Integration Accumulation Worksheet Answer Key – AP Calculus AB

Problem 1 Left and right Riemann sums

The width is 0.5. Left sum: 0.5[f(1) + f(1.5) + f(2) + f(2.5)] = 0.5(4 + 5.25 + 7 + 9.25) = 0.5(25.5) = 12.75. Right sum: 0.5(5.25 + 7 + 9.25 + 12) = 0.5(33.5) = 16.75. The exact area is 44/3 ≈ 14.67, between the two.

Problem 2 Trapezoidal sum from a table

Each trapezoid has width 2. Sum: (2/2)[5 + 2(8) + 2(9) + 2(7) + 4] = 5 + 16 + 18 + 14 + 4 = 57 gallons.

Problem 3 Antiderivative of a polynomial

Integrate term by term: 2x³ − 2x² + 7x + C. The answer is 2x³ − 2x² + 7x + C.

Problem 4 Antiderivative with ln and exponential

The antiderivatives are 3 ln|x|, e^x and cos x (because the derivative of cos x is −sin x). The answer is 3 ln|x| + e^x + cos x + C.

Problem 5 Meaning of a definite integral

The integral of a rate gives the accumulated change. Between 2 and 6 hours after opening, the truck sells a total of 148 burritos.

Problem 6 Evaluating with the Fundamental Theorem

An antiderivative of 3x^1/2 is 2x^3/2. So the integral is 2(4)^3/2 − 2(1)^3/2 = 2(8) − 2 = 14.

Problem 7 Trigonometric definite integral

An antiderivative is 2 sin x + x. Evaluate: (2 sin π + π) − (2 sin 0 + 0) = π − 0 = π.

Problem 8 Properties of definite integrals

(a) By additivity, ∫₂⁵ = 12 − (−3) = 15.

(b) Reversing the limits changes the sign: −12.

(c) 4(12) − 2(5) = 48 − 10 = 38.

Problem 9 Substitution with a power

Let u = x² + 5, so du = 2x dx. Then ∫ u⁴ du = u⁵/5 + C = (x² + 5)⁵/5 + C.

Problem 10 Substitution with an exponential

Let u = x³, so du = 3x² dx and x² dx = du/3. The integral is (1/3)∫ e^u du = (1/3)e^x3 + C.

Problem 11 Definite integral with substitution

Let u = x² + 9, so x dx = du/2. The limits become u = 9 when x = 0 and u = 16 when x = √7. Then (1/2)∫₉¹⁶ u^1/2 du = (1/2)(2/3)(16^3/2 − 9^3/2) = (1/3)(64 − 27) = 37/3.

Problem 12 Derivative of an accumulation function

G(1) is an integral over an interval of width zero, so it equals 0. By FTC Part 1, G′(x) = √(x³ + 8), so G′(2) = √16 = 4.

Problem 13 Chain rule with an integral

Apply FTC Part 1 with the chain rule: H′(x) = (3x² + 1)·2x. At x = 2: (13)(4) = 52.

Problem 14 Accumulation from a graph

From 0 to 2 the area under the segment from height 2 to 4 is (2 + 4)/2·2 = 6, so g(2) = 6. From 2 to 4 the area is (4 + 0)/2·2 = 4, so g(4) = 10. From 4 to 6 the region lies below the axis: triangle area (1/2)(2)(3) = 3, counted as −3, so g(6) = 7. Because g′ = f changes from positive to negative at x = 4, the maximum is at x = 4, where g = 10.

Problem 15 Integrating after long division

Divide: x² + 5x + 7 = (x + 1)(x + 4) + 3, so the integrand is x + 4 + 3/(x + 1). Integrate: x²/2 + 4x + 3 ln|x + 1| + C.

Problem 16 Completing the square

Complete the square: x² + 6x + 13 = (x + 3)² + 4. The integral of 1/(u² + 4) is (1/2)arctan(u/2), so the answer is (1/2) arctan((x + 3)/2) + C.

Problem 17 Logarithm substitution

Let u = ln x, du = dx/x. When x = 1, u = 0, and when x = e², u = 2. So the integral is ∫₀² u du = 2²/2 = 2.

Problem 18 Distance from a speed table

With three subintervals of width 4, the midpoints are 2, 6 and 10. The estimate is 4[v(2) + v(6) + v(10)] = 4(18 + 40 + 45) = 4(103) = 412 miles.

Problem 19 Displacement and distance

An antiderivative is s(t) = t³ − 6t² + 9t. Displacement = s(4) − s(0) = 4 − 0 = 4. Since v = 3(t − 1)(t − 3), the particle turns at t = 1 and 3. Positions: s(1) = 4, s(3) = 0, s(4) = 4. Distance = 4 + 4 + 4 = 12.

Problem 20 Analyzing an accumulation function

By FTC Part 1, g′ = f, which is zero at x = 1 and x = 3. The second derivative is g″ = f′(x) = 2x − 4. At x = 1: −2 < 0, a local maximum. At x = 3: 2 > 0, a local minimum. Since g(x) = x³/3 − 2x² + 3x, g(1) = 4/3 and g(3) = 0.