5.5 The Substitution Rule

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# 5.5 The Substitution Rule - PowerPoint PPT Presentation

5.5 The Substitution Rule. The Substitution Rule. If u=g(x) is differentiable function whose range is an interval I , and f is continuous on I , then The Substitution rule is proved using the Chain Rule for differentiation.

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## 5.5 The Substitution Rule

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### 5.5 The Substitution Rule

The Substitution Rule
• If u=g(x) is differentiable function whose range is an interval I, and f is continuous on I, then
• The Substitution rule is proved using the Chain Rule for differentiation.
• The idea behind the substitution rule is to replace a relatively complicated integral by a simpler integral. This is accomplished by changing from the original variable x to a new variable u that is a function of x.
• The main challenge is to think of an appropriate substitution.

Example

The variable of integration must match the variable in the expression.

Don’t forget to substitute the value for u back into the problem!

We computed du by straightforward differentiation of the expression for u.

Substitution rule examples

Example

The substitution u = 2x was suggested by the function to be integrated. The main problem in integrating by substitution is to find the right substitution which simplifies the integral so that it can be computed by the table of basic integrals.

This rewriting allows us to finish the computation using basic formulae.

Example

Solution

Next substitute back to the original variable.

The substitution rule for definite integrals

If g’ is continuous on [a,b], and fis continuous on the range of u=g(x) then

1

e

The substitution rule for definite integrals

Example

The area of the yellow domain is ½.

new limit

new limit

Example

Find new limits

Example

Don’t forget to use the new limits.

-a

a

Integrals of Even and Odd Functions

Problem

Solution

An odd function is symmetric with respect to the origin. The definite integral from -a to a, in the case of the function shown in this picture, is the area of the blue domain minus the area of the red domain. By symmetry these areas are equal, hence the integral is 0.