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Lesson 83 — Fundamental Theorem of Calculus

Fundamental Theorem of Calculus Part 1 and Part 2. The bridge between derivative and integral. Leibniz rule for variable limits. Newton and Leibniz, 17th century.

Used in: 3rd year of high school (17 years old) · Equiv. Math II Japanese ch. 6 · Equiv. Grade 12 German

\int_a^b f(x)\, dx = F(b) - F(a), \quad F'(x) = f(x)

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Rigorous notation, full derivation, hypotheses

Statement and proofs

FTC — Part 1: differentiate the integral

"FTC Part 1 states that the derivative of the function defined by an integral with a variable upper limit is equal to the integrand evaluated at the upper limit." — OpenStax Calculus Vol. 1, §5.3

Proof of FTC Part 1. By definition of derivative:

$G'(x) = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t)\, dt.$

By the Mean Value Theorem for Integrals, there exists $c_h$ between $x$ and $x+h$ such that $\int_x^{x+h} f(t)\, dt = f(c_h) \cdot h$ . Therefore:

$G'(x) = \lim_{h \to 0} f(c_h).$

Since $c_h \to x$ as $h \to 0$ and $f$ is continuous, $f(c_h) \to f(x)$ . Therefore $G'(x) = f(x)$ . $\square$

FTC — Part 2: calculate the integral

Proof of FTC Part 2. By FTC Part 1, $G(x) = \int_a^x f(t)\, dt$ satisfies $G' = f$ . Since $F' = f$ as well, $F - G$ has zero derivative on $(a, b)$ , so $F(x) = G(x) + C$ for some constant $C$ . Then:

$F(b) - F(a) = [G(b) + C] - [G(a) + C] = G(b) - G(a) = \int_a^b f - 0 = \int_a^b f. \quad \square$

Leibniz rule (variable limits)

Solved examples

Example— 1· FTC2 applied to polynomial (basic)

Problem. Calculate $\displaystyle\int_1^3 (x^2 + 2x - 1)\, dx$ .

Strategy. Find antiderivative by FTC Part 2: $F(x)$ with $F'(x) = x^2 + 2x - 1$ . Then $F(3) - F(1)$ .

Solution. $F(x) = \frac{x^3}{3} + x^2 - x.$

$\int_1^3 (x^2 + 2x - 1)\, dx = F(3) - F(1) = \left(9 + 9 - 3\right) - \left(\frac{1}{3} + 1 - 1\right) = 15 - \frac{1}{3} = \frac{44}{3}.$

Verification. $F'(x) = x^2 + 2x - 1$ . Check.

Source. APEX Calculus, §5.4, Example 5.4.2 — CC-BY-NC.

Example— 3· FTC1 to differentiate integral (intermediate)

Problem. Calculate $\displaystyle\frac{d}{dx}\int_1^{x^3} \cos(t^2)\, dt$ .

Strategy. FTC Part 1 combined with the chain rule: the upper limit is $v(x) = x^3$ .

Solution. $\frac{d}{dx}\int_1^{x^3} \cos(t^2)\, dt = \cos((x^3)^2) \cdot (x^3)' = \cos(x^6) \cdot 3x^2.$

Verification. The general Leibniz rule: $\frac{d}{dx}\int_a^{v(x)} f(t)\, dt = f(v(x)) \cdot v'(x)$ . Applied with $f(t) = \cos(t^2)$ , $v(x) = x^3$ : $\cos(x^6) \cdot 3x^2$ . Check.

Source. APEX Calculus, §5.4, Example 5.4.5 — CC-BY-NC.

Example— 4· Leibniz rule with both limits variable (intermediate)

Problem. Calculate $\displaystyle\frac{d}{dx}\int_x^{x^2} \frac{1}{t}\, dt$ for $x > 0$ .

Strategy. General Leibniz rule: $f(v(x))v'(x) - f(u(x))u'(x)$ with $v(x) = x^2$ , $u(x) = x$ , $f(t) = 1/t$ .

Solution. $\frac{d}{dx}\int_x^{x^2} \frac{1}{t}\, dt = \frac{1}{x^2} \cdot 2x - \frac{1}{x} \cdot 1 = \frac{2}{x} - \frac{1}{x} = \frac{1}{x}.$

Verification. Alternatively: $\int_x^{x^2} \frac{1}{t}\, dt = \ln(x^2) - \ln x = 2\ln x - \ln x = \ln x$ . Differentiating: $(\ln x)' = 1/x$ . Check.

Source. OpenStax Calculus Vol. 1, §5.3, Exercise 5.132 — CC-BY-NC-SA.

Example— 5· Area between curve and axis with sign change (challenge)

Problem. Calculate the total geometric area between $y = x^2 - 4$ and the $x$ -axis on $[-3, 3]$ .

Strategy. The function $f(x) = x^2 - 4$ is negative on $(-2, 2)$ and positive elsewhere. Split the interval at zeros to sum areas with correct signs.

Solution. Zeros: $x^2 = 4 \Rightarrow x = \pm 2$ .

On $[-3, -2]$ and $[2, 3]$ : $f \geq 0$ . On $[-2, 2]$ : $f \leq 0$ .

$A = \int_{-3}^{-2}(x^2-4)\, dx + \left|\int_{-2}^{2}(x^2-4)\, dx\right| + \int_2^3(x^2-4)\, dx.$

$F(x) = x^3/3 - 4x$ .

$F(-2) - F(-3) = (-8/3 + 8) - (-9 + 12) = 16/3 - 3 = 7/3$ .
$F(2) - F(-2) = (8/3 - 8) - (-8/3 + 8) = -32/3$ . Absolute value: $32/3$ .
$F(3) - F(2) = (9 - 12) - (8/3 - 8) = -3 + 16/3 = 7/3$ .

Total area: $7/3 + 32/3 + 7/3 = 46/3$ .

Verification. The integral $\int_{-3}^3 (x^2 - 4)\, dx = F(3) - F(-3) = (9-12) - (-9+12) = -3 - 3 = -6 \neq 46/3$ . The definite integral differs from geometric area when $f$ changes sign.

Source. Active Calculus, §4.4, Exercise 4.4.5 — CC-BY-NC-SA.

Exercise list

30 exercises · 7 with worked solution (25%)

Application 20Understanding 2Modeling 3Challenge 4Proof 1

Ex. 83.1Application
Calculate $\int_0^3 2x\, dx$ by FTC Part 2.
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Ex. 83.2Application
Calculate $\int_{-1}^2 x^3\, dx$ .
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Ex. 83.3Application
Calculate $\int_0^\pi \sin x\, dx$ .
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Ex. 83.4Application
Calculate $\int_0^2 e^x\, dx$ .
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Ex. 83.5Application
Calculate $\int_0^2 (x^2 - 4x + 1)\, dx$ .
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Ex. 83.6Application
Calculate $\int_1^e \frac{1}{x}\, dx$ .
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Ex. 83.7Application
If $G(x) = \int_0^x (t^2 + 1)\, dt$ , calculate $G'(x)$ by FTC Part 1.
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Ex. 83.8Application
Calculate $\dfrac{d}{dx}\displaystyle\int_0^{x^2} \sin t\, dt$ .
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Ex. 83.9Application
Calculate $\int_0^{\pi/4} \sec^2 x\, dx$ .
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Ex. 83.10Application
Calculate $\int_1^9 \sqrt{x}\, dx$ .
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Ex. 83.11Application
Calculate $\dfrac{d}{dx}\displaystyle\int_0^{x^3} \sqrt{1 + t^2}\, dt$ .
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Ex. 83.12Application
Calculate $\int_0^\pi (\cos x + \sin x)\, dx$ .
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Ex. 83.13Application
Calculate $\int_0^1 (x^4 - 2x^2 + 1)\, dx$ .
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Ex. 83.14UnderstandingAnswer key
If $G(x) = \displaystyle\int_a^x f(t)\, dt$ , what is $G'(x)$ by FTC Part 1?
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Ex. 83.15Understanding
If $F'(x) = f(x)$ , what is the correct expression for $\int_a^b f(x)\, dx$ by FTC Part 2?
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Ex. 83.16ApplicationAnswer key
Calculate $\dfrac{d}{dx}\displaystyle\int_x^1 t^3\, dt$ .
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Ex. 83.17Application
Calculate $\int_0^1 \frac{1}{1+x^2}\, dx$ .
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Ex. 83.18ModelingAnswer key
An object has velocity $v(t) = t^2 - 4t + 3$ m/s. Calculate the net displacement and total distance traveled from $t = 0$ to $t = 4$ s.
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Ex. 83.19ApplicationAnswer key
Calculate $\dfrac{d}{dx}\displaystyle\int_x^{x^2} e^{t^2}\, dt$ .
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Ex. 83.20Application
Calculate $\int_0^2 (2x^3 - 3x^2)\, dx$ .
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Ex. 83.21Modeling
The marginal cost of production at a factory is $C'(q) = 2q + 50$ reais per unit. Calculate the total cost of producing the first 100 units.
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Ex. 83.22ChallengeAnswer key
Define $G(x) = \int_1^x (2t - 1)\, dt$ . Calculate $G(x)$ explicitly, verify that $G'(x) = 2x - 1$ , and evaluate $G(1)$ and $G(3)$ .
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Ex. 83.23ApplicationAnswer key
Given that $\int_0^5 f(x)\, dx = 12$ and $\int_0^2 f(x)\, dx = 5$ , calculate $\int_2^5 f(x)\, dx$ .
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Ex. 83.24Challenge
Calculate the area of the region bounded by $y = x^2 - x$ and the $x$ -axis on $[0, 1]$ .
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Ex. 83.25Application
Calculate $\int_{-2}^2 (x^2 - 1)\, dx$ .
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Ex. 83.26Application
Calculate $\dfrac{d}{dx}\displaystyle\int_0^x \cos(t^2)\, dt$ without computing the antiderivative.
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Ex. 83.27ModelingAnswer key
The electrical power at a factory varies as $P(t) = 3 + 0{,}5t$ kW ( $t$ in hours). Calculate the energy consumed in the first 12 hours of operation and the cost at the rate of R$ 0.85 per kWh.
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Ex. 83.28Challenge
Calculate $\dfrac{d}{dx}\displaystyle\int_{\sin x}^{\cos x} t^2\, dt$ .
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Ex. 83.29Challenge
Calculate the mean value of $f(x) = x^2$ on $[0, 3]$ and find the point $c$ guaranteed by the Mean Value Theorem for Integrals.
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Ex. 83.30Proof
Prove FTC Part 2 from FTC Part 1: if $F' = f$ and $f$ is continuous on $[a,b]$ , then $\int_a^b f(x)\, dx = F(b) - F(a)$ .

Sources

Active Calculus — Boelkins · §4.4 · CC-BY-NC-SA. Physical motivation, discovery activity for both parts of the FTC.
APEX Calculus — Hartman et al. · §5.4 · CC-BY-NC. Proofs of FTC1 and FTC2, Leibniz rule, varied exercises.
OpenStax Calculus Volume 1 · §5.3 · CC-BY-NC-SA. Historical context Newton/Leibniz, examples of differentiation of integrals.