Calculus
The Fundamental Theorem of Calculus
Цели урока
- Understand the deep connection between derivatives and integrals
- Master the Newton-Leibniz formula for evaluating integrals
- Apply FTC I to differentiate integrals
- Use substitution in definite integrals with updated limits
- Work with the generalized formula for variable limits
Предварительные знания
- Antiderivatives and the indefinite integral
- The definite integral
- The chain rule
October 1666, Woolsthorpe. Newton writes in his "October tract": if $A(x)$ is the area under the curve from 0 to $x$, then $A'(x) = f(x)$. In parallel, his teacher Isaac Barrow had already shown geometrically in his 1664-66 Cambridge lectures: the tangent to the area graph has slope equal to the height of the original curve. Barrow did not see a universal theorem. Newton did, and tied area, distance, work, and charge into one formula: $\int_a^b f = F(b) - F(a)$.
- **GPS receiver**: integrates IMU accelerations between satellite fixes. FTC turns $a(t)$ into velocity and position between updates that arrive once per second
- **Electricity meter**: $E = \int_0^T P(t)\,dt = F(T) - F(0)$. The current meter reading is literally the antiderivative of power. FTC in hardware
- **Wolfram Alpha / Mathematica**: Integrate[f, {x, a, b}] first searches for an antiderivative via Risch, then applies $F(b) - F(a)$. Without FTC, numerical quadrature would be the only option
- **Heat equation in COMSOL**: $\int_\Omega \nabla \cdot q\,dV = \oint_{\partial\Omega} q \cdot dS$ is the vector-valued FTC (divergence theorem). The basis of any CFD simulation
Newton vs Leibniz: The Dispute of the Century
Newton worked out this connection around 1665-1666, during the plague years, but didn't publish it; his Method of Fluxions appeared much later. Leibniz reached the same results independently and got into print first, in the 1684 issue of Acta Eruditorum. The two camps fought one of the most famous priority disputes in the history of science. Today both are credited with the discovery, but Leibniz's notation ($\int$, $dx$) won out and is still in use. The theorem is the payoff of two thousand years of work, going back to Archimedes.
The Central Idea
The Central Idea
Scene: a car is driving down the highway. The speedometer shows velocity $v(t)$ at every instant; the odometer shows the distance traveled $s(t)$. How are the two related?
- **Velocity** is the derivative of position: $v(t) = s'(t)$
- **Position** is the integral of velocity: $s(t) = \int_0^t v(\tau)\,d\tau$
That's the Fundamental Theorem in a nutshell: **differentiation and integration are inverses of each other**.
**The Fundamental Theorem of Calculus (FTC)** connects the two central operations of calculus - finding tangent lines (differentiation) and computing areas (integration).
What does the Fundamental Theorem of Calculus connect?
FTC ties the two pillars of calculus: derivative undoes integral and vice versa, turning area computations into antiderivative evaluations.
FTC I: Derivative of an Integral
FTC I: Derivative of an Integral
Let $f$ be continuous on $[a, b]$. Define the function:
This is the "accumulated area" from $a$ to $x$. The theorem states:
**In words**: the derivative of an integral (with respect to the upper limit) equals the integrand evaluated at that point.
**Geometric meaning**: the rate at which area accumulates at $x$ equals the height $f(x)$ at that point.
FTC I in Action
Differentiating an integral with a variable upper limit
$\frac{d}{dx}\int_0^x t^3\,dt$ By FTC I, the answer is simply $f(x) = x^3$! **Verification** (the long way): $\int_0^x t^3\,dt = \frac{t^4}{4}\Big|_0^x = \frac{x^4}{4}$ $\frac{d}{dx}\left(\frac{x^4}{4}\right) = x^3$ ✓
What is $\frac{d}{dx}\int_1^x \cos t\,dt$?
By FTC I: the derivative of an integral with respect to the upper limit equals the integrand evaluated at $x$.
FTC II: The Newton-Leibniz Formula
FTC II: The Newton-Leibniz Formula
If $F$ is an antiderivative of a continuous function $f$ on $[a, b]$, then:
**Why it's a big deal**: instead of computing a tricky limit of Riemann sums, the recipe is simple - find an antiderivative and plug in the endpoints.
The notation $F(x)\Big|_a^b$ is read as "$F$ of $x$ evaluated from $a$ to $b$" and means $F(b) - F(a)$.
Algorithm for Evaluating a Definite Integral
- Find an antiderivative $F(x)$ of $f(x)$
- Compute $F(b)$
- Compute $F(a)$
- Answer: $F(b) - F(a)$
Area Under a Parabola
Classic example: ∫₀¹ x² dx
$\int_0^1 x^2\,dx$ **Step 1**: Antiderivative $F(x) = \frac{x^3}{3}$ **Step 2**: $F(1) = \frac{1^3}{3} = \frac{1}{3}$ **Step 3**: $F(0) = \frac{0^3}{3} = 0$ **Step 4**: $\int_0^1 x^2\,dx = \frac{1}{3} - 0 = \frac{1}{3}$ The area under the parabola $y = x^2$ from 0 to 1 is $\frac{1}{3}$!
Area Under a Sine Curve
Integral ∫₀^π sin x dx
$\int_0^{\pi} \sin x\,dx$ **Antiderivative**: $F(x) = -\cos x$ $= -\cos x\Big|_0^{\pi} = -\cos\pi - (-\cos 0)$ $= -(-1) - (-1) = 1 + 1 = 2$ **Answer**: the area under one arch of the sine curve is 2.
Exponential Integral
∫₀¹ eˣ dx
$\int_0^1 e^x\,dx$ **Antiderivative**: $F(x) = e^x$ $= e^x\Big|_0^1 = e^1 - e^0 = e - 1 \approx 1.718$
The +C must be added when evaluating a definite integral
The constant +C cancels out in the subtraction: (F(b) + C) - (F(a) + C) = F(b) - F(a)
A definite integral is a specific number, not a family of functions. Any antiderivative gives the same answer.
What is $\int_1^2 \frac{1}{x}\,dx$?
The antiderivative of $\frac{1}{x}$ is $\ln|x|$. So $\int_1^2 \frac{1}{x}\,dx = \ln 2 - \ln 1 = \ln 2 - 0 = \ln 2$.
Generalization: Variable Limits
Generalization: Variable Limits
What if both limits depend on $x$? Apply the chain rule:
**Mnemonic**: "upper times derivative of upper, minus lower times derivative of lower"
Upper Limit Is a Function
d/dx ∫₀^x² sin t dt
$\frac{d}{dx}\int_0^{x^2} \sin t\,dt$ Here $a(x) = 0$, $b(x) = x^2$, $f(t) = \sin t$ $a'(x) = 0$, $b'(x) = 2x$ **By the formula**: $= \sin(x^2) \cdot 2x - \sin(0) \cdot 0$ $= 2x\sin(x^2)$
Both Limits Are Functions
d/dx ∫ₓ^x³ eᵗ dt
$\frac{d}{dx}\int_x^{x^3} e^t\,dt$ $a(x) = x$, $b(x) = x^3$, $f(t) = e^t$ $a'(x) = 1$, $b'(x) = 3x^2$ **By the formula**: $= e^{x^3} \cdot 3x^2 - e^x \cdot 1$ $= 3x^2 e^{x^3} - e^x$
What is $\frac{d}{dx}\int_0^{x^3} t^2\,dt$?
By the generalized formula: $f(b(x)) \cdot b'(x) = (x^3)^2 \cdot 3x^2 = x^6 \cdot 3x^2 = 3x^8$.
Substitution in Definite Integrals
Substitution in Definite Integrals
When making the substitution $u = g(x)$ in a definite integral, **the limits change**:
**Advantage**: no need to convert back to the original variable! We compute directly with the new limits.
Substitution with New Limits
∫₀¹ 2x·eˣ² dx
$\int_0^1 2x \cdot e^{x^2}\,dx$ **Substitution**: $u = x^2$, $du = 2x\,dx$ **New limits**: • When $x = 0$: $u = 0^2 = 0$ • When $x = 1$: $u = 1^2 = 1$ $= \int_0^1 e^u\,du = e^u\Big|_0^1 = e^1 - e^0 = e - 1$ **No need** to substitute $u = x^2$ back!
Trigonometric Substitution
∫₀^(π/2) cos x · sin²x dx
$\int_0^{\pi/2} \cos x \cdot \sin^2 x\,dx$ **Substitution**: $u = \sin x$, $du = \cos x\,dx$ **New limits**: • When $x = 0$: $u = \sin 0 = 0$ • When $x = \frac{\pi}{2}$: $u = \sin\frac{\pi}{2} = 1$ $= \int_0^1 u^2\,du = \frac{u^3}{3}\Big|_0^1 = \frac{1}{3}$
**Important**: when making a substitution, ALWAYS update the limits of integration! Forgetting to do so will give a wrong answer.
When substituting $u = x^2$ in $\int_1^3 f(x^2) \cdot 2x\,dx$, what are the new limits?
When $x = 1$: $u = 1^2 = 1$. When $x = 3$: $u = 3^2 = 9$. New limits: from 1 to 9.
The Unity of the Theorem
The Unity of the Theorem
The two parts of the FTC are two sides of the same coin:
| FTC I | FTC II |
|---|---|
| $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$ | $\int_a^b F'(x)\,dx = F(b) - F(a)$ |
| Differentiation "undoes" integration | Integration "undoes" differentiation |
| Differentiating after integrating | Integrating after differentiating |
| Yields a function | Yields a number |
Part 1 of the FTC states that $\frac{d}{dx}\int_a^x f(t)\,dt$ equals:
Differentiating the integral with variable upper limit recovers the integrand at $x$ (assuming $f$ continuous).
The Foundation of All Calculus
The FTC is the foundation for everything that follows:
- Integration Techniques — Substitution, integration by parts - next lesson
- Improper Integrals — Extension to infinite limits
- Differential Equations — Solving by integration
- Measure Theory — Generalization from the Riemann to the Lebesgue integral
Итоги
- **FTC I**: $\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$ - differentiation "undoes" integration
- **FTC II**: $\int_a^b f(x)\,dx = F(b) - F(a)$ - the Newton-Leibniz formula
- **Core idea**: differentiation and integration are inverse operations
- **Variable limits**: $f(b(x))b'(x) - f(a(x))a'(x)$
- **Substitution**: with $u = g(x)$, new limits are $g(a)$ and $g(b)$
Вопросы для размышления
- Why is the FTC considered one of the most important theorems in the history of mathematics?
- What is the geometric reason that the derivative of the area equals the height?
- Why is the constant +C not needed when evaluating a definite integral?
- What is the advantage of substitution with updated limits?