Question

Evaluate the double integral.

∫∫ e^(x/y) dA R= {(x, y) |0≤x≤y^3, 1≤y≤2}
R

Please explain

Answer 1

first integrate x keeping y as constant and then integrate y. dA = dxdy

Answer 2

i wasnt sure how to do that

Answer 3

∫ e^(kx) dx .. were k is 1/y --- evaluate this and put the boundary values
=> then integrate with dy

Answer 4

i didnt understand this explanation on how to keep y as a constant and integrate

Answer 5

and wouldn't dA= dydx?

Answer 6

first evaluate only this
∫ e^(kx) dx

Answer 7

that would give me e^2k - e^1k?

Answer 8

again put k=1/y and integrate the function with dy

Answer 9

dA could be dxdy or dydx....
since the limits on the integral for dx(I don't know what they're called)....

notice that e^(x/y) can be integrated with substitution.. why? because the exponent isn't just a single variable x, it has a constant 1/y beside it...
so let u=x/y du=1/ydx dx=ydu
now we have
\[\int\limits_{1}^{2}\int\limits_{0}^{y^3}e^{u}ydudy\]
\[\int\limits\limits_{1}^{2}e^{y^2}y-ydy\]
\[\int\limits\limits_{1}^{2}e^{y^2}y-ydy \]
now we do substitution again in integrating e^(y^2)y. so let u=y^2 du=2ydy ydy=du/2and doing so, we'll have....
\[1/2(e^2-e-4+1)\]
\[1/2(e^2-e-3)\]<---answer
you could simplify if you want with a calculator

Answer 10

the answer is 1/2e^4 - 1/2 e - 3/2 I'm not sure where you went wrong to get 1/2e^2

Answer 11

oh yeah I forgot the exponent of e y^2. After I got the antiderivative I carelessly forgot to substitute back y^2 to u which is why when I evaluated the definite integral, I wasn't able to square the exponent of e which is 2. If I did I would have the e^4 there... so at this part
\[\int\limits\limits_{1}^{2}e^{y^2}y-ydy=1/2(e^{2^2}-e^{1^2}-(4-1))+c\]

Answer 12

which will then end to the answer which is \[(1/2)e^4-(1/2)e-(3/2)\]

Answer 13

Thanks for clearing that up!

Answer 14

sorry I was a bit careless though