# Integration

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Integration

U-Substitution

Slide2Use pattern recognition to find an indefinite integral.Use a change of variables to find an indefinite integral.Use the General Power Rule for Integration to find an indefinite integral.

Objectives

Slide3Antidifferentiation of a Composite Function

Let g be a function whose range is an interval I and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, thenIf u = g(x), then du = g’(x)dx and

Slide4Pattern Recognition

In this section you will study techniques for integrating composite functions.

This is split into two parts, pattern recognition and change of variables.

u-substitution is similar to the techniques used for the chain rule in differentiation.

Slide5Pattern Recognition for Finding the Antiderivative

FindLet g(x)=5x and we have g’(x)=5dx So we have f(g(x))=f(5x)=cos5xFrom this, you can recognize that the integrand follows the f(g(x))g’(x) pattern. Using the trig integration rule, we getYou can check this by differentiating the answer to obtain the original integrand.

Slide6Recognizing Patterns

Look at the more complex part of the function- the stuff ‘inside’.

Does the stuff outside look like the derivative of the stuff inside?

Slide7Recognizing Patterns

Slide8Recognizing Patterns

Slide9Change of Variables

FindLet u=5x and we have du=5dx You can check this by differentiating the answer to obtain the original integrand.

Slide10Change of VariablesA.K.A. U-Substitution

Example 3

Slide11Pattern Recognition

The integrands in Example 1 fit the f(g(x))g'(x) pattern exactly—you only had to recognize the pattern. You can extend this technique considerably with the Constant Multiple Rule Many integrands contain the essential part (the variable part) of g'(x) but are missing a constant multiple. In such cases, you can multiply and divide by the necessary constant multiple, as shown in Example 3.

Slide12Multiplying and Dividing by a Constant

Example 3

Slide13Multiplying and Dividing by a Constant Alternate method

Example 3

Slide14Change of VariablesExample 4

Slide15Example 4 – Change of Variables

Find Solution:First, let u be the inner function, u = 2x – 1. Then calculate the differential du to be du = 2dx.Now, using and substitute to obtain

Slide16Example 4 – Solution

cont’d

Slide17Change of Variables

Example 5

Slide18Change of Variables

Example 5

Slide19Change of Variables Example

Example 6

Slide20Example 7 – Substitution and the General Power Rule

Slide21

Example 7 – Substitution and the General Power Rule

cont’d

Slide22Guidelines for Making a Change of Variables

Choose a substitution u=g(x). Usually it is best to choose the INNER part of a composite function, such as a quantity raised to a power.

Compute du=g’(x)dx

Rewrite the integral in terms of the variable u

Find the

resulting

integral in terms of u

Replace u by g(x) to obtain an

antiderivative

in terms of x.

Check your answer by differentiating.

Slide23General Power Rule for Integration

Theorem: The General Power Rule for IntegrationIf g is a differentiable function of x, then 1Equivalently, if u=g(x), then

Slide24Change of Variables for Definite Integrals

If the function u=g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, thenBasically, this gives us permission to find the area under the curve with these types of integration problems.

Slide25Definite Integrals and Change of Variables

Example 8

Slide26Example 8 – Change of Variables

EvaluateSolution:To evaluate this integral, let u = x2 + 1. Then, you obtain Before substituting, determine the new upper and lower limits of integration.

Slide27Example 8 – Solution

Now, you can substitute to obtain

cont’d

Slide28Example 8 – Solution

Try rewriting the antiderivative in terms of the variable x and evaluate the definite integral at the original limits of integration, as shown.Notice that you obtain the same result.

cont’d

Slide29Definite Integrals and Change of Variables

Example 9:This problem requires a little more work:

Slide30Integration of Even and Odd Functions

Even with a change of variables, integration can be difficult.Occasionally, you can simplify the evaluation of a definite integral over an interval that is symmetric about the y-axis or about the origin by recognizing the integrand to be an even or odd function (see Figure 4.40).

Figure 4.40

Slide31Integration of Even and Odd Functions

Slide32Example 10 – Integration of an Odd Function

EvaluateSolution:Letting f(x) = sin3x cos x + sin x cos x produces f(–x) = sin3(–x) cos (–x) + sin (–x) cos (–x) = –sin3x cos x – sin x cos x = –f(x).

Slide33Example 10 – Solution

So, f is an odd function, and because f is symmetric about the origin over [–π/2, π/2], you can apply Theorem 4.16 to conclude that

Figure 4.41

cont’d

Slide34Homework:

p.304 1,3,9,14,16,20,25,27,31,33,34,41,45,51,54,59,69,79,81,95,103

These are the minimum to help you PASS this next quiz/test. I suggest you complete all problems in the sections relating to what was covered in the notes.

Students taking the AP Exam in the spring should also

look at

107, 108, 111-112, and the word problems (see me for assistance when you get to them).

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## Integration

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