Question

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{1}^{2} \int_{2}^{4} d x d y $$

Short answer

The area of the region R, determined by the double integral, is 2. Changing the order of integration did not result in a different area, demonstrating the integral equivalence.

Step-by-step solution

Sketch the Region

The double integral \(\int_{1}^{2} \int_{2}^{4} d x d y\) covers a rectangular region in the xy-plane. We can visualize this as a rectangle with corners at (2,1), (2,2), (4,1), and (4,2)

Compute the Double Integral with Original Order

The current order of integration is dx dy. Sequentially perform the integration, first integrate with respect to x from 2 to 4, then integrate with respect to y from 1 to 2. This yields the area \( \int_{1}^{2} \int_{2}^{4} d x d y = (4-2)*(2-1) = 2 \)

Change the Order of Integration

The order dx dy can be switched to dy dx. The new double integral will be \(\int_{2}^{4} \int_{1}^{2} d y d x\)

Compute the Double Integral with New Order

Now repeat the computation of the area with the new order. This means we first integrate with respect to y from 1 to 2, then with respect to x from 2 to 4. This gives \(\int_{2}^{4} \int_{1}^{2} d y d x = (2-1)*(4-2) = 2 \)

Comparison of the results

You will notice that the area obtained in both cases is 2, which demonstrates that changing the order of integration did not affect the result.