   Chapter 14, Problem 9PS

Chapter
Section
Textbook Problem

Evaluating an IntegralIn Exercises 9 and 10, evaluate the integral. (Hint: See Exercise 63 in Section 14.3.) ∫ 0 ∞ x 2 e − x 2 d x

To determine

To calculate: The value of integral 0x2ex2dx.

Explanation

Given: The integral 0x2ex2dx.

Formula used: The gauss error function is aspecial function defined as

erf(x)=2πex2dx

Where, erf()=1 and erf(0)=0.

Integration by parts:

udv=uvuv

Calculation: The integral is 0x2ex2dx that can be rewritten as 0x(xex2)dx.

Now, let u=x and dv=xex2.

So, u=1 and v=ex22.

Now, by use integration by parts the integral is:

0x2ex2dx=[x(ex22)1

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