Posted by K on Monday, November 29, 2010 at 2:44am.
I asked this question on last thursday and u told me that the bottom limit for the 1st and the right limit for the 2nd area is missing? The question that i posted had no other information and i don't know why you said that the limits are missing. There is no other limits that are givin in this question. Does this question even have a limit? I don't know. I'm very confused now. I need help on this question. Please Help. Thank you.
Q. Suppose that 0 < c < pi/2. For what value of c is the area of the region enclosed by the curves y = cos x, y = cos(x  c), and x = 0 equal to the area of the region enclosed by the curves y = cos(x  c), x = pi, and y = 0?

Calculus  MathMate Please help  MathMate, Monday, November 29, 2010 at 9:01am
Thank you for reposting.
I must have misinterpreted the question, but a sketch of the curves made it crystal clear:
http://img406.imageshack.us/img406/1995/1291016654.png
We have two distinct areas enclosed by:
1. cos(x), cos(xc),and x=0
2. cos(xc), cos(x) and x=π
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 π+c/2 to x=0
2. from c/2 to π
We are able to find the limits because of the symmetry of the curves cos(x) and cos(xc).
Finally, the integrals:
1. ∫(cos(x)cos(xc))dx
2. ∫(cos(xc)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].

Calculus  MathMate Please help  K, Monday, November 29, 2010 at 3:10pm
Thank You. I think i got the idea. i will try to solve it.

Calculus  MathMate Please help  K, Monday, November 29, 2010 at 6:35pm
I got two areas 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???
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