AP Calculus AB › Unit 8
The last AB unit puts integrals to work. You will find average value, model motion and net change, compute the area between curves, and find volumes with cross sections, disks, and washers. Exam questions often describe a rate in context and ask how much accumulates, or give a region and ask for a volume. Students commonly confuse average value with average rate of change. With washers, many integrate with respect to the wrong variable or square the difference of the radii instead of subtracting the squared radii.
These are taken from the Regular set, with the answer and explanation shown.
1. A particle moves along the x-axis with acceleration a(t) = 6t. At t = 0 its velocity is 3 and its position is 2. What is its position at t = 2?
Answer: A. v(t) = 3 + ∫₀ᵗ 6s ds = 3t² + 3. Then x(2) = 2 + ∫₀² (3t² + 3) dt = 2 + 8 + 6 = 16.
Common trap: Choosing 14 gives the displacement only. You still add the starting position x(0) = 2.
2. The region R is bounded by y = x², the line y = 2 − x, and the x-axis, with x ≥ 0. Which single integral gives the area of R?
Answer: A. Slice horizontally. For 0 ≤ y ≤ 1, the left edge is x = √y (from y = x²) and the right edge is x = 2 − y (from the line). Area = ∫₀¹ (2 − y − √y) dy.
Common trap: Choosing ∫₀¹ (2 − x − x²) dx is the area between the line and the parabola, above the parabola. R lies below the parabola, between it and the x-axis.
3. The region between y = √x and y = x² is revolved about the x-axis. What is the volume?
Answer: A. The curves meet at x = 0 and 1, with √x on top. Washers: R = √x, r = x². V = π∫₀¹ (x − x⁴) dx = π(1/2 − 1/5) = 3π/10.
Common trap: Choosing 9π/70 computes π∫₀¹ (√x − x²)² dx. Squaring the gap gives a smaller, wrong solid; washers use R² − r².