over the region $R$ bounded by the lines $y = x$, $y = 0$, and $x = a$ (where $a > 0$). Show that the result is the same if the order of integration is reversed. Next, show that
\[
I = \int_0^a \int_0^x f(y) \, de{y} \de{x} = \int_0^a (a-y) f(y) \, \de{y}.
\]

Spoiler

The first integral can be verified to be
\[
I = \sin a - \frac{\sin 2a}{2}.
\]

To reverse the order of integration, use
\[
\int_0^a \int_0^x f(y) \, de{y} \de{x} = \int_0^a \int_y^a f(y) \, \de{x} \, de{y}.
\]

4. Reversing the integration order

(a) Evaluate
\[
\iint_R y \, \de{x} \de{y}
\]

over the region $R$ bounded by the lines $y = x$, $y = 2 - x$, and $y = 0$. Write down the integrals which must be evaluated if the order of integration is reversed.

(b) Change the order of integration in the repeated integral
\[
\int_0^1 \int_x^{2-x} \frac{x}{y} \de{y} \de{x},
\]

and evaluate the result.

Spoiler

The value of the integral should be $1/3$. To reverse the integration, we must split into two:
\[
I = \left(\int_0^1 \int_0^x + \int_1^2 \int_0^{2-x}\right) y \, \de{y} \, \de{x}.
\]

For the second integral, the value should be $-1 + 2 \log 2$.

5. An integral with multiple bounding curves

Sketch the region $R$ which is in the positive quadrant and is bounded by the curves
\[
xy = 2, \quad y = \frac{x^2}{4}, \quad y = 4.
\]

By integrating first with respect to $x$ and then $y$, find the area $R$. Check that the result is the same if you reverse the order of integration.

Spoiler

The value of the integral should be $28/3 - 2\log 4$.