Posted by **Ben** on Tuesday, December 4, 2007 at 11:04pm.

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

- Math -
**drwls**, Tuesday, December 4, 2007 at 11:27pm
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

- Math -
**Reiny**, Wednesday, December 5, 2007 at 12:10am
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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