The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k2 1, if P(k) is true, then P(k + 1) is true: Suppose that xke-x dx = k! for some positive integer k. Next, examine the integral k+'e* dx = lim *+1e-x dx. To evaluate * + le-x dx, we use integration by parts with u = , dv = e-* dx = du = dx, v = -e-X. So k+ 1e-x dx = dx + (k + 1 dx Applying the limit as t approaches infinity to this expression gives the following. lim + (k + 1 0 + (k + 1) dx. Further, (k + 1) dx = (k + 1) |, by the inductive hypothesis. Thus, ' x* + le-x dx = (k + 1)!. Hence P(k + 1) is true, which completes the inductive step. Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction complete.

The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer k2 1, if P(k) is true, then P(k + 1) is true: Suppose that xke-x dx = k! for some positive integer k. Next, examine the integral k+'e* dx = lim +1e-x dx. To evaluate + le-x dx, we use integration by parts with u = , dv = e-* dx = du = dx, v = -e-X. So k+ 1e-x dx = dx + (k + 1 dx Applying the limit as t approaches infinity to this expression gives the following. lim + (k + 1 0 + (k + 1) dx. Further, (k + 1) dx = (k + 1) |, by the inductive hypothesis. Thus, ' x* + le-x dx = (k + 1)!. Hence P(k + 1) is true, which completes the inductive step. Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction complete.

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

100%

The selected statement is true because both sides of the equation equal the same quantity.
Show that for each integer k > 1, if P(k) is true, then P(k + 1) is true: Suppose that
xke-x dx = k!
for some positive integer k. Next, examine the integral
k + 1
dx = lim
t - 00
k + 1.-x
te
'eX dx.
+ 1.-x
To evaluate
dx, we use integration by parts with
u =
dv = e-X dx
du =
dx, v = -eX.
So
k + le-X dx =
dx
nt
+ (k + 1)
dx.
or
Applying the limit as t approaches infinity to this expression gives the following.
])-)-•
lim
+ (k + 1)
= 0 + (k + 1)
dx.
Further, (k + 1)
dx = (k + 1)
by the inductive hypothesis. Thus,
| xk + le-X dx = (k + 1)!. Hence P(k + 1) is true, which completes the inductive step.
Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.

(a) Evaluate the integral
9x"e
-x
dx
for n = 0, 1, 2, and 3.
n = 0
9.
t
n = 1
-e
n = 2
n = 3
(b) Guess the value of
x"e-X dx
when n is an arbitrary positive integer.
(c) Prove your guess using mathematical induction.
Proof (by mathematical induction): Let P(k) be the following equation.
xke-Xdx :
= k!
We will show that P(k) is true for every integer k > 0.
Show that P(1) is true: Select P(1) from the choices below.
xke-*dx = k
00
xeXdx = 1!
00
ke-Xdx = 1!
00
xe-ldx = e
00
xe Xdx = k!

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