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OpenStudy (anonymous):

find the exact area under f(x)=x+cosx on [0,pi]

OpenStudy (anonymous):

integrate

OpenStudy (anonymous):

Answer:Pi^2/2

OpenStudy (anonymous):

the cosine function at 0 takes value of 1 and at pi takes -1 so the area under f(x)=cosx from 0 to pi is 0 now you only need to get the f(x)=x from 0 to pi and that is just a triangle you dont even need integration .-)

OpenStudy (anonymous):

yes it is \[\frac{\pi^2}{2}\]

OpenStudy (anonymous):

\[\int_0^{\pi} x + \cos(x)dx\] \[F(x)=\frac{x^2}{2}-\sin(x)\] \[F(\pi)-F(0)=\frac{\pi^2}{2}\]

OpenStudy (anonymous):

I prefer my solution :-)

OpenStudy (anonymous):

yes snappier

OpenStudy (anonymous):

Moral: always draw a rough diagram first, to see what's going on

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