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:

âˆ«0âˆžxneâˆ’xdx.

Here, n is an arbitrary positive integer.

Calculation:

Show the integral as shown below:

I=âˆ«0âˆžxneâˆ’xdx (1)

Calculate the value of the integral for n=0.

Substitute 0 for n in Equation (1).

I=âˆ«0âˆžx0eâˆ’xdx=âˆ«0âˆžeâˆ’xdx=[eâˆ’xâˆ’1]0âˆž=[âˆ’eâˆ’âˆž+eâˆ’0]

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

Thus, the value of the integral âˆ«0âˆžxneâˆ’xdx for n=0 is 1.

Calculate the value of the integral for n=1.

Substitute 1 for n in Equation (1).

I=âˆ«0âˆžx1eâˆ’xdx=âˆ«0âˆžxeâˆ’xdx (2)

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

Differentiate Equation (3).

du=dx

Consider the value of the function dv=eâˆ’xdx (4)

Integrate Equation (4).

âˆ«dv=âˆ«eâˆ’xdxv=âˆ’eâˆ’x

Show the formula of integration by parts.

âˆ«0âˆžudv=(uv)0âˆžâˆ’âˆ«0âˆžvdu (5)

Substitute x for u, eâˆ’xdx for dv, âˆ’eâˆ’x for v and dx for dv in Equation (5).

âˆ«0âˆžxeâˆ’xdx=[âˆ’xeâˆ’x]0âˆžâˆ’âˆ«0âˆž(âˆ’eâˆ’x)dx=limtâ†’âˆž[âˆ’xeâˆ’x]0tâˆ’âˆ«0âˆž(âˆ’eâˆ’x)dx=limtâ†’âˆž[âˆ’teâˆ’t+0]+âˆ«0âˆž(eâˆ’x)dx=limtâ†’âˆž[âˆ’teâˆ’t+0]+(âˆ’eâˆ’x)0âˆž

âˆ«0âˆžxeâˆ’xdx=limtâ†’âˆž[âˆ’teâˆ’t]+(âˆ’eâˆ’âˆž+eâˆ’0)=limtâ†’âˆž[âˆ’teâˆ’t]+(0+1)=limtâ†’âˆž[âˆ’teâˆ’t]+1=limtâ†’âˆž[âˆ’tet]+1 (6)

Apply Lâ€™Hospitalâ€™s Rule in limtâ†’âˆž[âˆ’tet] part of Equation (6).

Differentiate numerator and denominator of limtâ†’âˆž[âˆ’teâˆ’t] of Equation (6).

âˆ«0âˆžxeâˆ’xdx=limtâ†’âˆž[âˆ’ddt(t)ddt(et)]+1=limtâ†’âˆž[âˆ’1et]+1=limtâ†’âˆž[âˆ’eâˆ’t]+1=âˆ’eâˆ’âˆž+1

âˆ«0âˆžxeâˆ’xdx=0+1=1!

Thus, the value of the integral âˆ«0âˆžxneâˆ’xdx for n=1 is 1.

Calculate the value of the integral for n=2.

Substitute 2 for n in Equation (1).

I=âˆ«0âˆžx2eâˆ’xdx=âˆ«0âˆžx2eâˆ’xdx (7)

Show the formula of integration.

âˆ«uneaudu=1auneauâˆ’naâˆ«unâˆ’1eaudu (8)

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

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

âˆ«0âˆžx2eâˆ’xdx=[1(âˆ’1)x2eâˆ’x]0âˆžâˆ’2(âˆ’1)âˆ«0âˆžx2âˆ’1eâˆ’xdx=[âˆ’x2eâˆ’x]0âˆž+2âˆ«0âˆžxeâˆ’xdx=limtâ†’âˆž[âˆ’x2eâˆ’x]0t+2âˆ«0âˆžxeâˆ’xdx=limtâ†’âˆž[âˆ’t2eâˆ’t+0]+2âˆ«0âˆžxeâˆ’xdx=limtâ†’âˆž[âˆ’t2et]+2âˆ«0âˆžxeâˆ’xdx (9)

Substitute 1 for âˆ«0âˆžxeâˆ’xdx in Equation (9).

âˆ«0âˆžx2eâˆ’xdx=limtâ†’âˆž[âˆ’t2et]+2(1)=limtâ†’âˆž[âˆ’t2et]+2 (10)

Apply Lâ€™Hospitalâ€™s Rule in limtâ†’âˆž[âˆ’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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