1. ## Evaluate

I have attached an equation that requires an evaluation, I need help working this out, please!

2. Originally Posted by norivea
I have attached an equation that requires an evaluation, I need help working this out, please!
$\int_0^{\pi} cos(x) dx$.

Sketch this function over this range - you will see that it is anti-symmetric
about $\pi/2$, so the integral is zero.

RonL

4. No, you'd get:

$
\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!).

Ok thank you. I am just uncertain where I went wrong in my working. Could you look over my working to see where I went wrong? I have attached my original working.

Thanks again

6. Originally Posted by norivea
Ok thank you. I am just uncertain where I went wrong in my working. Could you look over my working to see where I went wrong? I have attached my original working.

Thanks again
Why did you split the range of integration into $0-\pi/2$, and
$\pi/2-\pi$?

The sign of $\cos$ is positive on the first of your sub-ranges and
negative on the second, so they cancel rather than add (since $\cos$
is anti-symmetric about $\pi/2$).

RonL

7. $\int^{\pi}_0 \cos x dx$
Thus,
$\left \sin x \right|^{\pi}_0$
Thus,
$\sin \pi -\sin 0=0-0=0$

8. Originally Posted by norivea
Ok thank you. I am just uncertain where I went wrong in my working. Could you look over my working to see where I went wrong? I have attached my original working.
Thanks again
Hello,

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:
$\int_{0}^{\pi}|{\cos(x)| dx$ which differs from your orginal integral.

Greetings

EB