# Thread: Area under curves again...

1. ## Area under curves again...

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
$y=\sqrt{x}$
$y=1/2(x)$
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
$so area =\int_{0}^4 \sqrt{x}-\frac{1}{2}(x)dx + \int_{9}^4 \frac{1}{2}(x)-\sqrt{x}dx$
anti derivative of first integral $= \frac{2}{3}x^\frac{3}{2}-\frac{1}{4}x^2$
antiderivative of 2nd integral $= \frac{1}{4}x^2-\frac{2}{3}x^\frac{3}{2}$
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

2. Your first integral is correct.

The second should be $\displaystyle \int_4^9{\frac{1}{2}x - \sqrt{x}\,dx}$.