Lesson 20: Calculus Capstone: Review and the Big Theorems

The capstone lesson collects the course into a single map and stress-tests it with synthesis problems. The map has a pleasing shape: two questions (how fast? how much?) generate two operations (derivative, integral), and a handful of big theorems certify that the operations invert each other and behave as intuition demands. If you can state each theorem, recognise its hypotheses, and deploy it on demand, you own the course — the techniques of Lessons 4–19 are vocabulary; these theorems are the grammar.

The big theorems

• Intermediate Value Theorem (IVT). A continuous function on [a, b] takes every value between f(a) and f(b). Use: proving roots exist. f(x) = x^3 + x − 1 has f(0) = −1 and f(1) = 1, so a root lies in (0, 1) — no formula for cubics required. Refining: f(0.5) = −0.375, f(0.7) = 0.043, so the root lies in (0.5, 0.7); bisection is IVT turned into an algorithm.
• Extreme Value Theorem (EVT). A continuous function on a closed interval attains a maximum and a minimum. Use: optimisation guarantees. On [−2, 2], f(x) = x^3 − 3x has critical points x = ±1 (f'(x) = 3x^2 − 3); candidates: f(−2) = −2, f(−1) = 2, f(1) = −2, f(2) = 2 — maximum 2 (attained twice), minimum −2 (attained twice). The hypotheses are the theorem: drop "closed" and f(x) = x on (0, 1) attains neither a maximum nor a minimum — it approaches 1 and 0 without ever reaching them; drop "continuous" and a jump can strand the extreme values in the gap.
• Mean Value Theorem (MVT). If f is continuous on [a, b] and differentiable on (a, b), some c in (a, b) has f'(c) = (f(b) − f(a))/(b − a): somewhere, the instantaneous rate equals the average rate. For f(x) = x^3 − 3x on [0, 2]: average rate = (2 − 0)/2 = 1; f'(x) = 3x^2 − 3 = 1 gives x^2 = 4/3, c = 2/sqrt(3) ≈ 1.155 ∈ (0, 2) ✓. MVT is the licence for reading derivatives as rates over intervals — speed cameras and average-speed enforcement are its folk version.
• Fundamental Theorem (both parts). Accumulation functions differentiate to their integrands (Part 1); definite integrals evaluate by antiderivatives (Part 2). Everything in Lessons 7–15 is commentary on this.
• Squeeze theorem and L'Hôpital's rule. The limit toolkit of Lessons 1, 2 and 12: bounding a hard limit between two easy ones, and differentiating indeterminate quotients — each with hypotheses to check before use.

The course on one line each

Limits define; continuity qualifies; the derivative measures instantaneous rate (Lessons 1–3). Rules make derivatives computable; chains and implicit relations extend their reach (4–6). Integrals accumulate, the FTC evaluates them, and substitution/parts/partial fractions/trig substitution enlarge the computable family (7–9, 13). Applications convert slices into areas, volumes, work, forces, and balance points (10, 17–19). Differential equations turn rate laws into functions (11); L'Hôpital disciplines the indeterminate (12); improper integrals extend accumulation to infinity (14); series sum the infinite, and Taylor series turn functions into polynomials (15–16).

Synthesis: one problem, five lessons

A tank holds 50 L. Water flows in at R(t) = 10 e^(−0.1t) L/min (a strong start, fading) and drains at a constant 2 L/min. The net rate is V'(t) = 10 e^(−0.1t) − 2.

• When does the volume peak? (Lesson 4/6 thinking: sign of the derivative.) V'(t) = 0 when e^(−0.1t) = 0.2, i.e. t = 10 ln 5 ≈ 16.1 min; V increases before (inflow wins) and decreases after.
• Volume at t = 10? (Lesson 7: accumulate.) V(10) = 50 + ∫ from 0 to 10 of (10 e^(−0.1t) − 2) dt = 50 + 100(1 − e^(−1)) − 20 ≈ 50 + 63.2 − 20 = 93.2 L. (The inflow integral is Lesson 8's substitution u = −0.1t in disguise.)
• Peak volume? V(10 ln 5) = 50 + 100(1 − e^(−ln 5)) − 20 ln 5 = 50 + 100(0.8) − 32.2 ≈ 97.8 L.
• Average inflow rate over [0, 10]? (Lesson 10: average value.) (1/10)·100(1 − e^(−1)) = 10(1 − e^(−1)) ≈ 6.32 L/min.
• Long run? (Lessons 2/14: limits.) Inflow dies away, the drain persists, and the model eventually predicts an empty tank — at which point the model, not the calculus, expires (a real tank stops draining at 0). Stating where a model stops is part of using it.

No single technique carried this problem; the course did, acting as one instrument. That integration — derivative for timing, integral for totals, limits for the long run — is the actual subject you have been learning.

Common mistakes (the greatest hits)

• Applying theorems without hypotheses. MVT on |x| over [−1, 1]: the average rate is 0, but no c has f'(c) = 0 — differentiability fails at 0 and the theorem never promised anything. Checking hypotheses is not pedantry; it is the difference between a theorem and a wish.
• Confusing average value with average rate of change. Average value of f on [a, b] is (1/(b − a))∫ f; average rate is (f(b) − f(a))/(b − a). The MVT bridges them (for f'), but they answer different questions: "typical height" versus "overall change per unit."
• Forgetting that + C is where initial conditions live. In synthesis problems the constant is usually the given starting value; answers missing it fail every subsequent part.
• Letting the model run past its domain. Negative tank volumes, populations below zero, probabilities above 1 — the mathematics continues serenely past the point where the model stops describing the world. The last step of every applied problem is a sentence about validity.

Exam Experience

1. A continuous function f satisfies f(0) = −2 and f(3) = 5. Which conclusion is guaranteed?

• (A) f has a root in (0, 3)
• (B) f is differentiable on (0, 3)
• (C) f(1.5) = 1.5
• (D) f attains a maximum on [0, 3] at x = 3

Answer: (A). IVT: every value between −2 and 5, including 0, is attained. (D) is also an EVT-flavoured claim but the location is not guaranteed — the maximum exists, not necessarily at x = 3; (A) is the certain one. Continuity alone never gives differentiability (B), and (C) invents a value.

2. For f(x) = x^2 on [1, 3], the Mean Value Theorem point c is

• (A) 1.5
• (B) 2
• (C) 2.5
• (D) sqrt(5)

Answer: (B). Average rate = (9 − 1)/(3 − 1) = 4; f'(x) = 2x = 4 gives c = 2 ∈ (1, 3) ✓.

3. The average value of sin x on [0, π] is

• (A) 0
• (B) 1/π
• (C) 2/π
• (D) 1

Answer: (C). (1/π)∫ from 0 to π of sin x dx = (1/π)(2) = 2/π ≈ 0.637. (A) would be the average over a full period [0, 2π] — interval awareness is the tested point.

Free-response question. A reservoir contains 50 thousand litres at time t = 0. Water enters at rate 10 e^(−0.1t) thousand litres per hour and leaves at a constant 2 thousand litres per hour. (a) Write the function V(t) for the volume, evaluating the integral. (b) Find the time at which the volume is maximal, justifying with the sign of V'. (c) Find the maximum volume. (d) Find the average rate of inflow over the first 10 hours.

Scoring guideline.

• (a) 2 points. 1 point for the accumulation set-up V(t) = 50 + ∫ from 0 to t of (10 e^(−0.1s) − 2) ds. 1 point for the evaluation V(t) = 50 + 100(1 − e^(−0.1t)) − 2t. (Variable-of-integration hygiene — s inside, t on the limits — is part of the point.)
• (b) 1 point. V'(t) = 10 e^(−0.1t) − 2 = 0 gives t = 10 ln 5 ≈ 16.1 hours, with V' changing + to − (inflow falls through the outflow rate), so this is the maximum.
• (c) 1 point. V(10 ln 5) = 50 + 100(1 − 1/5) − 20 ln 5 = 130 − 20 ln 5 ≈ 97.8 thousand litres.
• (d) 1 point. (1/10)∫ from 0 to 10 of 10 e^(−0.1t) dt = (1/10)·100(1 − e^(−1)) = 10(1 − e^(−1)) ≈ 6.32 thousand litres per hour. (This is a 5-point question in total: 2 + 1 + 1 + 1.)

Going deeper: Rolle's theorem and a technique-selection guide

Rolle's theorem is the MVT with equal endpoint values: if f(a) = f(b) (with the usual continuity/differentiability hypotheses), some c has f'(c) = 0 — between two equal heights, a smooth graph must level off somewhere. For f(x) = x^2 − 4x + 3 on [0, 4]: f(0) = 3 = f(4), and indeed f'(x) = 2x − 4 vanishes at c = 2. Rolle's theorem is how the MVT is proved (tilt the graph until the endpoints match), and its moral is practical: equal values force a stationary point, a fact used everywhere from root-counting (a function with n roots has a derivative with at least n − 1) to numerical methods. Choosing a technique — the course as a decision tree. Facing a limit: try substitution; if indeterminate, classify the form (Lesson 12's conversions) before differentiating. Facing a derivative: identify the outermost structure (sum, product, quotient, nest) and let it choose the rule. Facing an integral: substitution first (cheapest), then identities, then parts, then partial fractions or trig substitution for their signature shapes; if nothing fits, the integral may be non-elementary — estimate it numerically or by series (Lessons 7 and 16), which is a legitimate destination, not a defeat. Facing a rate law: it is a differential equation; separate it (Lesson 11). Facing an infinite sum: classify with tests before dreaming of a closed form (Lesson 15). The examinations of this course reward the choosing as much as the computing.