Question

Evaluate the integral
Int((t)(e^-t))dt from 0 to 1
(Integration by parts)

Answer 1

\[\int\limits_0^1\ln(t)e^{-t}\text dt\]

Answer 2

No ln(t)
just t

Answer 3

\[\int\limits_{0}^{1}t(e ^{-t})dt\]

Answer 4

oh, like this?
\[\int\limits_0^1 (t)e^{-t}\text dt\]

Answer 5

yes

Answer 6

yeah ok i see now

Answer 7

Integration by parts\[\int\limits_a^b u\text d v=[uv]_a^b-\int\limits_a^b v\text du\]

Answer 8

yes but I'm having trouble evaluating it

Answer 9

choose
\(u=t\qquad\text du=\text dt\)
\[\text dv=e^{-t}\qquad v={-e^{-t}}\]

Answer 10

I did, I'm still having trouble - would you possibly be able to do the whole thing out for me?

Answer 11

\[\int\limits_a^b u\text d v=[uv]_a^b-\int\limits_a^b v\text du\]\[\downarrow\]\[\int\limits_0^1 te^{-t}\text dt=[-te^{-t}]_0^1+\int\limits_0^1 e^{-t}\text dt\]