Need some double integral help.

[ChargedTaco]

Evaluate.

double integral (e^(y^3)) dy dx

Where dy is evaluated from sqrt(x/3) to 1

...and dx is evaluated from 0 to 3.

I am lost.
I don't even know how to start.

 

[mjsd]

do u know how to do integrals?
double integral is just normal integrals in disguise, so when ou are doing the "y"-integral, you treat all "x"s in the integrand as just a constant, and while doing the "x"-integral you treat all "y"s as constant. Then, everything would then be exactly the same as a normal single integral. In many cases, whether you do the x or y integral first does not matter.
in your case i think you should do y first

 

[dextercioby]

The "y" integration doesn't look pretty. It's basically

$$\int_{a}^{b} e^{y^3}{}dy$$

 

[ChargedTaco]

Hey guys yes I do know how to do integrals...however I don't know how to do this one. If I integrate y first I have to integrate e^(y^3)dy...and i don't know how to do this.

If I integrate x first then I'll end up with having to
integrate 3e^(y^3)dy...and I again I don't know how to do this.

Thanks.

 

[ChargedTaco]

Here it is.

$$\int_{0}^{3} \int_{sqrt(x/3)}^{1} e^{y^3} {}dy} {}dx$$

Yay! I got the latex code...however this is the only thing I know about this problem.

 

[ChargedTaco]

Nevermind...I got it. Thanks guys.