   Chapter 6.1, Problem 14E ### 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. ∫ x 2 x − 3   d x

To determine

To calculate: The value of indefinite integral x2x3dx.

Explanation

Given Information:

The provided indefinite integral is x2x3dx.

Formula used:

The integration by parts udv=uvvdu.

Where, u and v are function of x.

(ax+b)ndx=(ax+b)n+1n+1,n1

Calculation:

Consider the indefinite integral x2x3dx

Here,

dv=(x3)12dx and u=x2

First find v,

dv=(x3)12dxdv=(x3)12dx

On further solving,

v=23(x3)32 …...…... (1)

Now 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,

x2x3dx=23x2(x3)3243x(x3)32dx

Use integration by parts on the second integral x(x3)32dx

Here,

dv=(x3)32dx and u=x

First find v,

dv=(x3)32dxdv=(x3)32dx

On further solving,

v=25(x3)52 …...…... (3)

Now find du,

u=x

Differentiate both side with respect x,

dudx=d(x)dxdudx=1

And,

du=1dx …...…..

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