TL;DR

Nine practice problems covering the three main ways the FTC appears on exams: finding function values by hand using antiderivatives, using the net change theorem with a calculator, and reading signed area off a graph. Full worked solutions and a quick answer key at the end.

Practice Q&A: Antiderivative Method (Problems 1 and 2)

These problems give you f'(x) and an initial condition, and ask you to find f at a different point by computing the full antiderivative.

Q: Given y' = 2 + 1/x² and y(1) = 6, find y(3).

A: The antiderivative of 2 + 1/x² is 2x − 1/x + C.

Apply the initial condition: y(1) = 2(1) − 1/1 + C = 1 + C = 6, so C = 5.

Therefore y(x) = 2x − 1/x + 5.

y(3) = 2(3) − 1/3 + 5 = 6 − 1/3 + 5 = 11 − 1/3 = 32/3.

Q: Given f'(x) = 2cos(x) and f(0) = 3, find f(π/4).

A: The antiderivative of 2cos(x) is 2sin(x) + C.

Apply the initial condition: f(0) = 2sin(0) + C = 0 + C = 3, so C = 3.

Therefore f(x) = 2sin(x) + 3.

f(π/4) = 2sin(π/4) + 3 = 2(√2/2) + 3 = √2 + 3 ≈ 4.414.

Practice Q&A: Net Change Theorem with Calculator (Problems 3 to 6)

These problems give you a derivative you cannot easily antidifferentiate by hand. The strategy is always the same: use f(b) = f(a) + ∫ₐᵇ f'(t) dt and evaluate the integral numerically.

Q: Water flows into a tank at a rate of dW/dt = (1/75)(600 + 20t − t²) gallons per hour. There are 150 gallons at t = 0. How many gallons are in the tank at t = 24?

A: W(24) = W(0) + ∫₀²⁴ (1/75)(600 + 20t − t²) dt = 150 + ∫₀²⁴ (1/75)(600 + 20t − t²) dt.

Evaluate the integral by calculator: ≈ 207.36.

W(24) ≈ 357.36 gallons.

Q: Given f'(x) = cos(x³) and f(0) = 2, find f(1).

A: f(1) = f(0) + ∫₀¹ cos(t³) dt = 2 + ∫₀¹ cos(t³) dt.

By calculator, ∫₀¹ cos(t³) dt ≈ 0.932.

f(1) ≈ 2.932.

Q: Given f'(x) = e^(−x²) and f(5) = 1, find f(2).

A: f(2) = f(5) + ∫₅² e^(−t²) dt.

Flip the limits: ∫₅² e^(−t²) dt = −∫₂⁵ e^(−t²) dt.

By calculator, ∫₂⁵ e^(−t²) dt ≈ 0.004 (the function e^(−t²) is extremely small for t > 2).

So f(2) = 1 − 0.004 ≈ 0.996.

Key insight: this is a case where the known point (t = 5) is to the right of the target (t = 2), so you integrate "backwards" and flip the sign.

Q: A particle has velocity v(t) = 5sin(t²). At t = 6, its position is x(6) = 4. Find x(7).

A: x(7) = x(6) + ∫₆⁷ 5sin(t²) dt = 4 + ∫₆⁷ 5sin(t²) dt.

By calculator, ∫₆⁷ 5sin(t²) dt ≈ −0.163.

x(7) ≈ 3.837.

Practice Q&A: Graph-Based FTC (Problems 7 to 9)

These problems give you the graph of f' (or g') and ask you to find function values. The integral ∫ₐᵇ f'(x) dx equals the net signed area under the graph of f'.

Q: The graph of f' is given. ∫₁⁴ f'(x) dx = 6.2 and f(1) = 3. Find f(4).

A: By the FTC, f(4) = f(1) + ∫₁⁴ f'(x) dx = 3 + 6.2 = 9.2.

This is the most direct version of the formula. The integral value is handed to you.

Q: The graph of f' is a semicircle on [−4, 4]. f(4) = 7. Find f(−4).

A: The semicircle has radius 4 and sits above the x-axis, so its area is (1/2)π(4²) = 8π.

The integral ∫₋₄⁴ f'(x) dx = 8π (positive, since the curve is above the axis).

By the FTC: f(4) = f(−4) + ∫₋₄⁴ f'(x) dx, so 7 = f(−4) + 8π.

Therefore f(−4) = 7 − 8π ≈ −18.13.

Key insight: you are solving for the starting value, not the ending value, so you subtract the integral.

Q: The graph of g' is given. g(−1) = −8. Find g(5).

A: g(5) = g(−1) + ∫₋₁⁵ g'(x) dx = −8 + (net signed area under g' from −1 to 5).

From the graph, the net area computes to 30.

g(5) = −8 + 30 = 22.

For graph problems, break the region into geometric shapes you can compute: triangles, rectangles, trapezoids, semicircles. Add areas above the x-axis and subtract areas below.

Answer Key