ewww that one is gonna put hair on your chest....

>.< i been trying to do it but no luck :3

are you looking for an exact solution or just an approximation?

im looking for steps on how to solve it because i have no idea how to take the integration of 5x^2+e^x3+5y dx. i think thats why im unable to solve it :3 unless it gets more difficult after i intergrate it again for dx
i'm looking for an exact solution. I tried plugging the equation into wolfram alpha but i got a really weird solution

This post was edited by liniscool on Sep 11 2013 01:26am

ok wait...that ^^^ is not your original integral kernel. Can you please double check it and let us know.

This post was edited by Azrad on Sep 11 2013 01:29am

i probably typed that wrong, but the original pictured i posted is the correct equation.

This post was edited by liniscool on Sep 11 2013 01:35am

exp(x^(3+5y)) = exp(5y)^(x^3)
v = exp(5y)
dv = 5exp(5y)dy

Your integral (lets call it I) is equal to
I = int(int(x^2*v^(x^3)dxdv)) [with correct intervals]
Which is also equal to (i didnt check, but should be justified with appropriate theorems... ) :
I = int(int(x^2*v^(x^3)dx)dv) = int(J(v)dv)

J(v) = int(x^2*v^(x^3)dx) = (1/3)*[v^(b^3) - v^(a^3)], a and b be being the endpoints of the interval of integration of J (i.e. a = 0, b = 1))
J(v) = (1/3)*(v-1)

Assuming this is right, you have then to integrate J over v between exp(0) and exp(5*1)

Hope it helps, I apologize for the presentation

This post was edited by HbSoe on Sep 11 2013 07:44pm

Do you mean :

exp(x^(3+5y)) = exp(x^(5y))^(x^3)

?