   Chapter 6.1, Problem 22E ### 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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# Finding an Indefinite Integral In Exercises 17-38, find the indefinite integral. (Hint: Integration by parts is not required for all the integrals.) ∫ 3 2   x 2 e x / 2   d x

To determine

To calculate: The value of indefinite integral 32x2ex2dx.

Explanation

Given Information:

The provided indefinite integral is 32x2ex2dx.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of x.

eaxdx=eaxa+C

Calculation:

Consider the indefinite integral 32x2ex2dx

The above indefinite integral can be written as,

32x2ex2dx=32x2ex2dx

Here,

dv=ex2dx and u=x2

First find v,

dv=ex2dxdv=ex2dx

On further solving,

v=2ex2 …...…... (1)

Find du:

u=x2

Differentiate both side with respect x;

dudx=d(x2)dxdudx=2x

And,

du=2xdx …...…... (2)

Apply integration by parts formula and substitute equation (1) and (2) in udv=uvvdu,

32x2ex2dx=32[(x2)(2ex2)(2x)(2ex2)dx]=32[2x2ex24xex2dx]

Consider the xex2dx

Here,

dv=ex2dx and u=x

First find v,

dv=ex2dxdv=ex2dx

On further solving,

v=2ex2 …...…... (3)

Find du:

u=x

Differentiate both side with respect x;

dudx=d(x)dxdudx=1

And,

du=1dx

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