The Fundamental Theorem

The fundamental theorem of calculus is a theorem that links the concept of differentiating a function (calculating the gradient) with the concept of integrating a function (calculating the area under the curve). The two operations are inverses of each other apart from a constant value which depends where one starts to compute area.

The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also known as an indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound of integration. This implies the existence of antiderivatives for continuous functions.

Conversely, the second part of the theorem, sometimes called the second fundamental theorem of calculus, states that the integral of a function f over some interval can be computed by using any one, say F, of its infinitely many antiderivatives. This part of the theorem has key practical applications, because explicitly finding the antiderivative of a function by symbolic integration avoids numerical integration to compute integrals.

Geometric meaning

The area shaded in red stripes is close to h times f(x). Alternatively, if the function A(x) were known, this area would be exactly A(x + h) − A(x). These two values are approximately equal, particularly for small h.

For a continuous function $y=f(x)$ whose graph is plotted as a curve, each value of x has a corresponding area function A(x), representing the area beneath the curve between 0 and x. The function A(x) may not be known, but it is given that it represents the area under the curve.

The area under the curve between x and x + h could be computed by finding the area between 0 and x + h, then subtracting the area between 0 and x. In other words, the area of this “strip” would be A(x + h) − A(x).

There is another way to estimate the area of this same strip. As shown in the accompanying figure, h is multiplied by f(x) to find the area of a rectangle that is approximately the same size as this strip. So:

$A(x+h)-A(x)\approx f(x)\cdot h$

In fact, this estimate becomes a perfect equality if we add the red portion of the "excess" area shown in the diagram. So:

$A(x+h)-A(x)=f(x)\cdot h+({\text{Red Excess}})$

Rearranging terms:

$f(x)={\frac {A(x+h)-A(x)}{h}}-{\frac {\text{Red Excess}}{h}}$.

As h approaches 0 in the limit, the last fraction can be shown to go to zero. This is true because the area of the red portion of excess region is less than or equal to the area of the tiny black-bordered rectangle. More precisely,

$\left|f(x)-{\frac {A(x+h)-A(x)}{h}}\right|={\frac {|{\text{Red Excess}}|}{h}}\leq {\frac {h(f(x+h_{1})-f(x+h_{2}))}{h}}=f(x+h_{1})-f(x+h_{2}),$

where $x+h_{1}$ and $x+h_{2}$ are points where f reaches its maximum and its minimum, respectively, in the interval [x, x + h]. By the continuity of f, the latter expression tends to zero as h does. Therefore, the left-hand side tends to zero as h does, which implies

$f(x)=\lim _{h\to 0}{\frac {A(x+h)-A(x)}{h}}.$

This implies $f(x)=A'(x)$. That is, the derivative of the area function A(x) exists and is the original function f(x); so, the area function is simply an antiderivative of the original function. Computing the derivative of a function and finding the area under its curve are "opposite" operations. This is the crux of the Fundamental Theorem of Calculus.

Physical intuition

Intuitively, the theorem states that the sum of infinitesimal changes in a quantity over time (or over some other variable) adds up to the net change in the quantity.

Imagine for example using a stopwatch to mark-off tiny increments of time as a car travels down a highway. Imagine also looking at the car's speedometer as it travels, so that at every moment you know the velocity of the car. To understand the power of this theorem, imagine also that you are not allowed to look out of the window of the car, so that you have no direct evidence of how far the car has traveled.

For any tiny interval of time in the car, you could calculate how far the car has traveled in that interval by multiplying the current speed of the car times the length of that tiny interval of time. (This is because distance = speed $\times$ time.)

Now imagine doing this instant after instant, so that for every tiny interval of time you know how far the car has traveled. In principle, you could then calculate the total distance traveled in the car (even though you've never looked out of the window) by summing-up all those tiny distances.

distance traveled = $\sum$ the velocity at any instant $\times$ a tiny interval of time

In other words,

distance traveled = $\sum v(t)\times \Delta t$

On the right hand side of this equation, as $\Delta t$ becomes infinitesimally small, the operation of "summing up" corresponds to integration. So what we've shown is that the integral of the velocity function can be used to compute how far the car has traveled.

Now remember that the velocity function is the derivative of the position function. So what we have really shown is that integrating the velocity recovers the original position function. This is the basic idea of the theorem: that integration and differentiation are closely related operations, each essentially being the inverse of the other.

In other words, in terms of one's physical intuition, the theorem states that the sum of the changes in a quantity over time (such as position, as calculated by multiplying velocity times time) adds up to the total net change in the quantity. Or to put this more generally:

• Given a quantity $x$ that changes over some variable $t$, and
• Given the velocity $v(t)$ with which that quantity changes over that variable

then the idea that "distance equals speed times time" corresponds to the statement

$dx=v(t)dt$

meaning that one can recover the original function $x(t)$ by integrating its derivative, the velocity $v(t)$, over $t$.

Formal statements

There are two parts to the theorem. The first part deals with the derivative of an antiderivative, while the second part deals with the relationship between antiderivatives and definite integrals.

First part

This part is sometimes referred to as the first fundamental theorem of calculus.

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by

$F(x)=\int _{a}^{x}f(t)\,dt.$

Then F is uniformly continuous on [a, b] and differentiable on the open interval (a, b), and

$F'(x)=f(x)$

for all x in (a, b).

Corollary

Fundamental theorem of calculus (animation)

The fundamental theorem is often employed to compute the definite integral of a function $f$ for which an antiderivative $F$ is known. Specifically, if $f$ is a real-valued continuous function on $[a,b]$ and $F$ is an antiderivative of $f$ in $[a,b]$ then

$\int _{a}^{b}f(t)\,dt=F(b)-F(a).$

The corollary assumes continuity on the whole interval. This result is strengthened slightly in the following part of the theorem.

Second part

This part is sometimes referred to as the second fundamental theorem of calculus or the Newton–Leibniz axiom.

Let $f$ be a real-valued function on a closed interval $[a,b]$ and $F$ an antiderivative of $f$ in $(a,b)$:

$F'(x)=f(x).$

If $f$ is Riemann integrable on $[a,b]$ then

$\int _{a}^{b}f(x)\,dx=F(b)-F(a).$

The second part is somewhat stronger than the corollary because it does not assume that $f$ is continuous.

When an antiderivative $F$ exists, then there are infinitely many antiderivatives for $f$, obtained by adding an arbitrary constant to $F$. Also, by the first part of the theorem, antiderivatives of $f$ always exist when $f$ is continuous.

Proof of the first part

For a given f(t), define the function F(x) as

$F(x)=\int _{a}^{x}f(t)\,dt.$

For any two numbers x₁ and x₁ + Δx in [a, b], we have

$F(x_{1})=\int _{a}^{x_{1}}f(t)\,dt$

and

$F(x_{1}+\Delta x)=\int _{a}^{x_{1}+\Delta x}f(t)\,dt.$

Subtracting the two equalities gives

$F(x_{1}+\Delta x)-F(x_{1})=\int _{a}^{x_{1}+\Delta x}f(t)\,dt-\int _{a}^{x_{1}}f(t)\,dt.\qquad (1)$

It can be shown that

$\int _{a}^{x_{1}}f(t)\,dt+\int _{x_{1}}^{x_{1}+\Delta x}f(t)\,dt=\int _{a}^{x_{1}+\Delta x}f(t)\,dt.$

(The sum of the areas of two adjacent regions is equal to the area of both regions combined.)

Manipulating this equation gives

$\int _{a}^{x_{1}+\Delta x}f(t)\,dt-\int _{a}^{x_{1}}f(t)\,dt=\int _{x_{1}}^{x_{1}+\Delta x}f(t)\,dt.$

Substituting the above into (1) results in

$F(x_{1}+\Delta x)-F(x_{1})=\int _{x_{1}}^{x_{1}+\Delta x}f(t)\,dt.\qquad (2)$

According to the mean value theorem for integration, there exists a real number $c\in [x_{1},x_{1}+\Delta x]$ such that

$\int _{x_{1}}^{x_{1}+\Delta x}f(t)\,dt=f(c)\cdot \Delta x.$

To keep the notation simple, we write just $c$, but one should keep in mind that, for a given function $f$, the value of $c$ depends on $x_{1}$ and on $\Delta x,$ but is always confined to the interval $[x_{1},x_{1}+\Delta x]$. Substituting the above into (2) we get

$F(x_{1}+\Delta x)-F(x_{1})=f(c)\cdot \Delta x.$

Dividing both sides by $\Delta x$ gives

${\frac {F(x_{1}+\Delta x)-F(x_{1})}{\Delta x}}=f(c).$

The expression on the left side of the equation is Newton's difference quotient for F at x₁.

Take the limit as $\Delta x$ → 0 on both sides of the equation.

$\lim _{\Delta x\to 0}{\frac {F(x_{1}+\Delta x)-F(x_{1})}{\Delta x}}=\lim _{\Delta x\to 0}f(c).$

The expression on the left side of the equation is the definition of the derivative of F at x₁.

$F'(x_{1})=\lim _{\Delta x\to 0}f(c).\qquad (3)$

To find the other limit, we use the squeeze theorem. The number c is in the interval [x₁, x₁ + Δx], so x₁ ≤ c ≤ x₁ + Δx.

Also, $\lim _{\Delta x\to 0}x_{1}=x_{1}$ and $\lim _{\Delta x\to 0}x_{1}+\Delta x=x_{1}.$

Therefore, according to the squeeze theorem,

$\lim _{\Delta x\to 0}c=x_{1}.$

The function f is continuous at x₁, the limit can be taken inside the function:

$\lim _{\Delta x\to 0}f(c)=f(x_{1}).$

Substituting into (3), we get

$F'(x_{1})=f(x_{1}).$

which completes the proof.

Proof of the corollary

Suppose F is an antiderivative of f, with f continuous on [a, b]. Let

$G(x)=\int _{a}^{x}f(t)\,dt$.

By the first part of the theorem, we know G is also an antiderivative of f. Since F′ − G′ = 0 the mean value theorem implies that F − G is a constant function, that is, there is a number c such that $G(x)=F(x)+c$ for all x in [a, b]. Letting x = a, we have

$F(a)+c=G(a)=\int _{a}^{a}f(t)\,dt=0,$

which means $c=-F(a)$. In other words, $G(x)=F(x)-F(a)$, and so

$\int _{a}^{b}f(x)\,dx=G(b)=F(b)-F(a).$

Proof of the second part

This is a limit proof by Riemann sums. Let f be (Riemann) integrable on the interval [a, b], and let f admit an antiderivative F on [a, b]. Begin with the quantity F(b) − F(a). Let there be numbers x₁, ..., x_n such that

$a=x_{0}<x_{1}<x_{2}<\cdots <x_{n-1}<x_{n}=b.$

It follows that

$F(b)-F(a)=F(x_{n})-F(x_{0}).$

Now, we add each F(x_i) along with its additive inverse, so that the resulting quantity is equal:

${\begin{aligned}F(b)-F(a)&=F(x_{n})+[-F(x_{n-1})+F(x_{n-1})]+\cdots +[-F(x_{1})+F(x_{1})]-F(x_{0})\\&=[F(x_{n})-F(x_{n-1})]+[F(x_{n-1})-F(x_{n-2})]+\cdots +[F(x_{2})-F(x_{1})]+[F(x_{1})-F(x_{0})].\end{aligned}}$

The above quantity can be written as the following sum:

$F(b)-F(a)=\sum _{i=1}^{n}[F(x_{i})-F(x_{i-1})].\qquad (1)$

Next, we employ the mean value theorem. Stated briefly,

Let F be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there exists some c in (a, b) such that

$F'(c)={\frac {F(b)-F(a)}{b-a}}.$

It follows that

$F'(c)(b-a)=F(b)-F(a).$

The function F is differentiable on the interval [a, b]; therefore, it is also differentiable and continuous on each interval [x_i−1, x_i]. According to the mean value theorem (above),

$F(x_{i})-F(x_{i-1})=F'(c_{i})(x_{i}-x_{i-1}).$

Substituting the above into (1), we get

$F(b)-F(a)=\sum _{i=1}^{n}[F'(c_{i})(x_{i}-x_{i-1})].$

The assumption implies $F'(c_{i})=f(c_{i}).$ Also, $x_{i}-x_{i-1}$ can be expressed as $\Delta x$ of partition $i$.

$F(b)-F(a)=\sum _{i=1}^{n}[f(c_{i})(\Delta x_{i})].\qquad (2)$

A converging sequence of Riemann sums. The number in the upper left is the total area of the blue rectangles. They converge to the definite integral of the function.

We are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the mean value theorem, describes an approximation of the curve section it is drawn over. Also $\Delta x_{i}$ need not be the same for all values of i, or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with n rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we get closer and closer to the actual area of the curve.

By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann integral. We know that this limit exists because f was assumed to be integrable. That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity.

So, we take the limit on both sides of (2). This gives us

$\lim _{\|\Delta x_{i}\|\to 0}F(b)-F(a)=\lim _{\|\Delta x_{i}\|\to 0}\sum _{i=1}^{n}[f(c_{i})(\Delta x_{i})].$

Neither F(b) nor F(a) is dependent on $\|\Delta x_{i}\|$, so the limit on the left side remains F(b) − F(a).

$F(b)-F(a)=\lim _{\|\Delta x_{i}\|\to 0}\sum _{i=1}^{n}[f(c_{i})(\Delta x_{i})].$

The expression on the right side of the equation defines the integral over f from a to b. Therefore, we obtain

$F(b)-F(a)=\int _{a}^{b}f(x)\,dx,$

which completes the proof.

It almost looks like the first part of the theorem follows directly from the second. That is, suppose G is an antiderivative of f. Then by the second theorem, $G(x)-G(a)=\int _{a}^{x}f(t)\,dt$. Now, suppose $F(x)=\int _{a}^{x}f(t)\,dt=G(x)-G(a)$. Then F has the same derivative as G, and therefore F′ = f. This argument only works, however, if we already know that f has an antiderivative, and the only way we know that all continuous functions have antiderivatives is by the first part of the Fundamental Theorem. For example, if f(x) = e^−x2, then f has an antiderivative, namely

$G(x)=\int _{0}^{x}f(t)\,dt$

and there is no simpler expression for this function. It is therefore important not to interpret the second part of the theorem as the definition of the integral. Indeed, there are many functions that are integrable but lack elementary antiderivatives, and discontinuous functions can be integrable but lack any antiderivatives at all. Conversely, many functions that have antiderivatives are not Riemann integrable (see Volterra's function).

Examples

As an example, suppose the following is to be calculated:

$\int _{2}^{5}x^{2}\,dx.$

Here, $f(x)=x^{2}$ and we can use $F(x)={\frac {x^{3}}{3}}$ as the antiderivative. Therefore:

$\int _{2}^{5}x^{2}\,dx=F(5)-F(2)={\frac {5^{3}}{3}}-{\frac {2^{3}}{3}}={\frac {125}{3}}-{\frac {8}{3}}={\frac {117}{3}}=39.$

Or, more generally, suppose that

${\frac {d}{dx}}\int _{0}^{x}t^{3}\,dt$

is to be calculated. Here, $f(t)=t^{3}$ and $F(t)={\frac {t^{4}}{4}}$ can be used as the antiderivative. Therefore:

${\frac {d}{dx}}\int _{0}^{x}t^{3}\,dt={\frac {d}{dx}}F(x)-{\frac {d}{dx}}F(0)={\frac {d}{dx}}{\frac {x^{4}}{4}}=x^{3}.$

Or, equivalently,

${\frac {d}{dx}}\int _{0}^{x}t^{3}\,dt=f(x){\frac {dx}{dx}}-f(0){\frac {d0}{dx}}=x^{3}.$

As a theoretical example, the theorem can be used to prove that

$\int _{a}^{b}f(x)dx=\int _{a}^{c}f(x)dx+\int _{c}^{b}f(x)dx.$

Since,

${\begin{aligned}\int _{a}^{b}f(x)dx&=F(b)-F(a),\\\int _{a}^{c}f(x)dx&=F(c)-F(a),{\text{ and }}\\\int _{c}^{b}f(x)dx&=F(b)-F(c),\end{aligned}}$

the result follows from,

$F(b)-F(a)=F(c)-F(a)+F(b)-F(c).$

Licensing

Content obtained and/or adapted from:

• Fundamental theorem of calculus, Wikipedia under a CC BY-SA license