Question

I need help! :)
integral xe^x/5

Answer 1

What have you considered doing with this integral?

Answer 2

I did a I p p

So u=x
du=dx
dv=e^-x/5
v=-5^e-x/5

Answer 3

Ops de integral is xe^-x/5

Answer 4

Oh, that makes sense. And yes, I can agree with that setup. So you rewrote the integral then?
integral u dv = uv - integral v du

Answer 5

\[-5e^-x/5x-25e^-x/5 + C\]

Answer 6

\( -5 e^{-x/5} x - 25 e^{-x/5} + C \)
I think this is what you meant, and this appears to be correct to me. :)

Answer 7

really?
\[\int\limits_{5}^{\infty} xe^-x/5\]

because when i do my integral it does not work :(

Answer 8

Ohh, you have integration limits here.
So what did you end up with when you tried this?

Answer 9

actually i replace my infinity with b :)

Answer 10

The infinity part we should perhaps treat with a limit:
Put in a dummy variable that we send towards infinity... yeah
\( \displaystyle \lim_{b \to \infty} \int_{5}^{b} xe^{-x/5} \ dx \)
So that we evaluate this limit in the end:
\( \displaystyle \lim_{b \to \infty} \left( -5be^{-b/5} - 25e^{-b/5} \right) - F(5)\)
I made F(5) the antiderivative of 5 for convenience because that part is just substituting.

Answer 11

yes i have this :) but when i did my calcul it done 30/e

Answer 12

\[put-\frac{ x }{5 }=u,x=-5u,dx=-5du\]
when x=5,u=-1
when \[x=\infty,u=-\infty \]
\[I=\int\limits_{-1}^{-\infty}-5ue^u(-5du)\]
solve it

Answer 13

but the is 50/e and i dont understand why

Answer 14

I dont understand this integral :/

Answer 15

That first part goes to 0 entirely because e^(-x//5) has much more control. Or you could use L'Hopital's rule by putting the e^(-x/5) into the denominator so that you have infty/infty, and prove for yourself that it is 0.

F(5) = -5*5e^(-5/5) - 25e^(-5/5)
= -25 e^(-1) - 25 e^(-1)

Answer 16

OMG I forgot my x !!!! U are so good! THANK U

Answer 17

Ah, there you go! I'm glad to help! :D

Answer 18

Just as a side-note, I had encountered a more convenient way to approach integration by parts if you ever have a case where your u part is a polynomial that differentiates repeatedly into 0 and the dv is repetitive.
Tabular integration can be pretty useful, maybe not so important for this problem as its one-step but say you had x^3 e^(2x) you would have a lot of integration by parts!
http://www.uvu.edu/mathlab/docs/handouts/tabular_integration.pdf
u dv
x e^(-x/5)
1 -5e^(-x/5)
0 +5*5e^(-x/5)
-5e^(-x/5) * x - 1*(5*5e^(-x/5))
Just a neat tool to have, although you don't necessarily need it. :P

Answer 19

\[I=25\left[ u e^u)from-1~\to~-\infty -\int\limits_{-1}^{-\infty } 1.e^u~du\right]\]
\[=-25\left[ ue^u-e^u \right]from -\infty ~\to-1\]
\[=-25~e^u \left( u-1 \right)from~-\infty~\to-1\]
\[e ^{-\infty }=\frac{ 1 }{ e^\infty }\rightarrow 0\]
\[-25\left( 0-\left( e ^{-1} \left( -1-1 \right)\right) \right)=-\frac{ 50 }{ e }\]