Question

Reverse the order of integration in the following integrals. $$\int_{0}^{2} \int_{x^{2}}^{2 x} f(x, y) d y d x$$

Short answer

The reversed order integral for the given integral is: $$\int_{0}^{4} \int_{\sqrt{y}}^{\frac{1}{2}y} f(x, y) d x d y$$.

Step-by-step solution

Visualize the region

First, let's visualize the region described by the given integral. The region is bounded by the following: 1. \(x = 0\) 2. \(x = 2\) 3. \(y = x^2\) 4. \(y = 2x\) To find the intersection points between curves \(y=x^2\) and \(y=2x\), $$x^2 = 2x$$ $$x^2-2x=0$$ $$x(x-2)=0$$ So, \(x=0\) and \(x=2\) are the intersections. Region A is a triangular region with vertices at (0,0), (2,0), and (2,4).

Determine the new limits of integration

With the region now known, we can determine the new limits of integration in terms of y. For the outer integral, we need the range of y-values in the region A. y ranges from 0 to 4. Therefore, the outer integral (with respect to y) goes from 0 to 4. For the inner integral, we need the range of x-values as a function of y. In region A, the left boundary is formed by the curve \(y=x^2\) which in terms of x is \(x=\sqrt{y}\), and the right boundary is formed by the curve \(y=2x\) which in terms of x is \(x=\frac{1}{2}y\). Therefore, the inner integral (with respect to x) goes from \(\sqrt{y}\) to \(\frac{1}{2}y\).

Write down the reversed integral

Now that we have the limits of integration for the reversed order integral, we can write down the final answer. The reversed integral (integrate first with respect to x and then to y) is: $$\int_{0}^{4} \int_{\sqrt{y}}^{\frac{1}{2}y} f(x, y) d x d y$$