   Chapter 6.1, Problem 16E ### 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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# Integration by Parts In Exercises 5-16, use integration by parts to find the indefinite integral, See Examples 1, 2, 3, and 4. ∫ 3 x 3 e x   d x

To determine

To calculate: The value of indefinite integral 3x3exdx.

Explanation

Given Information:

The provided indefinite integral is 3x3exdx.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of x.

eaxdx=eaxa+C

Calculation:

Consider the indefinite integral 3x3exdx

The above indefinite integral can be written as,

3x3exdx=3x3exdx

Here,

dv=exdx and u=x3

First find v,

dv=exdx dv=exdx

On further solving,

v=ex …...…... (1)

Now find du,

u=x3

Differentiate both side with respect x,

dudx=d(x3)dxdudx=3x2

And,

du=3x2dx …...…... (2)

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

3x3exdx=3[x3ex3x2(ex)dx]=3[x3ex+3x2exdx]

Consider the indefinite integral x2exdx

Here,

dv=exdx and u=x2

First find v,

dv=exdxdv=exdx

On further solving,

v=ex …...…... (3)

Now find du,

u=x2

Differentiate both side with respect x,

dudx=d(x2)dxdudx=2x

And,

du=2xdx …...…... (4)

Apply integration by parts formula and substitute equation (3) and (4) in udv=uvvdu,

x2exdx=x2ex+2xexdx

Use integration by parts on the second integral xexdx

Here,

dv=exdx and u=x

First find v,

dv=exdxdv=exdx

On further solving,

v=ex …...…... (5)

Now find du,

u=x

Differentiate both side with respect x,

dudx=d(x)dxdudx=1

And,

du=1dx

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