Question

integral of [(7x)/(e^x)] as x goes from 1 to infinity

Answer 1

would most probly use integration by parts

Answer 2

\[\int\limits {7x/e^{x}} = 7 \int\limits {x \times e^{-x}}\]

Answer 3

\[\int\limits_{a}^{b} uv^{\prime} = [uv]_{a}^{b} - \int\limits_{a}^{b}vu^{\prime}\]

Answer 4

i just tried that an i keep getting undefined. could you possibly explain to me how you'd do it, and maybe i'll find my mistake

Answer 5

\[\int\limits_{1}^{\infty} 7x/e^{x} \]

Answer 6

\[7 \int\limits_{1}^{\infty} {x \times d(e^-x)/dx} = [xe^{-x}]_{1}^{\infty} - \int\limits_{1}^{\infty} {e^{-x}}{d{x}/dx}\]

Answer 7

\[[xe^{-x}]_{1}^{\infty} + {e^{-x}}{d{x}/dx} = 2e^-1\]
\[\therefore 7 \int\limits\limits_{1}^{\infty} {x \times d(e^-x)/dx} =14e^{-1}\]

Answer 8

so you picked x as your u and e^-x as your dv?
isn't v gonna be -e^-x then?

Answer 9

yup

Answer 10

my bad the last line shud b \[\int\limits_{1}^{\infty}{7xe^{-x}} = 14e^{-1}\]

Answer 11

did u get ?

Answer 12

i don't know but i keep getting undefined
here's a pic of what im doing

Answer 13

\[u = x, dv/dx = e^{-x} \therefore du/dx = 1, v = \int\limits {e^{-x}}dx =-e^{-x}\]
\[\int\limits\limits_{a}^{b} uv^{'} = [uv]_{a}^{b} - \int\limits\limits_{a}^{b}vu^{\prime}\]
\[7 \int\limits\limits_{1}^{\infty} {x \times e^{-x}} = [xe^{-x}]_{1}^{\infty} - \int\limits\limits_{1}^{\infty} {e^{-x} \times {1}}\]
\[[x \times e^{-x}]_{1}^{\infty} + [e^{-x}]_{1}^{\infty} = [(1)\times (e^{-1}) - (\infty \times e^{-\infty}] + [e^{-1} -e^{-\infty}] = e^{-1} + e^{-1}\]

Answer 14

\[7 \times[x \times e^{-x}]_{1}^{\infty} + [e^{-x}]_{1}^{\infty} =7 \times [(1)\times (e^{-1}) - (\infty \times e^{-\infty})] + [e^{-1} -e^{-\infty}] \]
\[7 \times [e^{-1} +e^{-1}] = 7 \times 2e^{-1} = 14e^{-1}\]

Answer 15

u got that watch out for the - signs when u expand ?

Answer 16

thanks.. :)

Answer 17

yw