Question

the integral of 6x^2(e^((-x)^3)) from one to infinity what does it converge at? and how do you get the answer?

Answer 1

no dx????

Answer 2

\[\Large\bf \int\limits_{x=1}^{\infty} \;e^{\color{#DD4747}{(-x)^3}}\color{#008353 }{6x^2\;dx}\]Making the following u-sub:\[\Large\bf \color{#DD4747}{u=(-x)^3}\]Taking the derivative of our sub and moving some stuff around...\[\Large \bf \color{#008353 }{-2du=6x^2\;dx}\]

Answer 3

thanks

Answer 4

thats all i should need

Answer 5

k :)

Answer 6

don't forget to either:
~change the limits of integration so they correspond to u
~or remember to undo your sub after you integrate before plugging in bounds.

Answer 7

whats the limit of e^-infinity

Answer 8

Remember your rules of exponents! :)
\[\Large e^{-99999}\quad=\quad \frac{1}{e^{99999}}\quad\approx\quad \frac{1}{1\;bajillion}\quad\approx\quad 0\]Does that make sense?

Answer 9

yes so now i have -2e^(-x^3) from one to infinity so that should be -2e^-1 right

Answer 10

or -2/e

Answer 11

\[\Large\bf \lim_{b\to\infty}-2e^{-b^3}-(-2e^{-1})\]I think you missed a minus sign on your lower limit.

Answer 12

isnt it upper minus lower?

Answer 13

never mind

Answer 14

thank you soo much

Answer 15

yay gj \c:/ np!