Question

integrate x^2 e^-x dx from 1 to infinite

Answer 1

\[e^{-x}(x^2+2x+2)\] might be a good start if i recall the pattern right

Answer 2

i am getting 2/e but the book says otherwise

Answer 3

-x^2.e^-x -2x.e^-x - x.e^-x

maybe the -e^-x is better

Answer 4

\[-e^{-inf}(inf^2+2inf+2)+ e^{-1}(1+2+2)\]
\[0+ e^{-1}(5)=\frac{5}{e}\]

Answer 5

Thank you. Can you please explain how u the first step? because i cant grasp the idea of completing the square in this situation

Answer 6

the first step is just the result of integration by parts; there is no need to complete the square afterwards since we are not wanting to simplify (x^2+2x+2), we simply want to use it

Answer 7

o i get it:)

Answer 8

thank you so much

Answer 9

\[\int x^2 e^{-x}dx=-x^2e^{-x}-\int x e^{-x}dx\]
and so on ;)

Answer 10

but allow for typos and its good lol

Answer 11

you are great!

Answer 12

i was so lost:)