Lesson 29: The Fundamental Theorem of Calculus

(a) Opening Question

You spent the last lesson computing definite integrals as limits of Riemann sums — slow, painful work. Try this one the slow way, then ask whether there is a faster path.

Compute ∫₁⁴ (3x² − 2x) dx.

If you set up a Riemann sum, you'd be summing rectangle areas forever. But notice: 3x² − 2x is the derivative of x³ − x². What if the area under a rate of change is just the total change in its antiderivative?

Try it: let F(x) = x³ − x². Compute F(4) − F(1) = (64 − 16) − (1 − 1) = 48 − 0 = 48.

That number, 48, is exactly the value of the definite integral — no rectangles, no limits. This is not a coincidence. It is the Fundamental Theorem of Calculus, the bridge between the two halves of the course: the derivative (Units 2–5) and the integral (Unit 6). This lesson makes that bridge precise, in both directions.

(b) Core Concepts

The Fundamental Theorem of Calculus (FTC) comes in two parts. They are different statements, and confusing them is one of the most common AP errors. Keep them straight: Part 2 evaluates a definite integral; Part 1 differentiates an accumulation function.

The Fundamental Theorem of Calculus, Part 2 (Evaluation)

FTC, Part 2. If f is continuous on [a, b] and F is any antiderivative of f (that is, F′(x) = f(x)), then

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∫ₐᵇ f(x) dx = F(b) − F(a)

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This is the computational workhorse. To evaluate a definite integral you do not need limits of Riemann sums — you find an antiderivative F (any one; the +C cancels), then subtract its values at the endpoints. The standard notation for that subtraction is the evaluation bar:

∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a)

Why the +C doesn't matter. If you used F(x) + C instead of F(x), the subtraction gives [F(b) + C] − [F(a) + C] = F(b) − F(a). The constant cancels, so we always drop it on a definite integral.

Worked evaluation. Compute ∫₁⁴ (3x² − 2x) dx. An antiderivative of 3x² − 2x is F(x) = x³ − x². Then:

∫₁⁴ (3x² − 2x) dx = [x³ − x²]₁⁴ = (4³ − 4²) − (1³ − 1²) = (64 − 16) − (1 − 1) = 48

Model the AP justification. When a problem requires it, write the reasoning in full: "Because f(x) = 3x² − 2x is continuous on [1, 4] and F(x) = x³ − x² satisfies F′(x) = f(x), by the Fundamental Theorem of Calculus ∫₁⁴ f(x) dx = F(4) − F(1) = 48." Naming the continuity hypothesis is what separates a full-credit justification from a bare answer.

The Fundamental Theorem of Calculus, Part 1 (Accumulation)

Part 1 looks at the integral in the other direction. Fix the lower limit and let the upper limit vary, building an accumulation function:

g(x) = ∫ₐˣ f(t) dt

Here t is a dummy variable of integration, and x is the genuine input: g(x) is the signed area accumulated from a out to x. The theorem says this accumulation function is differentiable, and its derivative is the original integrand:

FTC, Part 1. If f is continuous on an interval containing a, and g(x) = ∫ₐˣ f(t) dt, then g is differentiable and

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g′(x) = d/dx ∫ₐˣ f(t) dt = f(x)

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In words: differentiating undoes integrating. The derivative of "area-so-far" is the height of the curve at the right edge. This is why integration and differentiation are inverse processes.

The variable upper limit and the chain rule. Very often the upper limit is not just x but a function u(x). Then g(x) = ∫ₐ^{u(x)} f(t) dt is a composition, and you must apply the chain rule. The outer operation is FTC Part 1 (evaluate the integrand at the upper limit); the inner operation is u(x):

d/dx ∫ₐ^{u(x)} f(t) dt = f(u(x)) · u′(x)

The extra factor u′(x) is the single most-forgotten piece on this topic. For example:

d/dx ∫₂^{x²} cos t dt = cos(x²) · (2x) = 2x cos(x²)

Substitute the upper limit x² into the integrand (cos), then multiply by the derivative of the upper limit (2x).

Variable lower limit. If the lower limit varies, flip the integral first using ∫ₐᵇ = −∫ᵇₐ:

d/dx ∫ₓ^{a} f(t) dt = d/dx [−∫ₐˣ f(t) dt] = −f(x)

Accumulation functions from the graph of f

A top AP skill is analyzing g(x) = ∫ₐˣ f(t) dt when you are given only the graph of f (no formula). Because g′ = f (Part 1), every fact you learned about reading f from f′ in Unit 5 applies here — just shifted down one level. Read the graph of f as if it were the derivative of g:

The two layers to keep separate: the sign of f controls whether g rises or falls; the slope (increasing/decreasing) of f controls the concavity of g. To compute an actual value g(x), add up the signed areas geometrically.

Graph mini-analysis.

Let g(x) = ∫₀ˣ f(t) dt. On [0, 4], f ≥ 0, so g increases; the total accumulated area is the half-disk, g(4) = (1/2)π(2)² = 2π. On [4, 6], f < 0, so g decreases. Since f changes from positive to negative at x = 4, g has its maximum at x = 4, with g(4) = 2π ≈ 6.28. Then g(6) = 2π − 2 (subtracting the triangle's area (1/2)(2)(2) = 2 that lies below the axis).

Net change

FTC Part 2 has a powerful reading when the integrand is itself a rate of change. Let f = F′. Then:

∫ₐᵇ f′(x) dx = f(b) − f(a)

The integral of a rate of change over [a, b] equals the net change in the quantity over [a, b]. If r(t) is the rate (gallons per minute) at which water enters a tank, then ∫₀⁶ r(t) dt is the total gallons added from t = 0 to t = 6. If v(t) is velocity, ∫ₐᵇ v(t) dt is the net displacement. This interpretation is the backbone of nearly every AP "rate in / rate out" FRQ.

(c) Worked Examples

Example 1 — Evaluate a definite integral with FTC Part 2 (NO CALC)

Problem. Evaluate ∫₀^{π/4} sec²x dx, and write the FTC justification.

Strategy. Find an antiderivative of the integrand, then apply FTC Part 2 with the evaluation bar.

Solution. Since d/dx[tan x] = sec²x, an antiderivative is F(x) = tan x. The integrand sec²x is continuous on [0, π/4] (no asymptote until π/2), so FTC Part 2 applies:

∫₀^{π/4} sec²x dx = [tan x]₀^{π/4} = tan(π/4) − tan(0) = 1 − 0 = 1

Justification. Because f(x) = sec²x is continuous on [0, π/4] and F(x) = tan x satisfies F′(x) = sec²x, by the Fundamental Theorem of Calculus ∫₀^{π/4} sec²x dx = F(π/4) − F(0) = 1.

Example 2 — Differentiate accumulation functions, including a variable limit (NO CALC)

Problem. Find the derivative of each.

(i) g(x) = ∫₂ˣ (t³ + 1) dt (ii) h(x) = ∫₁^{x³} √(1 + t²) dt (iii) k(x) = ∫ₓ⁵ eᵗ² dt

Strategy. Use FTC Part 1. For a variable upper limit u(x), multiply by u′(x) (chain rule). For a variable lower limit, flip the sign.

Solution.

(i) Upper limit is plain x, so just substitute x into the integrand:

g′(x) = x³ + 1

(ii) Upper limit is u(x) = x³, so u′(x) = 3x². Substitute the upper limit into the integrand, then multiply by u′:

h′(x) = √(1 + (x³)²) · 3x² = 3x²·√(1 + x⁶)

(iii) The lower limit varies. Flip first: k(x) = ∫ₓ⁵ eᵗ² dt = −∫₅ˣ eᵗ² dt. Then

k′(x) = −e^{x²}

Justification. Part (ii) is the AP-critical case: forgetting the · 3x² factor is the classic error. The integrand √(1 + t²) is continuous everywhere, so FTC Part 1 (with the chain rule) applies.

Example 3 — Analyze g(x) = ∫₀ˣ f(t) dt from the graph of f (NO CALC)

Problem. The graph of f is the semicircle-and-line shown in the Core Concepts graph above: an upper semicircle of radius 2 on [0, 4] (so f ≥ 0, peak (2,2)) and a segment from (4, 0) to (6, −2) (so f < 0) on [4, 6]. Let g(x) = ∫₀ˣ f(t) dt. (a) Find g(4) and g(6). (b) On what interval is g increasing? (c) Where does g attain its absolute maximum on [0, 6]? Justify.

Strategy. Use areas for values (g = signed area under f). Use the sign of f for increasing/decreasing and extrema (g′ = f).

Solution.

(a) g(4) = ∫₀⁴ f(t) dt = area of the half-disk = (1/2)π(2)² = 2π ≈ 6.28. From 4 to 6, f < 0, contributing −(1/2)(2)(2) = −2. So g(6) = 2π − 2 ≈ 4.28.

(b) g′ = f, and f > 0 on (0, 4). So g is increasing on (0, 4) and decreasing on (4, 6).

(c) The only interior critical point is x = 4, where f changes from positive to negative, so g has a relative maximum there. Comparing with the endpoints, g(0) = 0 and g(6) = 2π − 2 ≈ 4.28, both less than g(4) = 2π. Justification: g has its absolute maximum at x = 4, because g′(x) = f(x) > 0 for 0 < x < 4 and g′(x) = f(x) < 0 for 4 < x < 6, so g increases up to x = 4 then decreases; and g(4) = 2π exceeds the values at both endpoints.

Example 4 — Net change interpretation (CALC)

Problem. Water flows into a tank at a rate r(t) = 4 + 6 sin(t/2) gallons per minute, for 0 ≤ t ≤ 8. How many gallons enter the tank during these 8 minutes?

Strategy. The total amount added is the net change of volume, which is the integral of the rate: ∫₀⁸ r(t) dt.

Solution. By the net-change interpretation of FTC Part 2,

total gallons = ∫₀⁸ [4 + 6 sin(t/2)] dt

This is a calculator part, so use fnInt:

TI-84: MATH → 9:fnInt( → fnInt(4 + 6sin(X/2), X, 0, 8) → 51.844

So approximately 51.844 gallons enter the tank. (By hand for comparison: an antiderivative is 4t − 12 cos(t/2), giving [4(8) − 12 cos 4] − [0 − 12 cos 0] = (32 − 12 cos 4) − (−12) = 44 − 12 cos 4 ≈ 51.844.) The units check: gallons-per-minute times minutes gives gallons.

(d) Common Mistakes

Forgetting the chain-rule factor u′(x) in FTC Part 1. Students write d/dx ∫₂^{x²} cos t dt = cos(x²) and stop. Why it's wrong: the upper limit is a composition u(x) = x², so the chain rule contributes u′(x) = 2x. Fix: the correct derivative is cos(x²)·2x. Whenever the upper limit is anything other than a bare x, multiply by its derivative.

Swapping the two parts. Students try to "evaluate" ∫ₐᵇ f by computing f(b) − f(a), or try to differentiate ∫ₐˣ f by finding an antiderivative first. Why it's wrong: Part 2 subtracts an antiderivative F at the endpoints (F(b) − F(a)); Part 1 says the derivative of the accumulation is just f at the upper limit. Fix: name your task first — evaluating a number (Part 2) or differentiating an accumulation (Part 1) — then apply the matching part.

Sign error from F(a) − F(b). FTC Part 2 is F(b) − F(a) — top limit minus bottom limit. Subtracting in the wrong order negates the answer. Fix: always write the upper-limit value first. A negative answer where you expected positive is the telltale sign you flipped it.

Reading g from f's graph as if g equals f. Students say "g has a maximum where f has a maximum." Why it's wrong: f is g′, not g. g has a maximum where g′ = f changes from + to − (an x-intercept of f), and g has an inflection point where f has a max or min. Fix: treat the given graph of f exactly as you treated f′ in Unit 5 — sign of f ⇒ increase/decrease of g; slope of f ⇒ concavity of g.

Dropping dx / writing +C on a definite integral. A definite integral is a number; there is no +C on it (the constant cancels in F(b) − F(a)). Fix: keep +C for indefinite integrals only; on a definite integral, the answer is a number with no C.

(e) Practice Problems

12 (short answer). [NO CALC] Evaluate ∫₁⁴ (3x² − 2x) dx using FTC Part 2, and write the full FTC justification (name the hypotheses).

13 (short answer, justification). [NO CALC] A student computes d/dx ∫₁^{x³} √(1 + t²) dt = √(1 + x⁶). Identify the error, give the correct derivative, and explain which part of the FTC (and which rule) was misapplied.

14 (short answer, accumulation from graph, justification). [NO CALC] The graph of f is the upper semicircle of radius 2 on [0, 4] (so f ≥ 0) together with the segment from (4, 0) to (6, −2) (so f < 0), as in Example 3. Let g(x) = ∫₀ˣ f(t) dt. Find g(6), state where g has its absolute maximum on [0, 6], and justify your answer using the sign of g′.

15 (short answer, net change interpretation). [CALC] Sand is added to a pile at a rate of r(t) = 10e^{−0.2t} kilograms per hour for 0 ≤ t ≤ 5. Write a definite integral for the total sand added over the 5 hours, evaluate it (round to three decimals), and explain the meaning of the answer in context.

(f) AP Exam Focus

This is the classic accumulation-function FRQ — given the graph of a function f, you analyze g(x) = ∫ₐˣ f(t) dt. It is a non-calculator question (AP Section II, Part B style). Budget about 15 minutes. Justifications must name the relevant theorem or the sign of g′ = f.

FRQ. Let f be the continuous function whose graph on [0, 7] is shown above (four line segments, with t-intercepts at t = 2 and t = 5 and a minimum at (4, −4)). Let g be the function defined by

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g(x) = ∫₀ˣ f(t) dt.

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(a) Find g(2), g(5), and g(7). (3 points)

(b) On the interval 0 < x < 7, find the x-coordinate of each relative extremum of g, and classify each as a relative maximum or minimum. Justify your answer. (3 points)

(c) On the interval 0 < x < 7, find the x-coordinate of each inflection point of the graph of g. Justify your answer. (2 points)

(d) Find the absolute minimum value of g on the closed interval [0, 7]. Justify your answer. (2 points)

Total: 10 points

Model Solution

(a) g(x) is the signed area under f from 0 to x (area above the t-axis counts positive, below counts negative). Using the triangle areas:

• g(2) = ∫₀² f(t) dt = +4 (triangle above the axis on [0, 2], area (1/2)(2)(4) = 4).
• g(5) = g(2) + ∫₂⁵ f(t) dt = 4 + (−4) + (−2) = −2 (two below-axis triangles on [2, 4] and [4, 5], areas 4 and 2).
• g(7) = g(5) + ∫₅⁷ f(t) dt = −2 + (+4) = 2 (triangle above the axis on [5, 7], area (1/2)(2)(4) = 4).

So g(2) = 4, g(5) = −2, g(7) = 2.

(b) Since g′(x) = f(x), the relative extrema of g occur where f changes sign.

• At x = 2, f changes from positive to negative, so g′ changes from positive to negative: g has a relative maximum at x = 2.
• At x = 5, f changes from negative to positive, so g′ changes from negative to positive: g has a relative minimum at x = 5.

Justification: g′ = f. At x = 2, f changes from positive to negative, so g has a relative maximum there; at x = 5, f changes from negative to positive, so g has a relative minimum there.

(c) Inflection points of g occur where g″ = f′ changes sign — that is, where f changes from increasing to decreasing or vice versa. From the graph, f is decreasing on (0, 4) and increasing on (4, 7), changing direction only at x = 4.

Justification: g″ = f′. The graph of f changes from decreasing to increasing at x = 4, so g″ = f′ changes from negative to positive there; therefore g has an inflection point at x = 4.

(d) Candidates for the absolute minimum of g on [0, 7] are the relative minimum x = 5 and the endpoints x = 0 and x = 7. Evaluate:

g(0) = 0,   g(5) = −2,   g(7) = 2

The least of these is g(5) = −2. Justification: On [0, 7], g is continuous, so by the Extreme Value Theorem its absolute minimum occurs at a critical point or an endpoint. The only interior critical point that is a relative minimum is x = 5, where g(5) = −2; comparing with g(0) = 0 and g(7) = 2, the absolute minimum value of g is −2, attained at x = 5.

Scoring Commentary — where students lose points

• (a) 1 point each for g(2) = 4, g(5) = −2, g(7) = 2. Trap: treating areas below the axis as positive. Below-axis regions contribute negative signed area — this is the single most common error and it cascades into g(5) and g(7).
• (b) 1 point for x = 2 (max), 1 point for x = 5 (min), 1 point for a sign-of-g′ justification. Trap: saying "g has a max at x = 2 because f has a max there" — false; g's extrema are at the zeros of f where f changes sign, not at f's peaks. The justification must reference that g′ = f changes sign.
• (c) 1 point for x = 4, 1 point for the justification via g″ = f′ changing sign. Trap: naming x = 2 or x = 5 (those are extrema of g, where f = 0, not inflection points) — inflection of g is where f turns around, here only at x = 4.
• (d) 1 point for the value −2, 1 point for justifying via the closed-interval/EVT candidate test. Trap: reporting the location x = 5 instead of the value −2 (the question asks for the minimum value), or failing to compare with the endpoints.

∫₁⁴ (3x² − 2x) dx = [x³ − x²]₁⁴ = (64 − 16) − (1 − 1) = 48 − 0 = 48.

d/dx ∫₁^{x³} √(1 + t²) dt = √(1 + (x³)²) · 3x² = 3x²·√(1 + x⁶).

total = ∫₀⁵ 10e^{−0.2t} dt.