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.

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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.
Transcribed Image Text: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!
Transcribed Image Text:(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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