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

Printable View

- May 18th 2006, 12:09 AMnoriveaEvaluate
I have attached an equation that requires an evaluation, I need help working this out, please!

- May 18th 2006, 01:58 AMCaptainBlackQuote:

Originally Posted by**norivea**

Sketch this function over this range - you will see that it is anti-symmetric

about $\displaystyle \pi/2$, so the integral is zero.

RonL - May 19th 2006, 01:42 PMnoriveaReply
So, is the answer:

- May 19th 2006, 01:48 PMTD!
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!). - May 19th 2006, 10:16 PMnoriveaReply
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 - May 19th 2006, 11:25 PMCaptainBlackQuote:

Originally 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 - May 20th 2006, 06:07 PMThePerfectHacker
$\displaystyle \int^{\pi}_0 \cos x dx$

Thus,

$\displaystyle \left \sin x \right|^{\pi}_0$

Thus,

$\displaystyle \sin \pi -\sin 0=0-0=0$ - May 20th 2006, 09:42 PMearbothQuote:

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