_{(-pi/2)}

^{(pi/2)}cos(x)dx = 2

_{(-pi/2)}

^{k }cos(x)dx + ∫

_{k}

^{(pi/2)}= 2

_{(-pi/2)}

^{k}cos(x)dx = (1/2) = 3*∫

_{k}

^{(}

^{pi/2)}cos(x)dx = 3*(1/2)

The region bounded by the x-axis and the part of the graph of y=cos x between x=-pi/2 and x=pi/2 is separated into two regions by the line x=k. If the area of the region for -pi/2≤x≤k is three times the area of the region for k≤x≤pi/2, what's k?

Answer: pi/6

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∫_{(-pi/2)}^{(pi/2)} cos(x)dx = 2

∫_{(-pi/2)}^{k }cos(x)dx + ∫_{k}^{(pi/2)} = 2

∫_{(-pi/2)}^{k} cos(x)dx = (1/2) = 3*∫_{k}^{(}^{pi/2)}cos(x)dx = 3*(1/2)

sin(k) - sin(-pi/2) = 3sin(pi/2) - 3sin(k)

sin(k) = -3[sin(k)]

sin(k) = -3[sin(k)]

sin(k) + 3[sin(k)] = 2

4[sin(k)] = 2

sin(k) = (1/2)

k = (pi/6) radians or 30º

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