   Chapter 13.6, Problem 13E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 1-16, use integration by parts to evaluate the integral. ∫ q 3 q 2 − 3   d q

To determine

To calculate: The value of the integral q3q23dq.

Explanation

Given Information:

The provided integral is q3q23dq.

Formula used:

The formula for integration by parts is given by:

udv=uvvdu

According to the power rule of integrals,

xndx=xn+1n+1+C

Differentiation formula,

d(xn)dx=nxn1

Calculation:

Consider the provided integral:

q3q23dq

Multiply and divide by 2 and rewrite the integral as

12q2q232qdq

Now, split the integrand into two parts

Set one part equal to u and another part equal to dv.

So, the values will be:

u=q2

And,

dv=q232qdq

Now use the differential formula d(xn)dx=nxn1 to differentiate u=q2

du=2qdq

And integrate v by the use of power rule of integrals,

v=q232qdq

Let q23=t,

Differentiate the equation,

2qdq=dt

Thus, the integral becomes,

v=q232qdq=tdt=23t3/2=23(q23)3/2

Now, use the formula for integration by parts,

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