Question

ok, i tried to do what you told me but i cant solve it for c because they cancel each others out!

the integral for the first one i got is [sin(c)cos(x)-cos(c)sin(x)+sin(x)+c]
and
the integral for the 2nd one i got is [-sin(c)cos(x)+cos(c)sin(x)-sin(x)+c]

I don't know what to do from here. can u pleses help??? To tell you the truth i don't really get what you are saying. I really need to get the correct answer. Please Help.
Here is your respond for my question from yerster day. Thank You.

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Thank you for reposting.
I must have misinterpreted the question, but a sketch of the curves made it crystal clear:
img406.imageshack.us/img406/1995/1291016654.png

We have two distinct areas enclosed by:
1. cos(x), cos(x-c),and x=0
2. cos(x-c), cos(x) and x=¥ð
The y=0 in the second area is extraneous and misleading for someone (like me) who hasn't taken the time to make a proper sketch.

Now for the limits for each of the areas,
1. from -¥ð+c/2 to x=0
2. from c/2 to ¥ð

We are able to find the limits because of the symmetry of the curves cos(x) and cos(x-c).

Finally, the integrals:
1. ¡ò(cos(x)-cos(x-c))dx
2. ¡ò(cos(x-c)-cos(x))dx

Hope this will get you along the way.

Hint: integrate and find each of the areas, equate the two areas and solve for c [if necessary].
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Answer 1

1. Integral ∫(cos(x)-cos(x-c))dx
(C=integration constant)
I1=∫[cos(x)-cos(x-c)]dx
=sin(x)-sin(x-c)+C

Evaluate I1=definite integral from x=-π+c/2 to x=0
[sin(0)-sin(0-c)]-[sin(-π+c/2)-sin(-π+c/2-c)]
=[0+sin(c)]-[-sin(c/2)-sin(c/2)]
=sin(c)+2sin(c/2)

2. Integral ∫(cos(x-c)-cos(x))dx
(C=integration constant)
I1=∫[cos(x-c)-cos(x)]dx
=sin(x-c)-sin(x)+C

Evaluate I2=definite integral from x=c/2 to x=%pi;
[sin(c/2)-sin(c/2-c)] - [sin(π)-sin(π-c)]
=[sin(c)-sin(-c/2)] - [0-sin(π-c)]
=sin(c)+2sin(c/2)

So to find c, we equate I1 and I2

I1=I2
sin(c)+2sin(c/2) = sin(c)+2sin(c/2)

Since it is an identity, we conclude that all values of c on the interval [0,π/2] will satisfy the condition that Area1 = Area2.

Note that the last sentence in the response was:
...equate the two areas and solve for c [<b>if necessary</b>]
and it turns out that it is not necessary because it is an identity.

Answer 2

nope, you are wrong. I got the correct answer. area1 and area2 are not equal to each other since their limits are DIFFERENT. for area1, the limits should be (from 0 to c/2) and for area2. the limits should be (from pi/2+c to pi). by solving it i got the value for c. Thank you anyways.