Question

\(\LARGE \int_{-\infty}^{0} x 5^{-x^2} dx\)

@Mimi_x3 please

Answer 1

use integral by parts formula

Answer 2

isn't it special in improper integrals?

Answer 3

u-sub

Answer 4

First, rewrite the improper integral as a limit of an integral, then use u-sub.

Answer 5

what do you mean limit of an integral @blockcolder ?

Answer 6

I mean like this:
\[\Large\lim_{t \rightarrow -\infty} \int_t^0 x5^{-x^2}\ dx\]

Answer 7

hmmm ohkayyyy...? then what?

Answer 8

Then use u-sub like the others said:
Let u=-x^2, du=-2xdx
\[\Large\lim_{t \rightarrow -\infty}\int_t^0 x5^{-x^2}\ dx=-\frac{1}{2}\lim_{t \rightarrow -\infty} \int_{-t^2}^0 5^u\ du\]

Answer 9

wait...why is the lower limit t^2 now?

Answer 10

hehe sorry just learning with the user here :p

Answer 11

I made a u-sub, so I plugged in the limits on x into my equation for u.

Answer 12

i dont get how it becam x^2 eeither @_@ please explain further

Answer 13

uhmm is it because t = x so when u became -x^2 it became -t62??

Answer 14

t^2*

Answer 15

Yep.
Continuing the integral:
\[\Large-\frac{1}{2}\lim_{t \rightarrow -\infty} \int_{-t^2}^0 5^u\ du=
-\frac{1}{2}\lim_{t \rightarrow -\infty} \left .\left ( \frac{5^u}{\ln5}\right ) \right |_{-t^2}^0\\
\Large=-\frac{1}{2 \ln5} \lim_{t \rightarrow -\infty}(1-5^{-t^2})=-\frac{1}{2 \ln5}\]
because \( \Large\displaystyle \lim_{t \rightarrow \infty} 5^{-t^2}=0\).

Answer 16

\(5^\infty\) is zero?? @blockcolder

Answer 17

Oops. Typo. That should be -infinity.
\(\displaystyle \Large\lim_{t \rightarrow -\infty}5^{-t^2}=0\)

Answer 18

so \(5^{-\infty}\) is zero?

Answer 19

Yep. :)

Answer 20

is that a property?

Answer 21

Kinda. For a>1, \(\lim_{x \rightarrow -\infty}a^x=0\).

Answer 22

i see..good work :D gonna practice more though