∫ (top − bottom) dx, finding the intersection points first.f_avg = (1/(b−a)) ∫ₐᵇ f dx.dy/dt = ky to get y = y₀e^{kt}, and use it for half-life, doubling time, and growth/decay word problems.This final lesson pulls together the integral applications that close out Calc I, adds the one differential-equation model the exam tests, and finishes with exam-day strategy. The recurring skill is translating a picture or a sentence into the right integral or equation — the calculus itself you already know.
If f(x) ≥ g(x) on [a, b], the area of the region between the curves is
A = ∫ₐᵇ (top − bottom) dx = ∫ₐᵇ [f(x) − g(x)] dx
Because top ≥ bottom, the integrand is positive and no absolute value is needed. Three setup rules:
f(x) = g(x); the solutions are the limits a and b.(top − bottom) on each piece.Worked example. Area between y = x and y = x². Intersections: x = x² gives x = 0, 1. At x = ½ the line gives 0.5, the parabola 0.25, so the line is on top:
A = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6
[GRAPH: y = x and y = x² on [0, 1.2] × [0, 1.2] - Line y = x through (0,0) and (1,1) - Parabola y = x² through (0,0) and (1,1), bowing below the line - Shade the lens-shaped region between them on [0, 1]; line is the upper boundary, parabola the lower]
The average value of f on [a, b] is the height of the rectangle whose area equals the area under f:
f_avg = (1/(b − a)) ∫ₐᵇ f(x) dx
The 1/(b − a) factor is essential — leaving it off gives the integral, not the average. For example, the average value of f(x) = x² on [0, 3] is (1/3)∫₀³ x² dx = (1/3)(9) = 3.
If r(t) is a rate (the derivative of some quantity), the integral of the rate is the net change in that quantity:
∫ₐᵇ r(t) dt = net change from t = a to t = b
Units carry through: if r(t) is in gallons per minute and t in minutes, ∫ₐᵇ r(t) dt is in gallons. If water flows in at r(t) = 2t gal/min for 0 ≤ t ≤ 5, the total added is ∫₀⁵ 2t dt = [t²]₀⁵ = 25 gallons — not 2t evaluated at a point.
Many quantities change at a rate proportional to how much is present — continuously compounded money, radioactive samples, unchecked populations. The mathematical statement is:
dy/dt = ky
Separating variables and integrating gives the general solution:
y = y₀ e^{kt}
where y₀ = y(0) is the initial amount and k is the growth/decay constant.
k > 0: growth (e^{kt} → ∞).k < 0: decay (e^{kt} → 0).Finding k from data. You are almost never handed k; you solve for it with a logarithm. Given y₀ and one later value y(t₁) = y₁:
y₁ = y₀ e^{k t₁} ⟹ k = (1/t₁) ln(y₁/y₀)
Half-life and doubling time. Set y to half (or double) the start:
Half-life (k < 0): t½ = −(ln 2)/k (equivalently k = −(ln 2)/t½)
Doubling time (k > 0): T_d = (ln 2)/k (equivalently k = (ln 2)/T_d)
Both depend only on k, never on y₀: the amount halves (or doubles) over every interval of the same length. A handy shortcut for clean multiples — after n half-lives, y = y₀·(1/2)ⁿ.
Worked example (decay). A 50 mg sample has a half-life of 6 hours. After 15 hours, 15/6 = 2.5 half-lives have passed, so y = 50·(1/2)^{2.5} ≈ 8.84 mg.
Structure. CLEP Calculus is about 44 questions in roughly 90 minutes, in two sections:
When the calculator earns its keep (Section 2): numerical integration (fnInt) for antiderivatives that have no elementary form, finding zeros or intersections, evaluating messy exponentials and logs for growth/decay, and reading tables of values. Set problems up by hand; let the calculator do arithmetic only.
Best-approximation items. Some Section 2 stems say "if the exact value is not a choice, select the best approximation." Your decimal from the calculator (e.g. 2.227) may not match a choice exactly — pick the closest.
Numeric entry. A few items are type-a-number-in-a-box rather than multiple choice (for example, "for what value of k is f continuous? k = ___"). There is no letter to guess — you must produce the number.
Rights-only scoring. Your raw score is simply the number correct; there is no penalty for a wrong answer. So never leave anything blank — with the clock running out, fill in every remaining question.
Pacing. About two minutes per question. If one stalls you, flag a guess and move on; a hard item is worth exactly as much as an easy one.
∫ (top − bottom) dx; find intersections first, then confirm which curve is on top.(1/(b−a)) ∫ₐᵇ f dx; the 1/(b−a) factor is mandatory.dy/dt = ky solves to y = y₀e^{kt}; k > 0 grows, k < 0 decays. Solving for k or a time always needs a logarithm.= −(ln 2)/k and doubling time = (ln 2)/k depend only on k; after n half-lives, y = y₀·(1/2)ⁿ.y = x and y = x² is1. C) 1/6. Intersections x = x² give x = 0, 1; the line is on top, so A = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6. Distractors: A) 1/2 and B) 1/3 are the two separate antiderivative terms, not their difference. D) 5/6 added the terms instead of subtracting. E) 1 integrated only x. Fix: area is ∫ (top − bottom), and here top − bottom = x − x².
y = x² and y = x + 2 is2. E) 9/2. Intersections x² = x + 2 give x = −1, 2; the line x + 2 is on top, so A = ∫_{−1}^{2} (x + 2 − x²) dx = 9/2. Distractors: A) 7/6 subtracted in the wrong order over part of the interval. B) 19/3 used limits 0 to 2, dropping the left region. C) 5/2 and D) 3 come from antiderivative slips. Fix: solve for both intersection points, then integrate (top − bottom) across the full span.
f(x) = 3x² + 2x on [1, 3] is3. E) 17. f_avg = (1/2)∫₁³ (3x² + 2x) dx = (1/2)[x³ + x²]₁³ = (1/2)(36 − 2) = 17. Distractors: A) 34 is the integral itself — the 1/(b−a) factor was dropped. B) 12 and C) 30 are antiderivative or arithmetic errors. D) 10 divided by the wrong length. Fix: average value always includes the 1/(b−a) divisor; here b − a = 2.
f(x) = sin x on [0, π] is4. B) 2/π. f_avg = (1/π)∫₀^π sin x dx = (1/π)[−cos x]₀^π = (1/π)(1 + 1) = 2/π. Distractors: A) 1/π halved the integral (which is 2). C) 0 confused net signed area (sin is positive on [0, π]) with the average. D) 1 forgot the 1/π. E) π/2 inverted the factor. Fix: ∫₀^π sin x dx = 2, then divide by the length π.
dy/dt = ky with y(0) = 4 and y(2) = 12. Then y(4) =5. B) 36. y = 4e^{kt}; from 12 = 4e^{2k}, e^{2k} = 3. Then y(4) = 4e^{4k} = 4(e^{2k})² = 4·9 = 36. Distractors: A) 20 modeled the growth linearly (4 + 4·4). C) 28 added increments. D) 108 tripled at each step (4·3·3·3) instead of squaring. E) 16 doubled once. Fix: exponential growth means each equal time interval multiplies by the same factor; e^{2k} = 3, so over 4 years the factor is 3² = 9.
k equals6. D) (ln 2)/5. Doubling time T_d = 5, and k = (ln 2)/T_d = (ln 2)/5. Distractors: A) 2/5 confused the doubling factor with k. B) 5/(ln 2) inverted the relationship. C) (ln 5)/2 used the wrong numbers. E) ln(2/5) is negative — wrong for growth. Fix: from 2 = e^{kT_d}, take the log: k = (ln 2)/T_d.
k = −0.045 per year. Its half-life, in years, is closest to7. D) 15.4. t½ = −(ln 2)/k = (ln 2)/0.045 ≈ 15.4 years. Distractors: A) 6.9 used ln 2 / 0.1. B) 22.2 used 1/k without ln 2. C) 30.8 doubled the wrong value. E) 45.0 used 2/k. Fix: half-life is (ln 2)/|k|; with |k| = 0.045, that is about 15.4 years.
r(t) = 2t gallons per minute for 0 ≤ t ≤ 5. The total amount of water added during these 5 minutes is8. B) 25 gal. Accumulated change is the integral of the rate: ∫₀⁵ 2t dt = [t²]₀⁵ = 25 gallons. Distractors: A) 10 evaluated the rate 2t at t = 5, not the accumulation. C) 50 used rate-times-interval (2·5·5) instead of integrating. D) 5 used only the interval length. E) 12.5 halved incorrectly. Fix: total amount = ∫ r(t) dt, not the rate at an endpoint.
9. A) 8.8 mg. After 15/6 = 2.5 half-lives, y = 50·(1/2)^{2.5} ≈ 8.84 mg. Distractors: B) 6.3 used 3 half-lives. C) 12.5 used 2 half-lives. D) 17.7 used 1.5. E) 25.0 used a single half-life. Fix: count half-lives as t / t½; here 2.5, then multiply by (1/2)^{2.5}.
10. A) dy/dt = ky. "Rate of change proportional to the amount present" is exactly dy/dt = ky. Distractors: B) dy/dt = kt is proportional to time. C) dy/dt = k/y is inversely proportional. D) dy/dt = k(y − t) mixes in time. E) dy/dt = ky + t adds an extra term. Fix: "proportional to the amount y" means the right side is a constant times y alone.
11. A) Mark an answer on every remaining question, because a wrong answer costs nothing. CLEP uses rights-only scoring — the raw score is just the number correct, with no deduction for wrong answers — so a guess can only help. Distractors: B) and C) leave points on the table; blanks and wrongs both score zero, but a guess has a positive chance of being right. D) sacrifices two guaranteed chances for one. E) is not permitted; the section time is fixed. Fix: with no guessing penalty, always fill in every answer before time expires.
r(t) = 100e^{−0.1t} liters per hour for 0 ≤ t ≤ 8, and ∫₀⁸ r(t) dt ≈ 550.7. The value 550.7 represents the12. C) The total number of liters removed during the 8 hours. The integral of a rate over an interval is the accumulated change, and (liters/hour)(hours) = liters. Distractors: A) the rate at t = 8 is r(8) = 100e^{−0.8}, a single value, not the integral. B) the average rate would divide by 8. D) "remaining" would require the starting amount, which is not given. E) time-to-empty is a different calculation. Fix: ∫ₐᵇ r(t) dt is total accumulated change, carrying the units rate×time.
1. C) 1/6. Intersections x = x² give x = 0, 1; the line is on top, so A = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6. Distractors: A) 1/2 and B) 1/3 are the two separate antiderivative terms, not their difference. D) 5/6 added the terms instead of subtracting. E) 1 integrated only x. Fix: area is ∫ (top − bottom), and here top − bottom = x − x².
2. E) 9/2. Intersections x² = x + 2 give x = −1, 2; the line x + 2 is on top, so A = ∫_{−1}^{2} (x + 2 − x²) dx = 9/2. Distractors: A) 7/6 subtracted in the wrong order over part of the interval. B) 19/3 used limits 0 to 2, dropping the left region. C) 5/2 and D) 3 come from antiderivative slips. Fix: solve for both intersection points, then integrate (top − bottom) across the full span.
3. E) 17. f_avg = (1/2)∫₁³ (3x² + 2x) dx = (1/2)[x³ + x²]₁³ = (1/2)(36 − 2) = 17. Distractors: A) 34 is the integral itself — the 1/(b−a) factor was dropped. B) 12 and C) 30 are antiderivative or arithmetic errors. D) 10 divided by the wrong length. Fix: average value always includes the 1/(b−a) divisor; here b − a = 2.
4. B) 2/π. f_avg = (1/π)∫₀^π sin x dx = (1/π)[−cos x]₀^π = (1/π)(1 + 1) = 2/π. Distractors: A) 1/π halved the integral (which is 2). C) 0 confused net signed area (sin is positive on [0, π]) with the average. D) 1 forgot the 1/π. E) π/2 inverted the factor. Fix: ∫₀^π sin x dx = 2, then divide by the length π.
5. B) 36. y = 4e^{kt}; from 12 = 4e^{2k}, e^{2k} = 3. Then y(4) = 4e^{4k} = 4(e^{2k})² = 4·9 = 36. Distractors: A) 20 modeled the growth linearly (4 + 4·4). C) 28 added increments. D) 108 tripled at each step (4·3·3·3) instead of squaring. E) 16 doubled once. Fix: exponential growth means each equal time interval multiplies by the same factor; e^{2k} = 3, so over 4 years the factor is 3² = 9.
6. D) (ln 2)/5. Doubling time T_d = 5, and k = (ln 2)/T_d = (ln 2)/5. Distractors: A) 2/5 confused the doubling factor with k. B) 5/(ln 2) inverted the relationship. C) (ln 5)/2 used the wrong numbers. E) ln(2/5) is negative — wrong for growth. Fix: from 2 = e^{kT_d}, take the log: k = (ln 2)/T_d.
7. D) 15.4. t½ = −(ln 2)/k = (ln 2)/0.045 ≈ 15.4 years. Distractors: A) 6.9 used ln 2 / 0.1. B) 22.2 used 1/k without ln 2. C) 30.8 doubled the wrong value. E) 45.0 used 2/k. Fix: half-life is (ln 2)/|k|; with |k| = 0.045, that is about 15.4 years.
8. B) 25 gal. Accumulated change is the integral of the rate: ∫₀⁵ 2t dt = [t²]₀⁵ = 25 gallons. Distractors: A) 10 evaluated the rate 2t at t = 5, not the accumulation. C) 50 used rate-times-interval (2·5·5) instead of integrating. D) 5 used only the interval length. E) 12.5 halved incorrectly. Fix: total amount = ∫ r(t) dt, not the rate at an endpoint.
9. A) 8.8 mg. After 15/6 = 2.5 half-lives, y = 50·(1/2)^{2.5} ≈ 8.84 mg. Distractors: B) 6.3 used 3 half-lives. C) 12.5 used 2 half-lives. D) 17.7 used 1.5. E) 25.0 used a single half-life. Fix: count half-lives as t / t½; here 2.5, then multiply by (1/2)^{2.5}.
10. A) dy/dt = ky. "Rate of change proportional to the amount present" is exactly dy/dt = ky. Distractors: B) dy/dt = kt is proportional to time. C) dy/dt = k/y is inversely proportional. D) dy/dt = k(y − t) mixes in time. E) dy/dt = ky + t adds an extra term. Fix: "proportional to the amount y" means the right side is a constant times y alone.
11. A) Mark an answer on every remaining question, because a wrong answer costs nothing. CLEP uses rights-only scoring — the raw score is just the number correct, with no deduction for wrong answers — so a guess can only help. Distractors: B) and C) leave points on the table; blanks and wrongs both score zero, but a guess has a positive chance of being right. D) sacrifices two guaranteed chances for one. E) is not permitted; the section time is fixed. Fix: with no guessing penalty, always fill in every answer before time expires.
12. C) The total number of liters removed during the 8 hours. The integral of a rate over an interval is the accumulated change, and (liters/hour)(hours) = liters. Distractors: A) the rate at t = 8 is r(8) = 100e^{−0.8}, a single value, not the integral. B) the average rate would divide by 8. D) "remaining" would require the starting amount, which is not given. E) time-to-empty is a different calculation. Fix: ∫ₐᵇ r(t) dt is total accumulated change, carrying the units rate×time.