∫ₐᵇ f(x) dx = F(b) − F(a), where F is any antiderivative of f.g(x) = ∫ₐˣ f(t) dt with FTC Part 1, including a variable upper limit u(x) that forces the chain-rule factor u′(x).g(x) = ∫ₐˣ f(t) dt off the graph of f — where g rises, falls, and reaches its extremes.The Fundamental Theorem of Calculus (FTC) links the two halves of calculus — the derivative and the integral — and it does so in two directions. Keep them straight: Part 2 evaluates a definite integral; Part 1 differentiates an accumulation function. Mixing them up is the single most common error on this topic.
If
fis continuous on[a, b]andFis any antiderivative off(that is,F′ = f), then
∫ₐᵇ f(x) dx = F(b) − F(a)
This replaces the slow limit-of-Riemann-sums work with a two-step recipe: find one antiderivative F, then subtract its endpoint values. The notation for that subtraction is the evaluation bar:
∫ₐᵇ f(x) dx = [F(x)]ₐᵇ = F(b) − F(a) (upper limit first, then subtract the lower)
The +C never appears on a definite integral — it cancels in the subtraction [F(b)+C] − [F(a)+C].
Worked evaluation. Compute ∫₁³ (3x² − 4x) dx. An antiderivative is F(x) = x³ − 2x²:
∫₁³ (3x² − 4x) dx = [x³ − 2x²]₁³ = (27 − 18) − (1 − 2) = 9 − (−1) = 10
The order matters: top value minus bottom value. Reversing it flips the sign.
Now fix the lower limit and let the upper limit vary. This builds an accumulation function:
g(x) = ∫ₐˣ f(t) dt
Here t is a dummy variable and x is the real input; g(x) is the signed area accumulated from a out to x. The theorem says differentiating undoes integrating:
If
fis continuous on an interval containinga, andg(x) = ∫ₐˣ f(t) dt, then
g′(x) = f(x)
The derivative of "area so far" is just the height of the curve at the right edge. The lower limit a is a constant and has no effect on g′.
The variable upper limit — chain rule. Very often the upper limit is a function u(x), not a bare x. Then g(x) = ∫ₐ^{u(x)} f(t) dt is a composition, and you multiply by u′(x):
d/dx ∫ₐ^{u(x)} f(t) dt = f(u(x)) · u′(x)
The factor u′(x) is the most-forgotten piece of this topic. Example:
d/dx ∫₀^{x²} sin t dt = sin(x²) · 2x = 2x sin(x²)
Substitute the upper limit x² into the integrand sin, then multiply by the derivative of the upper limit, 2x.
The variable lower limit. If the lower limit varies, flip the integral first using ∫ₐᵇ = −∫ᵇₐ:
d/dx ∫ₓ³ cos t dt = d/dx [−∫₃ˣ cos t dt] = −cos x
g from the graph of fA frequent CLEP item gives you only the graph of f and asks about g(x) = ∫ₐˣ f(t) dt. Because g′ = f, treat the graph you are given exactly as if it were the derivative of g:
What f does (the graph you see) |
What happens to g(x) = ∫ₐˣ f(t) dt |
|---|---|
f(x) > 0 (above the axis) |
g is increasing (g′ = f > 0) |
f(x) < 0 (below the axis) |
g is decreasing |
f changes + → − |
g has a relative maximum |
f changes − → + |
g has a relative minimum |
the g-value itself |
the signed area under f from a to x |
To get an actual value g(x), add up signed areas geometrically: regions above the axis count positive, below count negative.
[GRAPH: y = f(t) on [0, 6] × [−3, 3] - Upper semicircle of radius 2 centered at (2, 0), spanning t = 0 to t = 4, lying ABOVE the axis (f ≥ 0), peak at (2, 2). Its area is a half-disk = (1/2)π(2)² = 2π. - Line segment from (4, 0) down to (6, −2), lying BELOW the axis (f < 0). The region under it is a triangle of base 2, height 2, area 2. - t-intercepts marked at t = 0 and t = 4]
Let g(x) = ∫₀ˣ f(t) dt. On [0, 4], f ≥ 0, so g climbs; the accumulated area is the half-disk, g(4) = 2π ≈ 6.28. On [4, 6], f < 0, so g falls, subtracting the triangle's area 2. Thus g(6) = 2π − 2 ≈ 4.28. Because f switches from positive to negative at x = 4, g reaches its maximum at x = 4.
When the integrand is itself a rate of change (f = F′), FTC Part 2 reads as accumulation:
∫ₐᵇ f′(x) dx = f(b) − f(a) (integral of a rate = net change in the quantity)
If r(t) is a flow rate in liters per hour, then ∫₀⁸ r(t) dt is the total liters delivered from t = 0 to t = 8. Units carry through: (liters/hour)(hours) = liters. This is the backbone of every "rate in / rate out" word problem.
On the calculator-permitted section, evaluate a definite integral numerically with fnInt:
fnInt(integrand, variable, lower, upper) e.g. fnInt(e^(−X²), X, 0, 2) ≈ 0.882
Set the integral up by hand; let the calculator do only the arithmetic. Many antiderivatives (like e^(−x²)) have no elementary form, so numerical integration is the only route in the second section.
∫ₐᵇ f = F(b) − F(a), upper minus lower; no +C on a definite integral.d/dx ∫ₐˣ f(t) dt = f(x); the constant lower limit is irrelevant to g′.u(x) demands the chain-rule factor: f(u(x))·u′(x). A variable lower limit flips the sign.g′ = f, read g off the graph of f: f > 0 ⇒ g rising, f < 0 ⇒ g falling, and g peaks where f crosses from + to −.∫₁³ (3x² − 4x) dx =1. D) 10. With F(x) = x³ − 2x², FTC Part 2 gives F(3) − F(1) = 9 − (−1) = 10. Distractors: A) 26 integrated only the 3x² term ([x³]₁³ = 26), dropping −4x. B) 9 is F(3) alone — the lower limit was never subtracted. C) −10 reversed the order to F(1) − F(3). E) 20 comes from an antiderivative slip. Fix: always compute F(b) − F(a), upper value first, and antidifferentiate every term.
∫₁^e (1/x) dx =2. E) 1. ∫₁^e (1/x) dx = [ln|x|]₁^e = ln e − ln 1 = 1 − 0 = 1. Distractors: A) e − 1 used x as the antiderivative instead of ln|x|. B) 0 assumed the endpoints cancel. C) ln(e − 1) applied the log to the difference of the limits. D) e reported F(e) without subtracting. Fix: the antiderivative of 1/x is ln|x|; evaluate the log at each limit separately.
g(x) = ∫₂ˣ (t³ + 1) dt, then g′(x) =3. B) x³ + 1. By FTC Part 1 the derivative of ∫₂ˣ (t³+1) dt is the integrand evaluated at the upper limit x. The constant lower limit 2 plays no role. Distractors: A) integrated instead of applying FTC Part 1. C) 3x² differentiated the integrand. D) dropped the +1. E) x³ − 8 subtracted the lower-limit value — that is evaluating (Part 2), not differentiating. Fix: d/dx ∫ₐˣ f(t) dt = f(x); just substitute the upper limit into the integrand.
d/dx ∫₀^{x²} sin t dt =4. C) 2x sin(x²). Upper limit u = x², so u′ = 2x; FTC Part 1 with the chain rule gives sin(x²)·2x. Distractors: A) forgot the chain-rule factor 2x. B) used the antiderivative cos rather than the integrand sin. D) combined both slips. E) 2x sin x failed to substitute x² into the integrand. Fix: substitute the upper limit into the integrand, then multiply by the derivative of that limit.
∫₁⁴ (1/√x) dx =5. D) 2. ∫₁⁴ x^{−1/2} dx = [2√x]₁⁴ = 2·2 − 2·1 = 4 − 2 = 2. Distractors: A) 1 used the antiderivative √x, missing the factor 2. B) 4 reported 2√4 without subtracting F(1). C) 3 came from an arithmetic slip on the constant. E) 1/2 differentiated instead of integrating. Fix: rewrite 1/√x as x^{−1/2}; its antiderivative is 2x^{1/2}.
d/dx ∫ₓ³ cos t dt =6. C) −cos x. The lower limit varies: ∫ₓ³ cos t dt = −∫₃ˣ cos t dt, so the derivative is −cos x. Distractors: A) cos x ignored the sign flip from the variable lower limit. B) sin 3 − sin x evaluated the integral (Part 2) instead of differentiating. D) −sin x used the antiderivative sin rather than the integrand cos. E) cos 3 − cos x again evaluated instead of differentiated. Fix: a variable lower limit means flip the limits first, which introduces a minus sign.
Questions 7, 10, and 11 refer to the graph below.
[GRAPH: y = f(t) on [0, 6] × [−3, 3] - Upper semicircle of radius 2 centered at (2, 0) on [0, 4], lying ABOVE the axis (f ≥ 0), peak (2, 2); half-disk area = 2π - Line segment from (4, 0) down to (6, −2), lying BELOW the axis (f < 0); triangular region area = 2 - t-intercepts at t = 0 and t = 4]
Let g(x) = ∫₀ˣ f(t) dt.
g(6) =7. C) 2π − 2. On [0, 4], f ≥ 0 contributes the half-disk +2π; on [4, 6], f < 0 contributes the triangle −2. So g(6) = 2π − 2. Distractors: A) 2π stopped at g(4), ignoring the below-axis triangle. B) 2π + 2 added the triangle as positive, missing the sign. D) 2 − 2π reversed both signs. E) 2π − 4 doubled the triangle's area. Fix: below-axis area is subtracted; the definite integral is signed area.
d/dx ∫₁^{sin x} eᵗ dt =8. E) e^{sin x} cos x. Upper limit u = sin x, u′ = cos x; FTC Part 1 with the chain rule gives e^{sin x}·cos x. Distractors: A) forgot the chain-rule factor cos x. B) evaluated the integral (Part 2) instead of differentiating. C) cos x kept only the chain factor. D) e^{cos x} sin x swapped sin and cos. Fix: substitute the upper limit into eᵗ, then multiply by the derivative of the upper limit.
∫₀² e^{−x²} dx is closest to9. A) 0.882. Numerically, fnInt(e^(−X²), X, 0, 2) ≈ 0.882. Distractors: B) 1.000 is a rough area guess. C) 7.389 is e², unrelated. D) 0.135 is e^{−2}, the integrand near x = 2, not the integral. E) 2.000 confuses the interval width with the area. Fix: e^{−x²} has no elementary antiderivative — use numerical integration; the answer is well under 1.
g increasing?10. B) (0, 4). g′ = f, and f > 0 on (0, 4) (the semicircle sits above the axis), so g increases there. Distractors: A) (0, 2) used only where f is rising, confusing g increasing with f increasing. C) (4, 6) is where f < 0, so g decreases. D) (0, 6) includes the decreasing stretch. E) (2, 6) mixes both. Fix: g increases exactly where its derivative f is positive — the whole region above the axis.
[0, 6], g attains its absolute maximum at11. B) x = 4. f changes from positive to negative at x = 4, so g switches from rising to falling — a maximum. Comparing values, g(4) = 2π ≈ 6.28 exceeds g(0) = 0 and g(6) = 2π − 2 ≈ 4.28. Distractors: A) x = 2 is where f peaks, not where f crosses zero. C) x = 6 and D) x = 0 are endpoints with smaller g-values. E) x = 5 is an arbitrary interior point. Fix: g has its maximum where f crosses from + to −, i.e. at a zero of f, not at a peak of f.
g(x) = ∫₀ˣ f(t) dt, where f is continuous and f(x) > 0 for every x. Which of the following statements CANNOT be true?12. A) g has a relative maximum at x = 3. Since g′(x) = f(x) > 0 everywhere, g is strictly increasing and can have no relative maximum, so this statement cannot be true. Distractors: B) g(5) > g(2) must be true because g is increasing. C) concavity depends on f′, which can be positive somewhere, so concave-up is possible. D) g(0) = ∫₀⁰ f = 0 is always true. E) g increasing everywhere follows directly from g′ = f > 0. Fix: g′ = f; if f is always positive, g climbs forever and has no interior maximum.
1. D) 10. With F(x) = x³ − 2x², FTC Part 2 gives F(3) − F(1) = 9 − (−1) = 10. Distractors: A) 26 integrated only the 3x² term ([x³]₁³ = 26), dropping −4x. B) 9 is F(3) alone — the lower limit was never subtracted. C) −10 reversed the order to F(1) − F(3). E) 20 comes from an antiderivative slip. Fix: always compute F(b) − F(a), upper value first, and antidifferentiate every term.
2. E) 1. ∫₁^e (1/x) dx = [ln|x|]₁^e = ln e − ln 1 = 1 − 0 = 1. Distractors: A) e − 1 used x as the antiderivative instead of ln|x|. B) 0 assumed the endpoints cancel. C) ln(e − 1) applied the log to the difference of the limits. D) e reported F(e) without subtracting. Fix: the antiderivative of 1/x is ln|x|; evaluate the log at each limit separately.
3. B) x³ + 1. By FTC Part 1 the derivative of ∫₂ˣ (t³+1) dt is the integrand evaluated at the upper limit x. The constant lower limit 2 plays no role. Distractors: A) integrated instead of applying FTC Part 1. C) 3x² differentiated the integrand. D) dropped the +1. E) x³ − 8 subtracted the lower-limit value — that is evaluating (Part 2), not differentiating. Fix: d/dx ∫ₐˣ f(t) dt = f(x); just substitute the upper limit into the integrand.
4. C) 2x sin(x²). Upper limit u = x², so u′ = 2x; FTC Part 1 with the chain rule gives sin(x²)·2x. Distractors: A) forgot the chain-rule factor 2x. B) used the antiderivative cos rather than the integrand sin. D) combined both slips. E) 2x sin x failed to substitute x² into the integrand. Fix: substitute the upper limit into the integrand, then multiply by the derivative of that limit.
5. D) 2. ∫₁⁴ x^{−1/2} dx = [2√x]₁⁴ = 2·2 − 2·1 = 4 − 2 = 2. Distractors: A) 1 used the antiderivative √x, missing the factor 2. B) 4 reported 2√4 without subtracting F(1). C) 3 came from an arithmetic slip on the constant. E) 1/2 differentiated instead of integrating. Fix: rewrite 1/√x as x^{−1/2}; its antiderivative is 2x^{1/2}.
6. C) −cos x. The lower limit varies: ∫ₓ³ cos t dt = −∫₃ˣ cos t dt, so the derivative is −cos x. Distractors: A) cos x ignored the sign flip from the variable lower limit. B) sin 3 − sin x evaluated the integral (Part 2) instead of differentiating. D) −sin x used the antiderivative sin rather than the integrand cos. E) cos 3 − cos x again evaluated instead of differentiated. Fix: a variable lower limit means flip the limits first, which introduces a minus sign.
7. C) 2π − 2. On [0, 4], f ≥ 0 contributes the half-disk +2π; on [4, 6], f < 0 contributes the triangle −2. So g(6) = 2π − 2. Distractors: A) 2π stopped at g(4), ignoring the below-axis triangle. B) 2π + 2 added the triangle as positive, missing the sign. D) 2 − 2π reversed both signs. E) 2π − 4 doubled the triangle's area. Fix: below-axis area is subtracted; the definite integral is signed area.
8. E) e^{sin x} cos x. Upper limit u = sin x, u′ = cos x; FTC Part 1 with the chain rule gives e^{sin x}·cos x. Distractors: A) forgot the chain-rule factor cos x. B) evaluated the integral (Part 2) instead of differentiating. C) cos x kept only the chain factor. D) e^{cos x} sin x swapped sin and cos. Fix: substitute the upper limit into eᵗ, then multiply by the derivative of the upper limit.
9. A) 0.882. Numerically, fnInt(e^(−X²), X, 0, 2) ≈ 0.882. Distractors: B) 1.000 is a rough area guess. C) 7.389 is e², unrelated. D) 0.135 is e^{−2}, the integrand near x = 2, not the integral. E) 2.000 confuses the interval width with the area. Fix: e^{−x²} has no elementary antiderivative — use numerical integration; the answer is well under 1.
10. B) (0, 4). g′ = f, and f > 0 on (0, 4) (the semicircle sits above the axis), so g increases there. Distractors: A) (0, 2) used only where f is rising, confusing g increasing with f increasing. C) (4, 6) is where f < 0, so g decreases. D) (0, 6) includes the decreasing stretch. E) (2, 6) mixes both. Fix: g increases exactly where its derivative f is positive — the whole region above the axis.
11. B) x = 4. f changes from positive to negative at x = 4, so g switches from rising to falling — a maximum. Comparing values, g(4) = 2π ≈ 6.28 exceeds g(0) = 0 and g(6) = 2π − 2 ≈ 4.28. Distractors: A) x = 2 is where f peaks, not where f crosses zero. C) x = 6 and D) x = 0 are endpoints with smaller g-values. E) x = 5 is an arbitrary interior point. Fix: g has its maximum where f crosses from + to −, i.e. at a zero of f, not at a peak of f.
12. A) g has a relative maximum at x = 3. Since g′(x) = f(x) > 0 everywhere, g is strictly increasing and can have no relative maximum, so this statement cannot be true. Distractors: B) g(5) > g(2) must be true because g is increasing. C) concavity depends on f′, which can be positive somewhere, so concave-up is possible. D) g(0) = ∫₀⁰ f = 0 is always true. E) g increasing everywhere follows directly from g′ = f > 0. Fix: g′ = f; if f is always positive, g climbs forever and has no interior maximum.