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

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- May 18th 2006, 01:09 AMnoriveaEvaluate
I have attached an equation that requires an evaluation, I need help working this out, please!

- May 18th 2006, 02:58 AMCaptainBlackQuote:

Originally Posted by**norivea**

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

about , so the integral is zero.

RonL - May 19th 2006, 02:42 PMnoriveaReply
So, is the answer:

- May 19th 2006, 02:48 PMTD!
No, you'd get:

You could have found this without integrating with CaptainBlack's argument, remember what an integral represents (signed area!). - May 19th 2006, 11:16 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 20th 2006, 12:25 AMCaptainBlackQuote:

Originally Posted by**norivea**

?

The sign of is positive on the first of your sub-ranges and

negative on the second, so they cancel rather than add (since

is anti-symmetric about ).

RonL - May 20th 2006, 07:07 PMThePerfectHacker

Thus,

Thus,

- May 20th 2006, 10:42 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:

which differs from your orginal integral.

Greetings

EB