Question

I've got an integration-by-parts monstrosity. I have the first part right (u-substitution), but when I have to use different variables, I mess up. Here's my problem: f(x) = k |x|^3 * e^(-x^2), where the integral from -infinity to infinity of the function should equal 1.

Answer 1

I've gotten the u-sub as u = x^2 and du = 2x dx; so I can reduce this function down to 1= 2k [integral from 0 to b] x^3 * e^u du

Answer 2

my attempt for integration by parts include w = u, dw = 1, dv = e^u, and v = e^u

Answer 3

but I am just not having success. The answer, however, is k =1. I'm totally not getting that.

Answer 4

anyone else getting stuck?

Answer 5

My work:\[u = x^2; du = 2x;\]
\[1= 2\int\limits_{0}^{b} kx^3 e^-x^2 dx\]
\[1= 2k\int\limits\limits_{0}^{b} x^3 (e^(-x^2)) dx\]

Answer 6

how about bringing the 2 back into the integral and using the sub \[dv=2xe^{-x^2}dx\]\[u=x^2\]I see interesting things happening I think, but I am multi-tasking

Answer 7

\[1= k\int\limits\limits_{0}^{b} u* e^(u) du\]

Answer 8

hmm...

Answer 9

now, w = u, dw = 1, dv = e^u, and v = e^u, right? So...

Answer 10

Hello NoelGreco!

Answer 11

\[\int\limits_{0}^{b}u*e^u = u*e^u -\int\limits_{0}^{b}(1)e^u du\]

Answer 12

but how do I link that back to my original reduced eqn and solve for k?

Answer 13

1= k[\[u*e^u - \int\limits_{0}^{b} e^u du\]] ???

Answer 14

that's similar to what I'm getting

Answer 15

but u* e^u - inifinity does NOT get k = 1. It gets madness.

Answer 16

\[\int_{-\infty}^\infty k|x|^3e^{-x^2}dx=k\int_0^\infty x^2(2xe^{-x^2})dx\]\[u=x^2\implies du=2xdx\]\[dv=2xe^{-x^2}dx\implies v=-e^{-x^2}\]\[k\int_0^\infty x^2(2xe^{-x^2})dx=k\left(\cancel{-x^2e^{-x^2}|_0^\infty}^{\LARGE0}+\int2xe^{-x^2}dx\right)\]\[=k\left(-e^{-x^2}|_0^\infty\right)=k=1\]so it looks like k=1 :)

Answer 17

woah. Hold on, I gotta see what I did wrong.

Answer 18

I think the trick was in letting the 2 from the even integral trick into the integrand to make dv=2xe^(-x^2)

Answer 19

this...looks so much easier than mine. Did you do integration by parts?

Answer 20

OH

Answer 21

sure did

Answer 22

you actually gave me the idea, so it was a team effort :)

Answer 23

I've gotta run for a minute but I'll be back online to make sure I get this. Thanks so much for your help (again, lol)! :D

Answer 24

anytime, as usual :D