Math
posted by Ben .
Find the area between y=3sinx and y=3cosx over the interval [0,pi].

Find the indefinite integral of
3 sin x  3 cos x.
It is 3 sin x  3 cos x.
Finally, evaluate that at x = pi and subtract the value at x = 0, the two limts of integration.
I get (0 + 3)  (0 3(1)) = 6
Over part of the interval, the 3 cos x curve is higher than 3 sin x. That part of the area integral is counted as negative 
As drwls noted, the cosine curve is above the sine curve from x =0 to x=pi/4 and from pi/4 to pi, the sine curve is higher.
So the calculation has to be done in two parts
Area = integral(3cosx  3sinx)dx from 0 to pi/4 + integral(3sinx  3cosx)dx from pi/4 to pi.
I got 6√2 as a final answer
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