Recent content by ChargedTaco

Nevermind...I got it. Thanks guys.

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.

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...

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.:frown:

thanks, I'll edit it. oops, looks like I can't change the title.

Homework Statement An athlete puts a 16lbs shot at an angle of 45 degrees to horizontal from 6.5ft above the ground at initial speed of 44ft/sec. How far does the shot land? g=32ft/s^2 Homework Equations I can only calculated the range from the point it is fired to the point...