Chapter 7.8, Problem 60E

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

Chapter
Section

### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# (a) Evaluate the integral ∫ 0 ∞ x n e − x d x for n = 0, 1, 2, and 3.(b) Guess the value of ∫ 0 ∞ x n e − x d x when n is an arbitrary positive integer.(c) Prove your guess using mathematical induction.

Part (a):

To determine

To Evaluate: The integral 0xnexdx for n=0, 1, 2, and 3.

Explanation

Given Information:

The integral is shown as follows:

0xnexdx.

Here, n is an arbitrary positive integer.

Calculation:

Show the integral as shown below:

I=0xnexdx (1)

Calculate the value of the integral for n=0.

Substitute 0 for n in Equation (1).

I=0x0exdx=0exdx=[ex1]0=[e+e0]

I=[0+1]=1=0!

Thus, the value of the integral 0xnexdx for n=0 is 1.

Calculate the value of the integral for n=1.

Substitute 1 for n in Equation (1).

I=0x1exdx=0xexdx (2)

Consider the value of the function u=x (3)

Differentiate Equation (3).

du=dx

Consider the value of the function dv=exdx (4)

Integrate Equation (4).

dv=exdxv=ex

Show the formula of integration by parts.

0udv=(uv)00vdu (5)

Substitute x for u, exdx for dv, ex for v and dx for dv in Equation (5).

0xexdx=[xex]00(ex)dx=limt[xex]0t0(ex)dx=limt[tet+0]+0(ex)dx=limt[tet+0]+(ex)0

0xexdx=limt[tet]+(e+e0)=limt[tet]+(0+1)=limt[tet]+1=limt[tet]+1 (6)

Apply L’Hospital’s Rule in limt[tet] part of Equation (6).

Differentiate numerator and denominator of limt[tet] of Equation (6).

0xexdx=limt[ddt(t)ddt(et)]+1=limt[1et]+1=limt[et]+1=e+1

0xexdx=0+1=1!

Thus, the value of the integral 0xnexdx for n=1 is 1.

Calculate the value of the integral for n=2.

Substitute 2 for n in Equation (1).

I=0x2exdx=0x2exdx (7)

Show the formula of integration.

uneaudu=1auneaunaun1eaudu (8)

Apply the formula of integration along with limits [0,] in Equation (8).

Substitute 2 for n, x for u, dx for du, and 1 for a.

0x2exdx=[1(1)x2ex]02(1)0x21exdx=[x2ex]0+20xexdx=limt[x2ex]0t+20xexdx=limt[t2et+0]+20xexdx=limt[t2et]+20xexdx (9)

Substitute 1 for 0xexdx in Equation (9).

0x2exdx=limt[t2et]+2(1)=limt[t2et]+2 (10)

Apply L’Hospital’s Rule in limt[t2et] part of Equation (10)

Part (b):

To determine

To guess: the value of the integral 0xnexdx with n as an arbitrary positive integer.

Part (c):

To determine

To prove: the guess with mathematical induction.

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