I want to take the solve the double integral of $sin(x)$ where the bounds are $x \leq y \leq 2x$ and $1 \leq x \leq 5$.

Instead of using the given bounds, I want to solve by reversing the order of integration to avoid integration by parts.

I tried using the new bounds $1 \leq y \leq 10$ and $y \leq x \leq \frac {y}{2}$. I had no luck finding the correct bounds integrating using dxdy instead of dydx that yield the correct solution when reversing the order of integration.

I need help finding the correct bounds given that the correct value of the double integral is $-2.67$.

  • $\begingroup$ Make a picture of your region. If you want to reverse the order of integration, you need to break it up into 3 integrals: $\int_1^2\int_1^y f(x,y)\ dx \ dy + \int_2^5\int_{\frac 12 y}^y f(x,y)\ dx \ dy + \int_5^{10}\int_{\frac 12 y}^5 f(x,y)\ dx \ dy $ $\endgroup$ – Doug M May 9 '17 at 22:14

When you reverse the order of integration, the left- and right- bounding functions are piecewise continuous. You'd have to split the integral into 3 pieces in order to evaluate this. You're better off integrating by parts.

But for the record, your new integral would be

$$\int_1^2 \int_1^y \sin x \; dx \; dy+ \int_2^5 \int_{y/2}^y \sin x \; dx \; dy+\int_5^{10} \int_{y/2}^5 \sin x \; dx \; dy.$$

  • $\begingroup$ How do you know they are piecewise continuous and not continuous for all x,y in the interval? $\endgroup$ – user420360 May 9 '17 at 22:23
  • 1
    $\begingroup$ They're the boundaries of a closed, convex region. $\endgroup$ – B. Goddard May 9 '17 at 22:26

This should illustrate why you must divide the region into 3 pieces.

enter image description here


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