AP Calculus AB: Fundamental Theorem of Calculus Practice Questions
Test yourself on Fundamental Theorem of Calculus with 12 original AP Exams practice questions. Pick an answer to see instant feedback and a full explanation.
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1. If F(x) = ∫₀ˣ (t² + 1) dt, what is F'(x)?
Explanation. By Part 1 of the FTC, if F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). Here f(t) = t² + 1, so F'(x) = x² + 1. No antiderivative computation is needed.
2. Evaluate ∫₁⁴ (2x) dx.
Explanation. By Part 2 of the FTC, ∫₁⁴ 2x dx = [x²]₁⁴ = 16 − 1 = 15.
3. If G(x) = ∫₂^{x²} sin(t) dt, find G'(x).
Explanation. Use the chain rule with FTC. If the upper limit is u(x) = x², then G'(x) = sin(u)·u' = sin(x²)·2x.
4. If H(x) = ∫ₓ⁵ eᵗ dt, what is H'(x)?
Explanation. When x is the lower limit, reverse the bounds: H(x) = −∫₅ˣ eᵗ dt, so H'(x) = −eˣ.
5. Evaluate ∫₀^{π/2} cos(x) dx.
Explanation. An antiderivative of cos x is sin x. So the integral = sin(π/2) − sin(0) = 1 − 0 = 1.
6. Let f be continuous and F(x) = ∫₃ˣ f(t) dt. If f(3) = 5 and f(7) = 2, what is F'(7)?
Explanation. By FTC Part 1, F'(x) = f(x), so F'(7) = f(7) = 2. The value f(3) is irrelevant here.
7. If the velocity of a particle is v(t) = 6t and its position at t = 0 is s(0) = 4, what is s(2)?
Explanation. s(2) = s(0) + ∫₀² 6t dt = 4 + [3t²]₀² = 4 + 12 = 16. This uses the net change (accumulation) interpretation of the FTC.
8. Evaluate ∫₋₁¹ (3x² ) dx.
Explanation. An antiderivative of 3x² is x³. So ∫₋₁¹ 3x² dx = [x³]₋₁¹ = 1 − (−1) = 2.
9. If P(x) = ∫_{x}^{x²} ln(t) dt for x > 1, find P'(x).
Explanation. For ∫_{a(x)}^{b(x)} f(t) dt the derivative is f(b)·b' − f(a)·a'. Here b = x² (b' = 2x), a = x (a' = 1): P'(x) = ln(x²)·2x − ln(x)·1.
10. A function g satisfies ∫₂⁶ g'(x) dx = 11 and g(2) = 4. What is g(6)?
Explanation. By FTC Part 2, ∫₂⁶ g'(x) dx = g(6) − g(2) = 11, so g(6) = 11 + 4 = 15.
11. Let F(x) = ∫₀ˣ f(t) dt where f is positive on (0, 5) and negative on (5, 8). At what x does F have a local maximum on (0, 8)?
Explanation. F'(x) = f(x). F increases where f > 0 (before x = 5) and decreases where f < 0 (after x = 5). The sign change from + to − at x = 5 gives a local maximum.
12. Evaluate d/dx [∫₀ˣ √(1 + t⁴) dt].
Explanation. By FTC Part 1, the derivative of the accumulation function equals the integrand evaluated at x: √(1 + x⁴). No simplification of the radical is possible or needed.
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FAQ
What is the difference between Part 1 and Part 2 of the FTC?
Part 1 says the derivative of an accumulation function F(x) = ∫ₐˣ f(t) dt equals f(x); it connects integration and differentiation. Part 2 says ∫ₐᵇ f(x) dx = F(b) − F(a) where F is any antiderivative; it lets you evaluate definite integrals.
How do I handle variable limits like ∫₀^{x²} f(t) dt?
Use the chain rule: if the upper limit is u(x), the derivative is f(u(x))·u'(x). If x is the lower limit, you get a minus sign. With both limits as functions, use f(b)·b' − f(a)·a'.
What real-world idea does FTC Part 2 capture on the AP exam?
It expresses net change: the integral of a rate of change gives the total change in the quantity. For example, ∫ₐᵇ v(t) dt = displacement, and final value = initial value + ∫ of the rate.