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This is an exam problem that should be solvable in less than 30 minutes:

$$\int _{0}^{1}\!\int _{{x}^{2}}^{1}\!{x}^{3}\sin \left( {y}^{3}\right) {dy}\,{dx}$$

I have tried switching the order of integration and the the boundaries like so:

$$\int _{0}^{1}\!\int _{\sqrt {y}}^{1}\!{x}^{3}\sin \left( {y}^{3}\right) {dx}\,{dy}$$

But I always end up with having to evaluate something of the form:

$$\int \!\sin \left( {y}^{3}\right) {dy}$$

Which even using software looks like a difficult one to evaluate and gives an absurdly long answer. Any pointers would help, I don't necessarily need all of the steps but if you can it would be very helpful.

Many thanks!

qwr
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user33355
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1 Answers1

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You made a mistake in the change of order. It should be $$\int _{0}^{1}\!\int _{0}^{\sqrt y}\!{x}^{3}\sin \left( {y}^{3}\right) {dx}\,{dy}$$

for $\{(x,y):0\leqslant x\leqslant 1\;,x^2\leqslant y\leqslant 1\}=\{(x,y):0\leqslant y\leqslant 1\;, 0\leqslant x\leqslant \sqrt y\}$. (Make a drawing!)

Then you get

$$\int _{0}^{1}\!\int _{0}^{\sqrt y}\!{x}^{3}dx\sin \left( {y}^{3}\right) {dy}=\int _{0}^{1}\left.\frac{x^4}{4}\right\vert_0^{\sqrt y}\sin \left( {y}^{3}\right) {dy}=\int _{0}^{1}\frac{y^2}{4}\sin \left( {y}^{3}\right) {dy}$$

And things are easy now.

Pedro
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  • Ohh.. now that you mention it.. wow. I think I'll have to draw the domain of integration each time from now on. Thank you! – user33355 Feb 20 '14 at 00:41