Double integral e^(ysqrtx)dxdy

[Digitalism]

Homework Statement

∫∫e^(y√x)dxdy from 1 to 4 then from 0 to 2

Homework Equations

∫ e^x = e^x
u substitution

The Attempt at a Solution

I am just curious if this is equal to double integral e^(y\sqrt{x})dydx from 0 to 2 then from 1 to 4. In other words can I change the order of integration without screwing up my function? If so I can solve it. If not I have tried U substitution:

u = y√x
du = y/2√x} dx

which changes my equation to double integral (2√x/y)e^(u)dudy from 1 to 4 then from 0 to 2 which is equal to (2u/y^2)e^(u)dudy from 1 to 4 then from 0 to 2 but I don't see how integrating that will give me 2e^u(u-1)/y^2 which was wolfram alpha's indefinite integration for e^y\sqrt{x}. help?

 

[SammyS]

Homework Statement

∫∫e^(y√x)dxdy from 1 to 4 then from 0 to 2

Homework Equations

∫ e^x = e^x
u substitution

The Attempt at a Solution

I am just curious if this is equal to double integral e^(y\sqrt{x})dydx from 0 to 2 then from 1 to 4. In other words can I change the order of integration without screwing up my function? If so I can solve it. If not I have tried U substitution:

u = y√x
du = y/2√x} dx

which changes my equation to double integral (2√x/y)e^(u)dudy from 1 to 4 then from 0 to 2 which is equal to (2u/y^2)e^(u)dudy from 1 to 4 then from 0 to 2 but I don't see how integrating that will give me 2e^u(u-1)/y^2 which was wolfram alpha's indefinite integration for e^y\sqrt{x}. help?

Assuming you integral is :

\displaystyle \int_0^2\int_1^4 {e^{y\sqrt{x}}}\,dx\,dy​

It should be fine to switch the order of integration.

If you're trying to do the integration in the given order, then remember you need to treat y as a constant as you integrate with respect to x. Remember then to evaluate that as a definite integral over x.

 

[Digitalism]

thanks! that's what I thought, so now I can solve it