Question

In Exercises \(43-50\), sketch the region \(R\) whose area is given by the iterated integral. Then switch the order of integration and show that both orders yield the same area. $$ \int_{0}^{1} \int_{y^{2}}^{\sqrt[3]{y}} d x d y $$

Short answer

Thus, switching the order of integration gives the same region, and hence the same exact value of the integral, thereby demonstrating the validity of the Fubini's theorem in this case.

Step-by-step solution

Sketch the region of integration

The first integral iterates from 0 to 1, represented on y-axis and for each y, the second integral iterates from \(y^2\) to \(\sqrt[3]{y}\), represented on x-axis. So, it outlines a region R in the xy-plane that lies between the curves \(y = x^2\), \(x = y^3\) and \(y = 1\), \(y = 0\).

Redefine the limits of the region in terms of x

We can also represent the region in terms of x. For this, we observe that x varies from 0 to 1, and for each x, y varies from \(x^2\) to \(x^{1/3}\).

Write the integral with switched order

Now we can change the order of integration and keep the specified limits. The iterated integral with the switched order of integration becomes \(\int_{0}^{1} \int_{x^2}^{x^{1/3}} dy dx\).

Show that both orders yield the same area

Given that both integrals \(\int_{0}^{1} \int_{y^{2}}^{\sqrt[3]{y}} dx dy\) and \(\int_{0}^{1} \int_{x^2}^{x^{1/3}} dy dx\) represent the same region R, they must have the same value. If you calculate both integrals they will provide the same numerical solution.