$$\int\limits_{0}^4\int\limits_{\sqrt{y}}^{2}{\sqrt{x^{3}+1}} dxdy$$

Question

blockcolder · Answer

Is that lower bound on x $\Large \sqrt{3}$?

anonymous · Answer

it's on $$\sqrt{y}$$

blockcolder · Answer

Right. 
Our first step here is to see that we really can't evaluate $\large\int \sqrt{1+x^3} dx$ so we can't evaluate the double integral as it is. One possible remedy that usually works is to reverse the order of integration.
But first, give me a sketch of the region we're integrating over.

anonymous · Answer

∫1+x3−−−−−√dx

anonymous · Answer

they don't give me a sketch...they just say evaluate the integral...i tried to do it as they gave me but once i integrate with respect to x it gets messy... it becomes $$2/9\int\limits_{0}^{4}27/4-((y \sqrt{y}+1)^3/2)/y    dy$$

anonymous · Answer

it's supposed to be raised to the 3/2 and not raised to 3 and then divided by 2

blockcolder · Answer

Yeah, that's because you can't directly evaluate $\Large \int \sqrt{1+x^3}\ dx$, as I said. However, you CAN evaluate $\Large \int \sqrt{1+x^3} dy$, and that's what we're gonna do.
If they don't provide a sketch, it's up to you to infer what the region is based on the bounds of the integrals.
For now, here's the sketch:
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