Suppose that 0 < c < ?/2. For what value of c is the area of the region enclosed

Question

Suppose that 0 < c < ?/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 = ?, and y = 0?

Explanation / Answer

Area of the region enclosed by the curves y = cos x, y = cos(x - c), and x = 0: Since these curves intersect when x = x - c ==> x = c/2, we have A = ?(0 to c/2) [cos x - cos(x - c)] dx = [sin x - sin(x - c)] {for x = 0 to c/2} = 2 sin(c/2) - sin c ----------- Area of the region enclosed by the curves y = cos(x - c), x = p, and y = 0: Since cos(x - c) = 0 when x = p/2 + c, we have A = ?(p/2 + c to p) [0 - cos(x - c)] dx = -sin(x - c) {for x = p/2 + c to p} = 1 - sin(p - c) = 1 - sin c. ----------------- Finally, 2 sin(c/2) - sin c = 1 - sin c ==> sin(c/2) = 1/2 ==> c/2 = p/6 ==> c = p/3.

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