# Lecture 20: The Fundamental Theorem of Calculus ## Calculus I — Differential Calculus --- ## Learning Objectives By the end of this lecture, students will be able to: 1. State and apply Part 1 of the Fundamental Theorem of Calculus (FTC1) 2. State and apply Part 2 of the Fundamental Theorem of Calculus (FTC2) 3. Evaluate definite integrals using antiderivatives 4. Differentiate integral functions using FTC1 5. Understand how the FTC connects differential and integral calculus --- ## Lecture Content ### I. The Big Picture Differential calculus (derivatives) and integral calculus (integrals and areas) appear at first to be entirely different subjects. The **Fundamental Theorem of Calculus** reveals that they are, in fact, inverse operations of each other. This is arguably the most important theorem in all of calculus. It has two parts: FTC1 shows that differentiation undoes integration, and FTC2 shows that integration undoes differentiation (with a net-change interpretation). ### II. FTC Part 1 — Differentiation of an Integral FTC Part 1 states that if f is continuous on [a, b], and we define the function g(x) = integral from a to x of f(t) dt, then g is continuous on [a, b], differentiable on (a, b), and g'(x) = f(x). In Leibniz notation, d/dx [integral from a to x of f(t) dt] = f(x). The meaning is profound: the derivative of the "area-so-far function" at x equals the integrand evaluated at x. The function g(x) is sometimes called the **accumulation function**, and it is itself an antiderivative of f. A key consequence is that FTC1 guarantees every continuous function has an antiderivative. ### III. FTC1 with the Chain Rule When the upper limit is a function of x rather than x itself, the chain rule enters: d/dx [integral from a to u(x) of f(t) dt] = f(u(x)) * u'(x). If both limits are functions of x, the result is d/dx [integral from v(x) to u(x) of f(t) dt] = f(u(x)) * u'(x) - f(v(x)) * v'(x). For example, d/dx [integral from 0 to x^2 of sin(t) dt] = sin(x^2) * 2x = 2x sin(x^2). As a more involved example, d/dx [integral from x to x^3 of e^{t^2} dt] = e^{(x^3)^2} * 3x^2 - e^{x^2} * 1 = 3x^2 e^{x^6} - e^{x^2}. A graph illustrating FTC Part 1. Panel A shows a continuous function f(t) with the area from a to x shaded, representing g(x) = integral from a to x of f(t) dt. A thin vertical strip of width dx at position x shows how adding a small slice changes the area by approximately f(x)*dx. Panel B shows g(x) plotted as a function, with its slope at a point x equal to f(x). An annotation connects the height of f at x to the slope of g at x. Title: "FTC1: The rate of change of accumulated area equals the integrand." ### IV. FTC Part 2 — The Evaluation Theorem FTC Part 2 states that if f is continuous on [a, b] and F is any antiderivative of f (meaning F' = f), then integral from a to b of f(x) dx = F(b) - F(a). The notation F(x) |_a^b = F(b) - F(a) (also written [F(x)]_a^b) is standard. This theorem provides a practical method to evaluate definite integrals without computing limits of Riemann sums. The choice of antiderivative does not matter because the constant C cancels: [F(b) + C] - [F(a) + C] = F(b) - F(a). ### V. Examples Using FTC2 To evaluate integral from 1 to 3 of x^2 dx, use the antiderivative F(x) = x^3/3. The result is 3^3/3 - 1^3/3 = 27/3 - 1/3 = 26/3. For integral from 0 to pi of sin x dx, the antiderivative is F(x) = -cos x. Evaluating gives -cos(pi) - (-cos(0)) = -(-1) + 1 = 2. For integral from 1 to e of (1/x) dx, the antiderivative is F(x) = ln x. The result is ln e - ln 1 = 1 - 0 = 1. For integral from 0 to 1 of (3x^2 - 2x + 1) dx, the antiderivative is x^3 - x^2 + x, and evaluating gives (1 - 1 + 1) - (0) = 1. ### VI. Net Change Interpretation FTC2 can be rewritten as integral from a to b of F'(x) dx = F(b) - F(a), which says that the integral of a rate of change gives the **net change** in the quantity over the interval. This interpretation has wide-ranging applications. If v(t) is velocity, then integral from a to b of v(t) dt = s(b) - s(a), the net displacement. If R(t) is a production rate, then the integral gives the total production from t = a to t = b. If I(t) is current, then the integral gives the total charge. An important distinction must be made: net displacement is not the same as total distance traveled. The total distance is integral from a to b of |v(t)| dt, which counts all movement as positive regardless of direction. A velocity-time graph where v(t) is positive on [a, c] and negative on [c, b]. The area above the t-axis on [a, c] is shaded blue and labeled "forward displacement." The area below the t-axis on [c, b] is shaded red and labeled "backward displacement." The net displacement (blue area minus red area) and total distance (blue area plus red area) are computed and displayed. Title: "Net displacement vs. total distance." ### VII. Summary: The FTC Connects Two Worlds Before the FTC, computing areas required laborious Riemann sum limits. The FTC reveals that to compute an integral, one need only find any antiderivative and subtract endpoint values. The two parts together encapsulate a powerful duality: FTC1 states d/dx [integral from a to x of f(t) dt] = f(x), showing that differentiation undoes integration; FTC2 states integral from a to b of f'(x) dx = f(b) - f(a), showing that integration undoes differentiation (up to boundary values). This duality is the central insight of calculus. A conceptual diagram showing the relationship between differentiation and integration as inverse processes. Two boxes labeled "f(x)" and "F(x)" are connected by two arrows: one pointing from F to f labeled "differentiate: F'(x) = f(x)" and one pointing from f to F labeled "integrate: integral of f(x) dx = F(x) + C." Below, FTC1 and FTC2 are stated in a banner connecting the two operations. Title: "The Fundamental Theorem of Calculus: differentiation and integration are inverse processes."