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December 20, 2014

December 20, 2014

Posted by **K** on Monday, November 29, 2010 at 2:44am.

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 9:01amThank 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(x-c),and x=0

2. cos(x-c), 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(x-c).

Finally, the integrals:

1. ∫(cos(x)-cos(x-c))dx

2. ∫(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].

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

- Calculus - MathMate Please help -
**K**, Monday, November 29, 2010 at 6:35pmI 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 dont know what to do from here. can u pleses help???

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