   Chapter 5.3, Problem 3CP ### 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

# Checkpoint 3 Worked-out solution available at LarsonAppliedCalculus.comFind ∫ 4 x e x 2 d x .

To determine

To calculate: The indefinite integral 4xex2dx.

Explanation

Given Information:

The provided indefinite integral is 4xex2dx

Formula used:

The exponent rule of integrals:

eudu=eu+C

Calculation:

Consider the indefinite integral:

4xex2dx

Let u=x2, then derivative will be,

du=d(x2)=2xdx

Rewrite the integral as:

22xex2dx

Substitute du for 2xdx and u for x2 in provided integration

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