Chapter 6, Problem 3TYS

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

Chapter
Section

### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views

# In Exercises 1–3, use integration by parts to find the indefinite integral. ∫ x 2 e − x / 3   d x

To determine

To calculate: The infinite integral of x2ex/3dx by using the method of integration by parts.

Explanation

Given Information:

The integral is x2ex/3dx.

Formula used:

The integration by part of two differentiable function is,

udv=uvvdu

The basic rule of integration for constant C is,

Cdx=0

The simple power rule of integration is,

xndx=xn+1n+1+C

Where, n1.

The chain rule of differentiation is,

ddx(f(g))=(f(g))ddx(g)

Calculation:

Consider the provided integral, x2ex/3dx.

Since, the integrand of the provided indefinite integral is xex+1 which is made up of two function.

Let the first function is x2 so,

u=x2

Differentiate both side with respect to x, use the simple power rule of differentiation.

du=ddx(x2)dx=3x2dx

Let the second function ex/3 so,

dv=ex/3dx

Integrate both side with respect to x, use the integration for exponential.

dv=ex/3dxv=3ex/3

Now, use the integration by parts method

udv=uvvdu

Substitute x2 for u, 3ex/3 for v, ex/3dx for dv, 2xdx for du and solve.

x2ex/3dx=x2(3ex/3)(3ex/3)2xdx=3x2ex/3+6xex/3dx

Further solve the integration by parts,

Consider, the integrand 6xex/3dx from the above expression and observe that the simplest portion of the integrand is 6x

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started