integration help (1 Viewer)

[shuning]

 

[shaon0]

S t.e^(-t) dt
Let u=t, dv=e^(-t) dt
S u dv=-t.e^(-t)+S e^(-t) dt
S t.e^(-t) dt=-e^(-t)*(t+1)+C

 

[shuning]

Where did dv come from?

 

[independantz]

Where did dv come from?

 

[shuning]

i got a different answer from subsitituion, can some1 tell me where i got wrong?

S te^(-t) dt
let u = e^(-t)
du/dt = -te^(-t)
dt = du/(-te^(-t))

therefore
S te^(-t) dt
= - S (te^(-t))/te^(-t) du
=- S 1 du
= -e^(-t)+ c

 

[roadrage75]

i got a different answer from subsitituion, can some1 tell me where i got wrong?

S te^(-t) dt
let u = e^(-t)
du/dt = -te^(-t)
dt = du/(-te^(-t))

therefore
S te^(-t) dt
= - S (te^(-t))/te^(-t) du
=- S 1 du
= -e^(-t)+ c

this is where you went wrong:

you said let u = e^(-t)

if this is the case, then u' = -e^(-t) not -te^(-t)..... i think you should use the method of those who posted above ---- integration by parts.

 

[independantz]

i got a different answer from subsitituion, can some1 tell me where i got wrong?

S te^(-t) dt
let u = e^(-t)
du/dt = -te^(-t)
dt = du/(-te^(-t))

This line should be, du/dt=-e^-t

 

[x.Exhaust.x]

Let I=te^-t
Let u=t, let dV=e^-t.dx,
du=dt, v=e^-t
Therefore: I=uv-int.v.du (Integration By Parts Formula)
=-te^t-int.e^-t
=-te^-t-e^-t
=-e^-t(t+1) + C

Sorry about the messy writing. I tried LaTeX but my working out couldn't seem to go under each part, but in one line only. Any fix?