Anyone here any good at integration by parts and stuff??

Question

unklerhaukus · Answer

maybe

anonymous · Answer

ok. I've done this sum, but I wanna check that i've done it right and got the right answer. So, solve this using integration by parts.

$$\int\limits_{0}^{2}(2-x)e^-x dx$$

anonymous · Answer

Generally with forms of $$
\int f(x)e^{\pm x}\;dx
$$You want to sub $u=f(x),dv=e^{\pm x}dx$

anonymous · Answer

Did that :)

anonymous · Answer

let me just rewrite it$$\int\limits_{0}^{2}(2-x)e^{-x} dx$$:)

anonymous · Answer

ok, but i also expanded it, and kinda solves the xe^-x part separately, and then jioned them together. Will that get me the same answer??

unklerhaukus · Answer

$$uv'+vu'=(uv)'\
∫udv+∫vdu=uv\
∫udv=uv\big|-∫vdu$$

anonymous · Answer

@tanvidais13 Just show us your work please.

anonymous · Answer

ahh ok. and sorry if it looks a bit weird, but i don't know how to make more than one thing come as a superscript here :P

1) $$\int\limits_{0}^{2} 2e^-x - xe^-x = \int\limits_{0}^{2} 2e^-x - \int\limits_{0}^{2}xe^-x$$
and then i worked from here, solving the two individually and combining then at the end. I only put in the values after i had integrated both of them.