find the area between the x-axis and the graph of the given function over the given interval:
y = sqrt(9-x^2) over [-3,3]
you need to do integration from -3 to 3.
First you find the anti-derivative
when you find the anti-derivative you plug in -3 to the anti-derivative and then plug in 3 and find the difference of : f(-3)-f(3). Hope this helps :)
but how to you find the antideriv?
Paste sqrt(9-x^2) into
http://integrals.wolfram.com/index.jsp
This is an arcsin derivative.
Substitute x = 3 sin(t). Then t goes from - pi/2 to pi/2.
sqrt[9 - x^2] = 3 sqrt[1 - sin^2(t)] =
3 |cos(t)|
and dx = 3 cos(t) dt
So, you have to integrate 9 cos^2(t) from t = -pi/2 to pi/2 (note that in the interval cos(t) is positive so you can remoce the absolute value signs).
Since we are integrating over an entire period of the cos^2 the integral would be the same if you replace cos^2 by sin^2. By replacing cos^2 by cos^2 + sin^2 you thus obtain twice the value of the integral but since sin^2 + cos^2 = 1 that's 9 pi. So, the integral is 9/2 pi.
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