2 Order of integration

All of the preceding Examples and Tasks have been integrals of the form

$I={\int \; <!--nolimits\; -->}_{x=a}^{x=b}{\int \; <!--nolimits\; -->}_{G_1\left(x\right)}^{G_2\left(x\right)}f\left(x,y\right)dydx$

These integrals represent taking vertical slices through the volume that are parallel to the $yz$ -plane. That is, vertically through the $xy$ -plane.

Just as for integration over rectangular regions, the order of integration can be changed and the region can be sliced parallel to the $xz$ -plane. If the inner integral is taken with respect to $x$ then an integral of the following form is obtained:

$I={\int \; <!--nolimits\; -->}_{y=c}^{y=d}{\int \; <!--nolimits\; -->}_{H_1\left(y\right)}^{H_2\left(y\right)}f\left(x,y\right)dxdy$

Example 13

The following integral was evaluated in Example 9.

$I={\int \; <!--nolimits\; -->}_{x=0}^1{\int \; <!--nolimits\; -->}_{y=0}^{x^2}2xsin\left(y\right)dydx=1-sin\left(1\right)$

Change the order of integration and confirm that the new integral gives the same result.

Figure 16

Solution

The integral is taken over the region which is bounded by the curve $y=x^2$ . Expressed as a function of $y$ this curve is $x=\sqrt{y}$ . Now consider this curve as bounding the region from the left, then the line $x=1$ bounds the region to the right. These then are the limit functions for the inner integral $H_1\left(y\right)=\sqrt{y}$ and $H_2\left(y\right)=1$ . Then the limits for the outer integral are $c=0\le y\le 1=d$ . The following integral is obtained

Task!

The double integral $I={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_x^1{\text{e}}^{y^2}dydx$ involves an inner integral which is impossible to integrate. Show that if the order of integration is reversed, the integral can be expressed as $I={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^y{\text{e}}^{y^2}dxdy$ . Hence evaluate the integral $I$ .

Answer