calc
posted by mikayla .
find the area between the xaxis and the graph of the given function over the given interval:
y = sqrt(9x^2) over [3,3]
you need to do integration from 3 to 3.
First you find the antiderivative
when you find the antiderivative you plug in 3 to the antiderivative 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(9x^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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