   Chapter 5.2, Problem 29E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Applying the General Power Rule In Exercises 9–34, find the indefinite integral. Check your result by differentiating. See Examples 1, 2, 3, and 5. ∫ 5 x 1 − x 2 3 d x

To determine

To calculate: The indefinite integral of 5x1x23dx and check the result by differentiating.

Explanation

Given Information:

The indefinite integral is 5x1x23dx.

Formula used:

General Power Rule:

If u is a differentiable function of x, then

undudxdx=undu=un+1n+1+C,n1

The Power Rule:

ddxxn=nxn1

Where n is a real number.

Calculation:

Consider indefinite integral,

5x1x23dx

This can be written as 5x(1x2)13dx.

Let u=1x2,

So,

du=2xdx

Now use the general power rule to get,

5x(1x2)13dx=52(u)13du

Now the integral will be:

52(u)13du=52u13+113+1+C=52u4343+C=5234u43+C=158u43+C

Substitute back the value of u to get,

5x1x23dx=158u43+C=158(1x2)<

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