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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 $1$.

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  • $\begingroup$ The region doesn't depend on the integrand! $\endgroup$
    – Hans Lundmark
    Nov 17, 2015 at 19:24
  • $\begingroup$ It doesn't matter, because the region of integration only represents the domain of the function, not the function itself $\endgroup$
    – Dylan
    Nov 17, 2015 at 19:29
  • $\begingroup$ @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? ? $\endgroup$
    – Ovi
    Nov 17, 2015 at 19:34
  • $\begingroup$ Yes. That region is the same no matter the function. Unless the function is undefined there, but that's another issue. $\endgroup$
    – Dylan
    Nov 17, 2015 at 19:52

1 Answer 1

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

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  • $\begingroup$ So the region is just bounded vertically by y = 2 and y = x, and horizontally by x = 2 and x = 0? $\endgroup$
    – Ovi
    Nov 17, 2015 at 19:34
  • $\begingroup$ Yepp. that is it $\endgroup$
    – user225425
    Nov 17, 2015 at 19:39
  • $\begingroup$ So the integrand really just might as not well be there and it would be the same problem haha? $\endgroup$
    – Ovi
    Nov 17, 2015 at 19:41
  • $\begingroup$ 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. $\endgroup$
    – Simon S
    Nov 17, 2015 at 19:44
  • $\begingroup$ If that's not clear, try evaluating as written. Then evaluate it after the swap (as is also required by your question). $\endgroup$
    – Simon S
    Nov 17, 2015 at 19:51

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