# Region of integration for a double integral

My book gives the integral $$\int_0^2 \int_x^2 x\sqrt{1+y^3} dydx$$

And the directions say to sketch the region of integration, and evaluate the integral by switching the order of integration.

Once I have the region sketched out I think I know how to find the area by changing it to $dxdy$, but I'm lost at how to draw the region. Up to now, when we had iterated integrals representing regions, they were always of the form $$\int_b^a \int_{g(x)}^{f(x)} dydx$$or $$\int_b^a \int_{g(y)}^{f(y)} dxdy$$

I know how to draw the region for these where the integrand is $1$, but I've never encountered one where the integrand isn't $1$.

• The region doesn't depend on the integrand! Nov 17, 2015 at 19:24
• It doesn't matter, because the region of integration only represents the domain of the function, not the function itself Nov 17, 2015 at 19:29
• @Dylan So I just pretend like the integrand doesnt exist and So the region is just bounded vertically by y = 2 and y = x, and horizontally by x = 2 and x = 0? ?
– Ovi
Nov 17, 2015 at 19:34
• Yes. That region is the same no matter the function. Unless the function is undefined there, but that's another issue. Nov 17, 2015 at 19:52

The region is the same. You can think about the change of integrand as change of density or weight if you want.

• So the region is just bounded vertically by y = 2 and y = x, and horizontally by x = 2 and x = 0?
– Ovi
Nov 17, 2015 at 19:34
• Yepp. that is it
– user225425
Nov 17, 2015 at 19:39
• So the integrand really just might as not well be there and it would be the same problem haha?
– Ovi
Nov 17, 2015 at 19:41
• No, the integrand does matter. The point of the exercise is this: as written the integral is all but impossible to evaluate. But by swapping the order of integration, you can integrate it easily. This suggests that the technique is a useful tool to add to your tool box for all double integrals in future. Nov 17, 2015 at 19:44
• If that's not clear, try evaluating as written. Then evaluate it after the swap (as is also required by your question). Nov 17, 2015 at 19:51