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CalculusCalculus: An Applied Approach (MindTap Course List)10th Edition

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Chapter 5.2, Problem 25E
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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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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 ∫(x2−6x)4(x−3)dx and check the result by differentiating.

Explanation

Given Information:

The indefinite integral is ∫(x2−6x)4(x−3)dx.

Formula used:

General Power Rule:

If u is a differentiable function of x, then

∫undudxdx=∫undu=un+1n+1+C,n≠−1

The Power Rule of derivative:

ddxxn=nxn−1

Where n is a real number.

Calculation:

Consider indefinite integral,

∫(x2−6x)4(x−3)dx

Let u=x2−6x,

So,

du=(2x−6 )dx=2(x−3) dx

Now use the general power rule to get,

∫(x2−6x)4(x−3)dx=12∫u4du

Now the integral will be:

12∫u4du=12.u4+14+1+C=12

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