Question

DOUBLE INTEGRATE: Ouside Integal (v=1, v=4) Inside Integral (u=0, u=1) [ ve^(uv)dudv ] ... i dont kno what to do with du??

Answer 1

\[\int_1^4\int_0^1ve^{uv}dudv\]?

Answer 2

yes

Answer 3

wait
there is also (1/2) in the beginning

Answer 4

do this first, treat \(v\) as a constant
\[\int_0^1ve^{uv}du\]

Answer 5

so e^uv , where e^v - 1 will be left

Answer 6

yes that is what i get

Answer 7

then
\[\int_1^4(e^v-1)dv=\int_1^4e^vdv-3\] etc

Answer 8

ok got it
it was easy

Answer 9

yeah, just hope it was right!

Answer 10

(1/2)(e^4 - e - 3) how? that is the answer

Answer 11

that is what i get, assuming you have a \(\frac{1}{2}\) outside

Answer 12

ok

Answer 13

anti derivative of \(e^v\) is \(e^v\) plug in 4 get \(e^4\) plug in 1 get \(e\) subtract get \(e^4-e\)