Difference between revisions of "Integration by Substitution"

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<p>The tricky part is trying to identify what you want to make your '''<i>u</i>''' to be.  Some times substitution will not be enough and you will have to use the rules for integration by parts.  That will be covered in a different section</p>
 
<p>The tricky part is trying to identify what you want to make your '''<i>u</i>''' to be.  Some times substitution will not be enough and you will have to use the rules for integration by parts.  That will be covered in a different section</p>
 +
 +
====Steps====
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:{|
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|<math>\int\limits_{x=a}^{x=b}f(x)dx</math>
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|<math>=\int\limits_{x=a}^{x=b} f(x)\ \frac{du}{du}\ dx</math>
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|style="padding-left: 20px"|(1)
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|i.e. <math>\frac{du}{du}=1</math>
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|-
 +
|
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|<math>=\int\limits_{x=a}^{x=b}{\left(f(x)\ \frac{dx}{du}\right)\left(\frac{du}{dx}\right)}\ dx</math>
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|style="padding-left: 20px"|(2)
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|i.e. <math>\frac{dx}{du}\cdot\frac{du}{dx}=1</math>
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|-
 +
|
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|<math>=\int\limits_{x=a}^{x=b}\left(f(x)\ \frac{dx}{du}\right)g'(x)\ dx</math>
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|style="padding-left: 20px"|(3)
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|i.e. <math>\frac{du}{dx}=g'(x)</math>
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|-
 +
|
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|<math>=\int\limits_{x=a}^{x=b}h(g(x))g'(x)dx</math>
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|style="padding-left: 20px"|(4)
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|i.e. Now equate <math>\left(f(x)\ \frac{dx}{du}\right)</math> with <math>h(g(x))</math>
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|-
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|
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|<math>=\int\limits_{x=a}^{x=b}h(u)g'(x)dx</math>
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|style="padding-left: 20px"|(5)
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|i.e. <math>g(x)=u</math>
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|-
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|
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|<math>=\int\limits_{u=g(a)}^{u=g(b)}h(u)du</math>
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|style="padding-left: 20px"|(6)
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|i.e. <math>du=\frac{du}{dx}dx=g'(x)dx</math>
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|-
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|
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|<math>=\int\limits_{u=c}^{u=d}h(u)du</math>
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|style="padding-left: 20px"|(7)
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|i.e. We have achieved our desired result
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|}
  
 
<p>'''Ex. 1'''</p>
 
<p>'''Ex. 1'''</p>
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<p><math>\int_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x</math></p>
 
<p><math>\int_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x</math></p>
−
 
−
 
  
 
==Resources==
 
==Resources==

Revision as of 12:31, 6 October 2021

Integration by Substitution

There is a theorem that will help you with substitution for integration. It is called Change of Variables for Definite Integrals.

what the theorem looks like is this

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_{a}^{b} f(x)\operatorname {d}x = \int_{\alpha}^{\beta} f(g(u))g\prime (u)\operatorname {d}u}


In order to get you must plug a into the function g and to get you must plug b into the function g.

The tricky part is trying to identify what you want to make your u to be. Some times substitution will not be enough and you will have to use the rules for integration by parts. That will be covered in a different section

Steps

(1) i.e.
(2) i.e.
(3) i.e. 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{du}{dx}=g'(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 =\int\limits_{x=a}^{x=b}h(g(x))g'(x)dx} (4) i.e. Now equate 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 \left(f(x)\ \frac{dx}{du}\right)} 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 h(g(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 =\int\limits_{x=a}^{x=b}h(u)g'(x)dx} (5) i.e. 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(x)=u}
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_{u=g(a)}^{u=g(b)}h(u)du} (6) i.e. 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=\frac{du}{dx}dx=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_{u=c}^{u=d}h(u)du} (7) i.e. We have achieved our desired result

Ex. 1

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_{0}^{2} x(x^2+1)^2 \operatorname {d}x}

Instead of making this a big polynomial we will just use the substitution method.

Step 1

Identify your u

Let 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^2+1}

Step 2


Identify 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 \operatorname {d}u}


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 \operatorname {d}u = 2x\operatorname {d}x}


Step 3

Now we plug in our limits of integration to our u to find our new 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 x = 0, u =0^2 + 1 = 1}

and 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 = 2, u = 2^2 + 1 = 5}

Now our integration problem looks something like this

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}{2} \int_{0}^{5} (x^2 + 1)^2 (2x)\operatorname {d}x}

Step 4

write your new integration problem


When we plug in our u it looks like

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}{2} \int_{0}^{5} (u)^2 \operatorname {d}u}


Step 5

Evaluate the Integral

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}{2} \left[\frac {1}{3} u^3 \right]_{0}^{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 \frac {1}{2} \left[\left(\frac {1}{3} * 5^3 \right) - \left(\frac {1}{3} * 0^3 \right)\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}{2} \left[\frac {1}{3} * 125 \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}{2} \left[\frac {125}{3}\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 {125}{6}}



As you can see this all simplified fairly nice. Using substitution will be hard, for most people, at first. Once you get the hang of doing this it should come to you faster and faster each time.

I'll give you some other problems to work on as well.

Ex. 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 \int_{0}^{\frac {\pi}{2}} \sin (x) \cos (x) \operatorname {d}x}

Ex. 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_{-1}^{2} \sqrt {x^2+4} (2x) \operatorname {d}x}

Resources

Example 1. Produced by Professor Zachary Sharon, UTSA

Example 2. Produced by TA Catherine Sporer, UTSA

Indefinite Integrals Using Substitution


Definite Integrals Using Substitution

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