   Chapter 5.2, Problem 25E ### 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. ∫ ( x 2 − 6 x ) 4 ( x − 3 ) d x

To determine

To calculate: The indefinite integral of (x26x)4(x3)dx and check the result by differentiating.

Explanation

Given Information:

The indefinite integral is (x26x)4(x3)dx.

Formula used:

General Power Rule:

If u is a differentiable function of x, then

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

The Power Rule of derivative:

ddxxn=nxn1

Where n is a real number.

Calculation:

Consider indefinite integral,

(x26x)4(x3)dx

Let u=x26x,

So,

du=(2x6 )dx=2(x3) dx

Now use the general power rule to get,

(x26x)4(x3)dx=12u4du

Now the integral will be:

12u4du=12.u4+14+1+C=12

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