# a)Evaluate the integral int_0^(oo) x^n e^(-x) dx for n =0, 1, 2, 3. b)Guess the value of int_0^(oo) x^n e^(-x) dx when n is an arbitrary nonnegative integer. c)Prove your guess using mathematical induction.

a)Evaluate the integral
b)Guess the value of ${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x}dx$ when n is an arbitrary nonnegative integer.
c)Prove your guess using mathematical induction.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Killaninl2
a)When n=0
${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x}dx={\int }_{0}^{\mathrm{\infty }}{e}^{-x}dx=1=0!$
When n=1
${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x}dx={\int }_{0}^{\mathrm{\infty }}x{e}^{-x}dx=\left[x\left(-{e}^{-x}\right)-\left(1\right)\left({e}^{-x}\right){\right]}_{0}^{\mathrm{\infty }}=1=1!$
When n=2
${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x}dx={\int }_{0}^{\mathrm{\infty }}{x}^{2}{e}^{-x}dx=\left[{x}^{2}\left(-{e}^{-x}\right)-\left(2x\right)\left({e}^{-x}\right)+\left(2\right)\left(-{e}^{-x}\right){\right]}_{0}^{\mathrm{\infty }}=2=2!$
When n=3
${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x}dx={\int }_{0}^{\mathrm{\infty }}{x}^{3}{e}^{-x}dx=\left[{x}^{3}\left(-{e}^{-x}\right)-\left(3{x}^{2}\right)\left({e}^{-x}\right)+\left(6x\right)\left(-{e}^{-x}\right)-\left(6\right)\left({e}^{-x}\right){\right]}_{0}^{\mathrm{\infty }}=6=3!$
###### Not exactly what you’re looking for?
Dawson Downs
b) In general, for non-negative integer,
${\int }_{0}^{\mathrm{\infty }}{x}^{n}{e}^{-x}dx=n!$ (1)
c)By part (a), the result (1) is true for n=0
Assume that the result (1) is true for n=k
i.e. ${\int }_{0}^{\mathrm{\infty }}{x}^{k}{e}^{-x}dx=k!$ (2)
We have to prove that the result (1) is true for n=k+1
i.e to prove that ${\int }_{0}^{\mathrm{\infty }}{x}^{k+1}{e}^{-x}dx=\left(k+1\right)!$
Put $u={x}^{k+1},dv={e}^{-x}$
$⇒du=\left(k+1\right){x}^{k},v=-{e}^{-x}$
By integration by parts,

Hence proved!