Free AP Calculus AB Lesson · 29 Practice Problems

AP Calculus AB: Definite Integrals

Definite integrals turn accumulation into a computable number, and the Fundamental Theorem of Calculus is the bridge that connects them back to derivatives. This free, complete lesson covers the Fundamental Theorem of Calculus, properties of definite integrals and the Mean Value Theorem for Integrals, evaluating definite integrals via substitution, Riemann sums, left/right/midpoint/trapezoidal approximation, comparing approximating sums, graphing f from f' using accumulation, interpreting integrals from tables and graphs, and average value. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

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1. The Fundamental Theorem of Calculus and Properties

If f is continuous on [a, b] and F' = f, the Fundamental Theorem of Calculus (FTC) says ∫_a^b f(x) dx = F(b) − F(a). A second form (FTC Part 1) says ddx∫_a^x f(t) dt = f(x). The Mean Value Theorem for Integrals guarantees at least one c in (a, b) where f(c)(b − a) = ∫_a^b f(x) dx.

Property	Statement
Constant multiple	∫_a^b kf(x) dx = k∫_a^b f(x) dx
Zero-width interval	∫_a^a f(x) dx = 0
Reversing limits	∫_a^b f(x) dx = −∫_b^a f(x) dx
Splitting the interval	∫_a^c f(x) dx + ∫_c^b f(x) dx = ∫_a^b f(x) dx

Worked Example: Evaluate ∫₋₁² (3x² − 2x) dx.

x³ − x² |₋₁² = (8 − 4) − (−1 − 1) = 4 − (−2) = 6

1. Evaluate ∫₀³ (4x − 1) dx.

Explanation: 2x² − x |₀³ = (18 − 3) − 0 = 15.

2. If ∫₁⁵ f(x) dx = 8 and ∫₃⁵ f(x) dx = 3, find ∫₁³ f(x) dx.

Explanation: Since ∫₁³ + ∫₃⁵ = ∫₁⁵: ∫₁³ = 8 − 3 = 5.

3. By the Mean Value Theorem for Integrals, if f is continuous on [2, 6] and ∫₂⁶ f(x) dx = 20, there must exist a c in (2, 6) such that f(c) equals what?

Explanation: f(c) = 206 − 2 = 5.

4. Find ddx[∫₁^x √(t² + 1) dt].

Explanation: By FTC Part 1, differentiating an integral with respect to its own upper limit just substitutes x for t: √(x² + 1).

2. Evaluating Definite Integrals via Substitution

When using u-substitution on a definite integral, either convert the limits of integration to u-values (and never switch back to x), or find the antiderivative in terms of x first and then substitute the original limits.

Worked Example: Evaluate ∫₀¹ 2(1 − 3x)² dx.

u = 1 − 3x, du = −3 dx. When x = 0, u = 1; when x = 1, u = −2.

2∫₁⁻² u²du−3 = −29[u³]₁⁻² = −29(−8 − 1) = 2

1. Evaluate ∫₀¹ 3(2x + 1)² dx using u = 2x + 1.

Explanation: u = 2x + 1, du = 2 dx. Limits: x = 0 → u = 1; x = 1 → u = 3. 3∫₁³u²(du/2) = 12[27 − 1] = 13.

2. Evaluate ∫₀² xe^x² dx using u = x².

Explanation: u = x², du = 2x dx. Limits: x = 0 → u = 0; x = 2 → u = 4. 12∫₀⁴e^u du = 12(e⁴ − 1).

3. Evaluate ∫₀^π/6 cos(3x) dx.

Explanation: 13sin(3x) |₀^π/6 = 13[sin(π/2) − sin(0)] = 13(1) = 13.

4. Evaluate ∫₀⁴ dxx + 1.

Explanation: ln|x + 1| |₀⁴ = ln 5 − ln 1 = ln 5.

3. Riemann Sums

A Riemann Sum approximates a definite integral by summing f(x_k) · Δx across subintervals; in the limit as the number of subintervals grows without bound, the sum becomes exact:

lim_n→∞ ∑_k=1ⁿ f(x_k) · Δx = ∫_a^b f(x) dx

Worked Example: Write lim_n→∞ ∑_k=1ⁿ (2 + 5kn)² · 5n as a definite integral.

Here x_k = 2 + 5k/n, so a = 2 and Δx = 5/n gives b = 7. Matching f(x_k) = x_k²:

∫₂⁷ x² dx

1. Write lim_n→∞ ∑_k=1ⁿ (1 + 3kn)³ · 3n as a definite integral.

Explanation: x_k = 1 + 3k/n gives a = 1 and Δx = 3/n, so b = 1 + 3 = 4. f(x) = x³: ∫₁⁴ x³ dx.

2. Write ∫₀⁵ (x² + 1) dx as the limit of a Riemann Sum, using equal-width subintervals with x_k = 5k/n.

Explanation: With Δx = 5/n and x_k = 5k/n, f(x_k) = (5k/n)² + 1, giving the sum in option A.

3. In a Riemann Sum, what does Δx represent?

Explanation: Δx is the width of each subinterval; the height comes separately from f(x_k).

4. Approximating Definite Integrals

Four common numerical approximations, using Δx for equal-width subintervals: the left sum L(n) uses left-endpoint heights, the right sum R(n) uses right-endpoint heights, the midpoint sum M(n) uses midpoint heights, and the trapezoidal sum T(n) averages consecutive heights.

Worked Example: Approximate ∫₀² x³ dx using L(4) (4 subintervals).

Δx = 0.5. Left heights at x = 0, 0.5, 1, 1.5:

L(4) = 0.5[0³ + 0.5³ + 1³ + 1.5³] = 0.5(4.5) = 2.25 = 94

1. Using the same function and subintervals, find R(4).

Explanation: R(4) = 0.5[0.5³ + 1³ + 1.5³ + 2³] = 0.5(12.5) = 254.

2. Using the same function, find M(4) with midpoints 0.25, 0.75, 1.25, 1.75.

Explanation: M(4) = 0.5[0.25³ + 0.75³ + 1.25³ + 1.75³] = 0.5(7.75) = 318.

3. Given f(0) = 5, f(2) = 9, f(4) = 11, f(6) = 8 (equal-width subintervals of 2), estimate ∫₀⁶ f(x) dx using T(3).

Explanation: T(3) = 5+92(2) + 9+112(2) + 11+82(2) = 14 + 20 + 19 = 53.

4. A function passes through (1, 4), (3, 10), (6, 7), with unequal subinterval widths. Estimate ∫₁⁶ f(x) dx using a left rectangular approximation.

Explanation: Widths are 2 (from 1 to 3) and 3 (from 3 to 6). Left heights: f(1) = 4 and f(3) = 10. Estimate: 2(4) + 3(10) = 8 + 30 = 38.

5. Using the same three points, estimate ∫₁⁶ f(x) dx using a trapezoidal approximation.

Explanation: 4+102(2) + 10+72(3) = 14 + 25.5 = 39.5.

5. Comparing Approximating Sums

If f is increasing on [a, b], L(n) ≤ ∫_a^b f(x) dx ≤ R(n); if decreasing, the inequality reverses. The trapezoid understates the true area when the graph is concave down, and overstates it when concave up. If concave up, M(n) ≤ ∫ ≤ T(n); if concave down, T(n) ≤ ∫ ≤ M(n).

Worked Example: If f is increasing on [a, b], how do L(n) and R(n) compare to the true integral?

L(n) ≤ ∫_a^b f(x) dx ≤ R(n).

1. If f is decreasing on [a, b], how do L(n) and R(n) compare to the true integral?

Explanation: Decreasing reverses the inequality from the increasing case: R(n) ≤ ∫ ≤ L(n).

2. If the graph of f is concave down on [a, b], how does T(n) compare to the true value of the integral?

Explanation: For a concave-down curve, the connecting chord (the trapezoid's top) lies below the curve, so the trapezoid underestimates the true area.

3. If the graph of f is concave up on [a, b], how do M(n) and T(n) relate to each other and to the true value?

Explanation: For concave-up graphs, the midpoint rectangle underestimates and the trapezoid overestimates, giving M(n) ≤ ∫ ≤ T(n).

6. Graphing f from f' Using Accumulation

If F(x) = ∫_a^x f(t) dt, then F'(x) = f(x) by FTC Part 1. This means all the usual derivative-based reasoning (increasing/decreasing, concavity, critical points) can be applied to F using f as its "derivative graph."

Worked Example: If F(x) = ∫₀^x f(t) dt, and f(t) > 0 for 0 < t < 3 but f(t) < 0 for t > 3, describe F's behavior.

F is increasing on (0, 3) and decreasing for t > 3, so F has a local maximum at t = 3.

1. If F(x) = ∫₂^x f(t) dt, and f changes from negative to positive at x = 5 (with f(5) = 0), what happens at x = 5?

Explanation: Since F' = f changes from negative to positive at x = 5, F has a local minimum there.

2. If f(t) > 0 and increasing on an interval, what does this tell us about the concavity of F(x) = ∫_a^x f(t) dt on that interval?

Explanation: F' = f is increasing, so F'' = f' > 0, meaning F is concave up.

3. Given F(0) = 2 and ∫₀³ f(t) dt = 5, where f = F', find F(3).

Explanation: By FTC: F(3) = F(0) + ∫₀³ f(t) dt = 2 + 5 = 7.

7. Interpreting Integrals from Tables and Graphs

When velocity is given only at discrete times, approximate total distance using a Riemann Sum built directly from the table.

Worked Example: Speed readings every 10 minutes: t = 0, 10, 20, 30, 40 (minutes); v = 20, 35, 50, 30, 45 (mph). Use L(4) to estimate distance traveled.

Each subinterval is 10 minutes = 1/6 hour.

L(4) = [20 + 35 + 50 + 30] × 16 = 1356 = 22.5 miles

1. Using the same table, find R(4).

Explanation: R(4) = [35 + 50 + 30 + 45] × 16 = 1606 ≈ 26.67 miles.

2. Using the same table, find T(4).

Explanation: T(4) = [27.5 + 42.5 + 40 + 37.5] × 16 = 147.56 ≈ 24.58 miles.

3. If a velocity function v(t) is strictly increasing throughout [0, 10], which gives a larger estimate of total distance: L(n) or R(n)?

Explanation: For an increasing function, right-endpoint heights are always at least as large as left-endpoint heights, so R(n) gives the larger estimate.

8. Average Value of a Function

The average value of f on [a, b] is:

1b − a ∫_a^b f(x) dx

The average value of a function is not the same as its average rate of change. Average value uses this integral formula; average rate of change uses [f(b) − f(a)]/(b − a). Read every question carefully to determine which is actually being asked.

Worked Example: Find the average value of f(x) = x² on [0, 3].

13∫₀³ x² dx = 13[x³3]₀³ = 13(9) = 3

1. Find the average value of f(x) = 4x on [1, 5].

Explanation: 14∫₁⁵4x dx = 14[2x²]₁⁵ = 14(50 − 2) = 12.

2. Find the average value of f(x) = 3x² on [0, 2].

Explanation: 12∫₀²3x² dx = 12[x³]₀² = 12(8) = 4.

3. Find the average value of f(x) = sin x on [0, π].

Explanation: 1π∫₀^πsin x dx = 1π[−cos x]₀^π = 1π[1 + 1] = 2π.

4. Is the average value of a function the same as its average rate of change?

Explanation: Average value integrates f itself over the interval; average rate of change compares only the endpoint values of f. They measure genuinely different things and generally give different numbers.

Common Mistakes to Avoid

Switching back to x-limits after converting to u-limits, or vice versa. Once you substitute the limits of integration in terms of u, evaluate entirely in u, never mix the two.
Forgetting that a definite integral of a negative function yields a negative number. "Area" is always positive, but the value of the definite integral itself can be negative when the curve dips below the x-axis, these are related but distinct ideas.
Mixing up which approximation over- or underestimates. The relationship depends on both monotonicity (for L/R) and concavity (for M/T); don't apply a rule from one case to the other.
Treating F(x) = ∫_a^x f(t) dt as if you need to actually compute the integral before analyzing F. FTC Part 1 lets you reason about F directly from the graph of f, exactly as you would reason about any function from its derivative's graph.
Confusing average value with average rate of change. They use entirely different formulas and generally produce different numbers, always double-check which one a question is actually asking for.

Frequently Asked Questions

Why does the Fundamental Theorem of Calculus matter so much?

It's the bridge between derivatives and integrals: it says that accumulating a rate of change over an interval (integration) and finding net change in the original quantity are the same operation, undoing differentiation. Without it, evaluating a definite integral would require going back to the limit definition every time.

Which approximation method should I trust the most?

None is universally best, it depends on the function's shape. The midpoint and trapezoidal sums tend to outperform the left and right sums for smooth curves, since they use more information about the curve's behavior between the endpoints, but a specific comparison always depends on concavity.

How is graphing f from f' using accumulation different from graphing a derivative-based curve sketch?

It's actually the same skill in reverse: instead of being given f and analyzing f' and f'' to sketch f, you're given f (playing the role of a derivative) and use FTC to build up F's actual y-values at key points, alongside the usual increasing/decreasing and concavity reasoning.

Where can I practice more problems like these?

The Definite Integrals quizzes in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.