   Chapter 6.2, Problem 6QY ### 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
1 views

# In Exercises 1–6, use integration by parts to find the indefinite integral. ∫ x 2 e − 2 x   d x

To determine

To calculate: The solution of indefinite integral x2e2xdx by the use of integration by parts.

Explanation

Given Information:

The provided expression is, x2e2xdx.

Formula used:

The formula for integration by parts is,

udv=uvvdu

Here, u and v be the differentiable functions x.

Integral Rule of exponential function:

enudu=enun+C

Here, n is the any number.

Calculation:

Consider the provided expression:

x2e2xdx

Let u=x2 and dv=e2xdx. So,

du=2xdx

And,

v=dv=e2xdx=e2x2=e2x2

Apply integration by parts to solve integral of x2e2xdx.

x2e2xdx=(x2)(e2x2)(e2x2)(2xdx)=x2e2x2+xe2xdx

Let u=x and dv=e2xdx. So,

du=dx

And,

v=dv=e2xdx=e2x2=e2x2

Apply integration by parts to solve integral of xe2xdx

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