(b-a)}{n}\sum_{i=1}^n i=\frac{2a(b-a)}{n}\cdot\frac{n(n+1)}{2}</math>
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For the third sum we have to use a formula
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:<math>\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}</math>
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to get
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:<math>\sum_{i=1}^n \frac{(b-a)^2i^2}{n^2}=\frac{(b-a)^2}{n^2}\frac{n(n+1)(2n+1)}{6}</math>
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Putting this together
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:<math>A_n=\frac{b-a}{n}\left(na^2+\frac{2a(b-a)}{n}\cdot\frac{n(n+1)}{2}+\frac{(b-a)^2}{n^2}\frac{n(n+1)(2n+1)}{6}\right)</math>
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Taking the limit as <math>n</math> tend to infinity gives
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:{|
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|<math>\int\limits_a^b x^2dx</math>
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|<math>=(b-a)\left(a^2+a(b-a)+\frac{1}{3}(b-a)^2\right)</math>
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|-
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|
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|<math>=(b-a)\left(a^2+ab-a^2+\frac{1}{3}(b^2-2ab+a^2)\right) </math>
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|-
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|
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|<math>=\frac{1}{3}(b-a)(b^2+ab+a^2)</math>
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|-
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|
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|<math>=\frac{1}{3}(b^3-a^3)</math>
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|}
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==Basic Properties of the Integral==
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From the definition of the integral we can deduce some basic properties. For all the following rules, suppose that <math>f</math> and <math>g</math> are continuous on <math>[a,b]</math> .
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===The Constant Rule===
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Constant Rule: <math>\int\limits_a^b c\cdot f(x)dx=c\int\limits_a^b f(x)dx</math>
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When <math>f</math>  is positive, the height of the function <math>c\cdot f</math> at a point <math>x</math> is <math>c</math> times the height of the function <math>f</math> .  So the area under <math>c\cdot f</math> between <math>a</math> and <math>b</math> is <math>c</math> times the area under <math>f</math> . We can also give a proof using the definition of the integral, using the constant rule for limits,
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:<math>\int\limits_a^b c\cdot f(x)dx=\lim_{n\to\infty}\sum_{i=1}^n c\cdot f(x_i^*)=c\cdot\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)=c\int\limits_a^b f(x)dx</math>
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'''Example'''
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We saw in the previous section that
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:<math>\int\limits_0^1 xdx=\frac{1}{2}</math>
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Using the constant rule we can use this to calculate that
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:<math>\int\limits_0^1 3xdx=3\int\limits_0^1 xdx=3\cdot\frac{1}{2}=1.5</math> ,
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:<math>\int\limits_0^1 -7xdx=-7\int\limits_0^1 xdx=(-7)\cdot\frac{1}{2}=-3.5</math> .
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'''Example'''
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We saw in the previous section that
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:<math>\int\limits_a^b x^2dx=\frac{b^3-a^3}{3}</math>
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We can use this and the constant rule to calculate that
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:<math>\int\limits_1^3 2x^2 dx=2\int\limits_1^3 x^2dx=2\cdot\frac{1}{3}\cdot(3^3-1^3)=\frac{2}{3}(27-1)=\frac{52}{3}</math>
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There is a special case of this rule used for integrating constants:
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If <math>c</math> is constant then <math>\int\limits_a^b c\ dx=c(b-a)</math>
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When <math>c>0</math> and <math>a<b</math> this integral is the area of a rectangle of height <math>c</math> and width <math>b-a</math> which equals <math>c(b-a)</math> .
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'''Example'''
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:<math>\int\limits_1^3 9dx=9(3-1)=9\cdot 2=18</math>
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:<math>\int\limits_{-2}^6 11dx=11(6-(-2))=11\cdot 8=88</math>
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:<math>\int\limits_2^{17} 0dx=0\cdot(17-2)=0</math>
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===The addition and subtraction rule===
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Addition and Subtraction Rules of Integration:
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<math>\int\limits_a^b \bigl(f(x)+g(x)\bigr)dx=\int\limits_a^b f(x)dx+\int\limits_a^b g(x)dx</math>
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<math>\int\limits_a^b \bigl(f(x)-g(x)\bigr)dx=\int\limits_a^b f(x)dx-\int\limits_a^b g(x)dx</math>
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As with the constant rule, the addition rule follows from the addition rule for limits:
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:{|
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|<math>\int\limits_a^b \bigl(f(x)+g(x)\bigr)dx</math>
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|<math>=\lim_{n\to\infty}\sum_{i=1}^n \Big(f(x_i^*)+g(x_i^*)\Big)</math>
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|-
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|
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|<math>=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)+\lim_{n\to\infty}\sum_{i=1}^n g(x_i^*)</math>
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|-
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|
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|<math>=\int\limits_a^b f(x)dx+\int\limits_a^b g(x)dx</math>
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|}
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The subtraction rule can be proved in a similar way.
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'''Example'''
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From above <math>\int\limits_1^3 9dx=18</math> and <math>\int\limits_1^3 2x^2dx=\frac{52}{3}</math> so
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:<math>\int\limits_1^3 (2x^2+9)dx=\int\limits_1^3 2x^2dx+\int\limits_1^3 9dx=\frac{52}{3}+18=\frac{106}{3}</math>
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:<math>\int\limits_1^3 (2x^2-9)dx=\int\limits_1^3 2x^2dx-\int\limits_1^3 9dx=\frac{52}{3}-18=-\frac{2}{3}</math>
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'''Example'''
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:<math>\int\limits_0^2 (4x^2+14)dx=4\int\limits_0^2 x^2dx+\int\limits_0^2 14dx=4\cdot\frac{1}{3}(2^3-0^3)+2\cdot 14=\frac{32}{3}+28=\frac{116}{3}</math>
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===The Comparison Rule===
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[[Image:Integral_comparison_1.png|thumb|300px|right|Figure 7: Bounding the area under <math>f(x)</math> on <math>[a,b]</math>]]
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Comparison Rule:
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*Suppose <math>f(x)\ge 0</math> for all <math>x\in[a,b]</math> . Then
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:<math>\int\limits_a^b f(x)dx\ge 0</math>
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*Suppose <math>f(x)\ge g(x)</math> for all <math>x\in[a,b]</math> . Then
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:<math>\int\limits_a^b f(x)dx\ge\int\limits_a^b g(x)dx</math>
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*Suppose <math>M\ge f(x)\ge m</math> for all <math>x\in[a,b]</math> . Then
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:<math>M(b-a)\ge\int\limits_a^b f(x)dx\ge m(b-a)</math>
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If <math>f(x)\ge 0</math> then each of the rectangles in the Riemann sum to calculate the integral of <math>f</math> will be above the <math>y</math> axis, so the area will be non-negative. If <math>f(x)\ge g(x)</math> then <math>f(x)-g(x)\ge 0</math> and by the first property we get the second property. Finally if <math>M\ge f(x)\ge m</math> then the area under the graph of <math>f</math> will be greater than the area of rectangle with height <math>m</math> and less than the area of the rectangle with height <math>M</math> (see [[:File:Integral_comparison_1.png|Figure 7]]).  So
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:<math>M(b-a)=\int\limits_a^b M\ge\int\limits_a^b f(x)dx\ge\int\limits_a^b m=m(b-a)</math>
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===Linearity with respect to endpoints===
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Additivity with respect to endpoints:
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Suppose <math>a<c<b</math> . Then
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:<math>\int\limits_a^b f(x)dx=\int\limits_a^c f(x)dx+\int\limits_c^b f(x)dx</math>
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 +
Again suppose that <math>f</math> is positive. Then this property should be interpreted as saying that the area under the graph of <math>f</math> between <math>a</math> and <math>b</math> is the area between <math>a</math> and <math>c</math> plus the area between <math>c</math> and <math>b</math> (see [[:File:Integral_linear_endpoints.png|Figure 8]]).
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[[Image:Integral_linear_endpoints.png|thumb|300px|right|Figure 8:  Illustration of the property of additivity with respect to endpoints]]
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{{Calculus/Def|text= '''Extension of Additivity with respect to limits of integration'''<br>
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When <math>a=b</math> we have that <math>\Delta x=\frac{b-a}{n}=0</math> so
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:<math>\int\limits_a^a f(x)dx=0</math>
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Also in defining the integral we assumed that <math>a<b</math> . But the definition makes sense even when <math>b<a</math> , in which case <math>\Delta x=\frac{b-a}{n}</math> has changed sign. This gives
 +
:<math>\int\limits_b^a f(x)dx=-\int\limits_a^b f(x)dx</math>
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With these definitions,
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:<math>\int\limits_a^b f(x)dx=\int\limits_a^c f(x)dx+\int\limits_c^b f(x)dx</math>
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whatever the order of <math>a,b,c</math> .
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===Even and odd functions===
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Recall that a function <math>f</math> is called odd if it satisfies <math>f(-x)=-f(x)</math> and is called even if <math>f(-x)=f(x)</math> .
 +
 +
Suppose <math>f</math> is a continuous odd function. Then for any <math>a</math>,
 +
 +
:<math>\int\limits_{-a}^a f(x)dx=0</math>
 +
 +
If <math>f</math> is a continuous even function then for any <math>a</math>,
 +
 +
:<math>\int\limits_{-a}^a f(x)dx=2\int\limits_0^a f(x)dx</math>
 +
 +
Suppose <math>f</math> is an odd function and consider first just the integral from <math>-a</math> to <math>0</math> . We make the substitution <math>u=-x</math> so <math>du=-dx</math> . Notice that if <math>x=-a</math> then <math>u=a</math> and if <math>x=0</math> then <math>u=0</math> . Hence
 +
:<math>\int\limits_{-a}^0 f(x)dx=-\int\limits_a^0 f(-u)du=\int\limits_0^a f(-u)du</math> .
 +
Now as <math>f</math> is odd, <math> f(-u)=-f(u)</math> so the integral becomes
 +
:<math>\int\limits_{-a}^0 f(x)dx=-\int\limits_0^a f(u) du</math> .
 +
Now we can replace the dummy variable <math>u</math> with any other variable. So we can replace it with the letter <math>x</math> to give
 +
:<math>\int\limits_{-a}^0 f(x)dx=-\int\limits_0^a f(u)du=-\int\limits_0^a f(x)dx</math> .
 +
 +
Now we split the integral into two pieces
 +
:<math>\int\limits_{-a}^a f(x)dx=\int\limits_{-a}^0 f(x)dx+\int\limits_0^a f(x)dx=-\int\limits_0^a f(x)dx+\int\limits_0^a f(x)dx=0</math> .
 +
 +
The proof of the formula for even functions is similar.
 +
 +
Prove that if <math>f</math> is a continuous even function then for any <math>a</math> ,
 +
:<math>\int\limits_{-a}^a f(x)dx=2\int\limits_0^a f(x)dx</math>.
 +
 +
From the property of linearity of the endpoints we have<br/>
 +
:<math>\int\limits_{-a}^a f(x)dx=\int\limits_{-a}^0 f(x)dx+\int\limits_0^a f(x)dx</math>
 +
Make the substitution <math>u=-x;\quad du=-dx</math> . <math>u=a</math> when <math>x=-a</math> and <math>u=0</math> when <math>x=0</math> . Then
 +
:<math>\int\limits_{-a}^0 f(x)dx=\int\limits_a^0 f(-u)(-du)=-\int\limits_a^0 f(-u)du=\int\limits_0^a f(-u)du=\int\limits_0^a f(u)du</math>
 +
where the last step has used the evenness of <math>f</math> . Since <math>u</math> is just a dummy variable, we can replace it with <math>x</math> . Then
 +
:<math>\int\limits_{-a}^a f(x)dx=\int\limits_0^a f(x)dx+\int\limits_0^a f(x)dx=2\int\limits_0^a f(x)dx</math>
 +
 +
==Resources==
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1193/The%20Definite%20Integral/Presentation11_Distance_Definite_Integral.pptx Distance Definite Integral] (Slides 23-38). PowerPoint file created by Professor Cynthia Roberts, UTSA.
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1193/The%20Definite%20Integral/Presentation11_Distance_Definite_Integral.pptx Distance Definite Integral] (Slides 23-38). PowerPoint file created by Professor Cynthia Roberts, UTSA.
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1193/The%20Definite%20Integral/Presentation12_DefiniteIntegral%20&%20Antiderivatives.pptx Definite Integral & Antiderivatives]. PowerPoint file created by Professor Cynthia Roberts, UTSA.
 
* [https://mathresearch.utsa.edu/wikiFiles/MAT1193/The%20Definite%20Integral/Presentation12_DefiniteIntegral%20&%20Antiderivatives.pptx Definite Integral & Antiderivatives]. PowerPoint file created by Professor Cynthia Roberts, UTSA.
 +
 +
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Definite%20Integral/MAT1214-5.3TheDefiniteIntegralPwPt.pptx The Definite Integral] PowerPoint file created by Dr. Sara Shirinkam, UTSA.
 +
 +
* [https://mathresearch.utsa.edu/wikiFiles/MAT1214/The%20Definite%20Integral/MAT1214-5.3TheDefiniteIntegralNotes.pdf The Definite Integral Notes] created by Professor Eduardo Duenez, UTSA.
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 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [https://en.wikibooks.org/wiki/Calculus/Definite_integral Definite integral, Wikibooks: Calculus] under a CC BY-SA license

Latest revision as of 10:25, 28 October 2021

Suppose we are given a function and would like to determine the area underneath its graph over an interval. We could guess, but how could we figure out the exact area? Below, using a few clever ideas, we actually define such an area and show that by using what is called the definite integral we can indeed determine the exact area underneath a curve.

Introduction

Figure 1: Approximation of the area under the curve from to .
Figure 2: Rectangle approximating the area under the curve from to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_3} with sample point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_3^*} .

The rough idea of defining the area under the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is to approximate this area with a finite number of rectangles. Since we can easily work out the area of the rectangles, we get an estimate of the area under the graph. If we use a larger number of smaller-sized rectangles we expect greater accuracy with respect to the area under the curve and hence a better approximation. Somehow, it seems that we could use our old friend from differentiation, the limit, and "approach" an infinite number of rectangles to get the exact area. Let's look at such an idea more closely.

Suppose we have a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} that is positive on the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} and we want to find the area Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} under Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} . Let's pick an integer Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} and divide the interval into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} subintervals of equal width (see Figure 1). As the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} has width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b-a} , each subinterval has width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{b-a}{n}} . We denote the endpoints of the subintervals by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0,x_1,\ldots,x_n} . This gives us

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i=a+i\Delta x\text{ for }i=0,1,\dots,n}
Figure 3: Riemann sums with an increasing number of subdivisions yielding better approximations.

Now for each Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i=1,\ldots,n} pick a sample point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*} in the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [x_{i-1},x_i]} and consider the rectangle of height Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_i^*)} and width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x} (see Figure 2). The area of this rectangle is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_i^*)\Delta x} . By adding up the area of all the rectangles for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i=1,\ldots,n} we get that the area Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} is approximated by

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=f(x_1^*)\Delta x+\dots+f(x_n^*)\Delta x}

A more convenient way to write this is with summation notation.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=\sum_{i=1}^n f(x_i^*)\Delta x}

For each number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} we get a different approximation. As Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} gets larger the width of the rectangles gets smaller which yields a better approximation (see Figure 3). In the limit of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n} as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} tends to infinity we get the area Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S} .

Definition of the Definite Integral

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a continuous function on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{b-a}{n}} . Then the definite integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*} are any sample points in the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [x_{i-1},x_i]} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_k=a+k\cdot\Delta x} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=0,\dots,n} .

It is a fact that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} then this limit always exists and does not depend on the choice of the points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*\in[x_{i-1},x_i]} . For instance they may be evenly spaced, or distributed ambiguously throughout the interval. The proof of this is technical and is beyond the scope of this section.

Notation

When considering the expression, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx} (read "the integral from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} to of the of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dx} "), the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is called the integrand and the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} is the interval of integration. Also Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} is called the lower limit and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} the upper limit of integration.

Figure 4: The integral gives the signed area under the graph.

One important feature of this definition is that we also allow functions which take negative values. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)<0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_i^*)<0} so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_i^*)\Delta x<0} . So the definite integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} will be strictly negative. More generally if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} takes on both positive and negative values then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx} will be the area under the positive part of the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} minus the area above the graph of the negative part of the graph (see Figure 4). For this reason we say that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx} is the signed area under the graph.

Independence of Variable

It is important to notice that the variable Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} did not play an important role in the definition of the integral. In fact we can replace it with any other letter, so the following are all equal:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx=\int\limits_a^b f(t)dt=\int\limits_a^b f(u)du=\int\limits_a^b f(w)dw}

Each of these is the signed area under the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} . Such a variable is often referred to as a dummy variable or a bound variable.

Left and Right Handed Riemann Sums

Figure 5: Right-handed Riemann sum
Figure 6: Left-handed Riemann sum

The following methods are sometimes referred to as L-RAM and R-RAM, RAM standing for "Rectangular Approximation Method."

We could have decided to choose all our sample points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*} to be on the right hand side of the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [x_{i-1},x_i]} (see Figure 5). Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*=x_i} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i} and the approximation that we called Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n} for the area becomes

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=\sum_{i=1}^n f(x_i)\Delta x}

This is called the right-handed Riemann sum, and the integral is the limit

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x}

Alternatively we could have taken each sample point on the left hand side of the interval. In this case Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*=x_{i-1}} (see Figure 6) and the approximation becomes

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=\sum_{i=1}^n f(x_{i-1})\Delta x}

Then the integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx=\lim_{n\to\infty}A_n=\lim_{n\to\infty}\sum_{i=1}^n f(x_{i-1})\Delta x}

The key point is that, as long as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous, these two definitions give the same answer for the integral.

Examples

Example 1
In this example we will calculate the area under the curve given by the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x} for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} between 0 and 1. First we fix an integer Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} and divide the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,1]} into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} subintervals of equal width. So each subinterval has width

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{1}{n}}

To calculate the integral we will use the right-handed Riemann sum. (We could have used the left-handed sum instead, and this would give the same answer in the end). For the right-handed sum the sample points are

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*=0+i\Delta x=\frac{i}{n}\quad i=1,\ldots,n}

Notice that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x_i^*)=x_i^*=\frac{i}{n}} . Putting this into the formula for the approximation,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=\sum_{i=1}^n f(x_{i}^*)\Delta x=\sum_{i=1}^n f\left(\frac{i}{n}\right)\Delta x=\sum_{i=1}^n \frac{i}{n}\cdot\frac{1}{n}=\frac{1}{n^2}\sum_{i=1}^n i}

Now we use the formula

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^n i=\frac{n(n+1)}{2}}

to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=\frac{1}{n^2}\cdot\frac{n(n+1)}{2}=\frac{n+1}{2n}}

To calculate the integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} we take the limit as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} tends to infinity,

Example 2
Next we show how to find the integral of the function between and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=b} . This time the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} has width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b-a} so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{b-a}{n}}

Once again we will use the right-handed Riemann sum. So the sample points we choose are

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_i^*=a+i\Delta x=a+\frac{i(b-a)}{n}}

Thus

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sum_{i=1}^n f(x_{i}^*)\Delta x}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sum_{i=1}^n f\left(a+\frac{(b-a)i}{n}\right)\Delta x}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{b-a}{n}\sum_{i=1}^n \left(a+\frac{(b-a)i}{n}\right)^2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{b-a}{n}\sum_{i=1}^n \left(a^2+\frac{2a(b-a)i}{n}+\frac{(b-a)^2i^2}{n^2}\right)}

We have to calculate each piece on the right hand side of this equation. For the first two,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^n a^2=a^2\sum_{i=1}^n 1=na^2}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^n \frac{2a(b-a)i}{n}=\frac{2a(b-a)}{n}\sum_{i=1}^n i=\frac{2a(b-a)}{n}\cdot\frac{n(n+1)}{2}}

For the third sum we have to use a formula

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}}

to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{i=1}^n \frac{(b-a)^2i^2}{n^2}=\frac{(b-a)^2}{n^2}\frac{n(n+1)(2n+1)}{6}}

Putting this together

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_n=\frac{b-a}{n}\left(na^2+\frac{2a(b-a)}{n}\cdot\frac{n(n+1)}{2}+\frac{(b-a)^2}{n^2}\frac{n(n+1)(2n+1)}{6}\right)}

Taking the limit as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} tend to infinity gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b x^2dx} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(b-a)\left(a^2+a(b-a)+\frac{1}{3}(b-a)^2\right)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =(b-a)\left(a^2+ab-a^2+\frac{1}{3}(b^2-2ab+a^2)\right) }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{3}(b-a)(b^2+ab+a^2)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\frac{1}{3}(b^3-a^3)}

Basic Properties of the Integral

From the definition of the integral we can deduce some basic properties. For all the following rules, suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} are continuous on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} .

The Constant Rule

Constant Rule: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b c\cdot f(x)dx=c\int\limits_a^b f(x)dx}

When Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is positive, the height of the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\cdot f} at a point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} times the height of the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} . So the area under Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\cdot f} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} times the area under Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} . We can also give a proof using the definition of the integral, using the constant rule for limits,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b c\cdot f(x)dx=\lim_{n\to\infty}\sum_{i=1}^n c\cdot f(x_i^*)=c\cdot\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)=c\int\limits_a^b f(x)dx}

Example

We saw in the previous section that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_0^1 xdx=\frac{1}{2}}

Using the constant rule we can use this to calculate that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_0^1 3xdx=3\int\limits_0^1 xdx=3\cdot\frac{1}{2}=1.5} ,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_0^1 -7xdx=-7\int\limits_0^1 xdx=(-7)\cdot\frac{1}{2}=-3.5} .

Example

We saw in the previous section that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b x^2dx=\frac{b^3-a^3}{3}}

We can use this and the constant rule to calculate that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^3 2x^2 dx=2\int\limits_1^3 x^2dx=2\cdot\frac{1}{3}\cdot(3^3-1^3)=\frac{2}{3}(27-1)=\frac{52}{3}}

There is a special case of this rule used for integrating constants:

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is constant then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b c\ dx=c(b-a)}

When Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c>0} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a<b} this integral is the area of a rectangle of height Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} and width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b-a} which equals .

Example

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-2}^6 11dx=11(6-(-2))=11\cdot 8=88}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_2^{17} 0dx=0\cdot(17-2)=0}

The addition and subtraction rule

Addition and Subtraction Rules of Integration:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b \bigl(f(x)+g(x)\bigr)dx=\int\limits_a^b f(x)dx+\int\limits_a^b g(x)dx} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b \bigl(f(x)-g(x)\bigr)dx=\int\limits_a^b f(x)dx-\int\limits_a^b g(x)dx}

As with the constant rule, the addition rule follows from the addition rule for limits:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b \bigl(f(x)+g(x)\bigr)dx} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lim_{n\to\infty}\sum_{i=1}^n \Big(f(x_i^*)+g(x_i^*)\Big)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)+\lim_{n\to\infty}\sum_{i=1}^n g(x_i^*)}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\int\limits_a^b f(x)dx+\int\limits_a^b g(x)dx}

The subtraction rule can be proved in a similar way.

Example

From above Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^3 9dx=18} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^3 2x^2dx=\frac{52}{3}} so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^3 (2x^2+9)dx=\int\limits_1^3 2x^2dx+\int\limits_1^3 9dx=\frac{52}{3}+18=\frac{106}{3}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^3 (2x^2-9)dx=\int\limits_1^3 2x^2dx-\int\limits_1^3 9dx=\frac{52}{3}-18=-\frac{2}{3}}

Example

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_0^2 (4x^2+14)dx=4\int\limits_0^2 x^2dx+\int\limits_0^2 14dx=4\cdot\frac{1}{3}(2^3-0^3)+2\cdot 14=\frac{32}{3}+28=\frac{116}{3}}

The Comparison Rule

Figure 7: Bounding the area under Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]}

Comparison Rule:

  • Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\ge 0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in[a,b]} . Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx\ge 0}
  • Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\ge g(x)} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in[a,b]} . Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx\ge\int\limits_a^b g(x)dx}
  • Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ge f(x)\ge m} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\in[a,b]} . Then
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M(b-a)\ge\int\limits_a^b f(x)dx\ge m(b-a)}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\ge 0} then each of the rectangles in the Riemann sum to calculate the integral of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} will be above the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} axis, so the area will be non-negative. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\ge g(x)} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)-g(x)\ge 0} and by the first property we get the second property. Finally if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M\ge f(x)\ge m} then the area under the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} will be greater than the area of rectangle with height Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} and less than the area of the rectangle with height Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M} (see Figure 7). So

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle M(b-a)=\int\limits_a^b M\ge\int\limits_a^b f(x)dx\ge\int\limits_a^b m=m(b-a)}

Linearity with respect to endpoints

Additivity with respect to endpoints:

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a<c<b} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx=\int\limits_a^c f(x)dx+\int\limits_c^b f(x)dx}

Again suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is positive. Then this property should be interpreted as saying that the area under the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} is the area between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} plus the area between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} (see Figure 8).

Figure 8: Illustration of the property of additivity with respect to endpoints

{{Calculus/Def|text= Extension of Additivity with respect to limits of integration
When Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=b} we have that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{b-a}{n}=0} so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^a f(x)dx=0}

Also in defining the integral we assumed that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a<b} . But the definition makes sense even when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b<a} , in which case has changed sign. This gives

With these definitions,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^b f(x)dx=\int\limits_a^c f(x)dx+\int\limits_c^b f(x)dx}

whatever the order of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a,b,c} .

Even and odd functions

Recall that a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is called odd if it satisfies Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(-x)=-f(x)} and is called even if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(-x)=f(x)} .

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a continuous odd function. Then for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^a f(x)dx=0}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a continuous even function then for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^a f(x)dx=2\int\limits_0^a f(x)dx}

Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is an odd function and consider first just the integral from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -a} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} . We make the substitution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=-x} so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-dx} . Notice that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-a} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=a} and if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=0} . Hence

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^0 f(x)dx=-\int\limits_a^0 f(-u)du=\int\limits_0^a f(-u)du} .

Now as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is odd, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(-u)=-f(u)} so the integral becomes

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^0 f(x)dx=-\int\limits_0^a f(u) du} .

Now we can replace the dummy variable Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} with any other variable. So we can replace it with the letter Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} to give

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^0 f(x)dx=-\int\limits_0^a f(u)du=-\int\limits_0^a f(x)dx} .

Now we split the integral into two pieces

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^a f(x)dx=\int\limits_{-a}^0 f(x)dx+\int\limits_0^a f(x)dx=-\int\limits_0^a f(x)dx+\int\limits_0^a f(x)dx=0} .

The proof of the formula for even functions is similar.

Prove that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a continuous even function then for any Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^a f(x)dx=2\int\limits_0^a f(x)dx} .

From the property of linearity of the endpoints we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^a f(x)dx=\int\limits_{-a}^0 f(x)dx+\int\limits_0^a f(x)dx}

Make the substitution Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=-x;\quad du=-dx} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=a} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=0} when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^0 f(x)dx=\int\limits_a^0 f(-u)(-du)=-\int\limits_a^0 f(-u)du=\int\limits_0^a f(-u)du=\int\limits_0^a f(u)du}

where the last step has used the evenness of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} is just a dummy variable, we can replace it with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-a}^a f(x)dx=\int\limits_0^a f(x)dx+\int\limits_0^a f(x)dx=2\int\limits_0^a f(x)dx}

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