the instruction says sketch the region enclosed by the given curves. decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. find the area of that region. I'm assuming when it says x=9, area up to x = 9

$\displaystyle y=\sqrt{x}$

$\displaystyle y=1/2(x)$

x=9

points of intersection, x = 0, x =4

y=\sqrt{x} is on top from x = 0 to 4

and on bottom from x = 4 to x = 9

find area using integrals from x = 0 to x = 4 then from x = 4 to x = 9

$\displaystyle so area =\int_{0}^4 \sqrt{x}-\frac{1}{2}(x)dx + \int_{9}^4 \frac{1}{2}(x)-\sqrt{x}dx$

anti derivative of first integral $\displaystyle = \frac{2}{3}x^\frac{3}{2}-\frac{1}{4}x^2$

antiderivative of 2nd integral $\displaystyle = \frac{1}{4}x^2-\frac{2}{3}x^\frac{3}{2}$

then

area = [F(4)-F(0)]+[F(9)-F(4)]

When I finish all the arithmetic, i get a bunch of square roots... and the answer is 59/12