AP Calculus AB Question of the Day

1. $\displaystyle \int_{0}^{2\pi}(\cos x-(-\cos x))\,dx$
2. $\displaystyle \int_{0}^{2\pi}((-\cos x)-\cos x)\,dx$
3. $\displaystyle \int_{0}^{2\pi}(\cos x+(-\cos x))\,dx$
4. $\displaystyle \int_{0}^{\pi/2}(\cos x-(-\cos x))dx+\int_{\pi/2}^{3\pi/2}((-\cos x)-\cos x)dx+\int_{3\pi/2}^{2\pi}(\cos x-(-\cos x))dx$ (correct answer)
5. $\displaystyle \int_{0}^{2\pi}(\cos^2 x)\,dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=π/2 and x=3π/2 because those are where the order changes for y=cos x and y=-cos x. From 0 to π/2, y=cos x is above y=-cos x. From π/2 to 3π/2, y=-cos x is above y=cos x in parts. From 3π/2 to 2π, y=cos x is above y=-cos x. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.