Suppose a student sets up the integral _ 0 ^ /3 [(2 )^2-(4 )^2 ] ,d to compute the…

MATHEMATICS · Single correct · AP · Calculus BC · Parametric, Polar, and Vector-Valued Functions · Parametric, Polar, and Vector-Valued Functions

Problem

Suppose a student sets up the integral ∫0π/3[(2cos⁡θ)2−(4cos⁡θ)2] dθ\int_{0}^{\pi/3}\left[(2\cos\theta)^2-(4\cos\theta)^2\right]\,d\theta to compute the area of the region bounded by the curves r=4cos⁡θr=4\cos\theta and r=2cos⁡θr=2\cos\theta for 0≤θ≤π30\le\theta\le\frac{\pi}{3}. Without evaluating the integral, which statement is most accurate about the value of the integral? (a) The integral is positive because an area cannot be negative. (b) The integral equals zero since the two curves meet at both θ=0\theta=0 and θ=π/3\theta=\pi/3. (c) The integral is negative because (4cos⁡θ)2(4\cos\theta)^2 exceeds (2cos⁡θ)2(2\cos\theta)^2 for every θ\theta in the interval. (d) The sign of the integral cannot be determined because the integrand might change sign on the interval.

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