I have attached an equation that requires an evaluation, I need help working this out, please!
No, you'd get:
$\displaystyle
\int\limits_0^\pi {\cos xdx} = \left[ {\sin x} \right]_0^\pi = \sin \pi - \sin 0 = 0 - 0 = 0
$
You could have found this without integrating with CaptainBlack's argument, remember what an integral represents (signed area!).
Why did you split the range of integration into $\displaystyle 0-\pi/2$, andOriginally Posted by norivea
$\displaystyle \pi/2-\pi$?
The sign of $\displaystyle \cos$ is positive on the first of your sub-ranges and
negative on the second, so they cancel rather than add (since $\displaystyle \cos$
is anti-symmetric about $\displaystyle \pi/2$).
RonL
Hello,Originally Posted by norivea
according to your working you didn't evaluate the integral but you calculated the area which is enclosed by the x-axis, the graph of the cosine function and the lines x = 0 and x = pi. So you calculated:
$\displaystyle \int_{0}^{\pi}|{\cos(x)| dx$ which differs from your orginal integral.
Greetings
EB