   Chapter 5.2, Problem 28E ### 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 x 2 − 4 x + 3 d x

To determine

To calculate: The indefinite integral of x2x24x+3dx and check the result by differentiating.

Explanation

Given Information:

The indefinite integral is x2x24x+3dx.

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,

x2x24x+3dx

This can be written as (x24x+3)12(x2)dx.

Let u=x24x+3,

So,

du=(2x4) dx=2(x2)dx

Now use the general power rule to get,

(x24x+3)12(x2)dx=12(u)12du

Now the integral will be:

12(u)12du=12u12+112+1+C=12u1212+C=u12+C

Substitute back the value of u to get,

x2<

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