The Fundamental Theorem of Calculus
State and prove both parts of the FTC. Part 1: d/dx[∫ₐˣ f(t)dt] = f(x). Part 2: ∫ₐᵇ f(x)dx = F(b)−F(a). Understand why the FTC is the central theorem connecting differentiation and integration.
A lesson of Calculus, in Integration and the Fundamental Theorem, about 10 minutes.
By the end of this lesson, you will be able to state and apply both parts of the Fundamental Theorem of Calculus, evaluate definite integrals using antiderivatives, and differentiate accumulation functions.
The Most Important Theorem in Calculus
The Fundamental Theorem of Calculus (FTC) connects the two central operations of calculus — differentiation and integration — showing they are inverse processes. This unity was discovered independently by Newton and Leibniz in the 17th century and transformed mathematics forever.
FTC Part 1: The Derivative of an Accumulation Function
If $f$ is continuous on $[a,b]$, then the function $G(x) = \int_a^x f(t),dt$ is differentiable on $(a,b)$ and:
$$G'(x) = \frac{d}{dx}\left[\int_a^x f(t),dt\right] = f(x)$$
Meaning: the derivative of an area function is the integrand. Integration and differentiation are inverse operations.
FTC Part 2: Evaluating Definite Integrals
If $F$ is any antiderivative of $f$ on $[a,b]$, then:
$$\int_a^b f(x),dx = F(b) - F(a)$$
Notation: $\left[F(x)\right]_a^b = F(b) - F(a)$
This transforms definite integrals from limits of Riemann sums into a simple computation.
Example 1 — Using FTC Part 2
$$\int_1^4 \sqrt{x},dx = \left[\frac{2x^{3/2}}{3}\right]_1^4 = \frac{2(8)}{3} - \frac{2(1)}{3} = \frac{16}{3} - \frac{2}{3} = \frac{14}{3}$$
Example 2 — Trigonometric
$$\int_0^{\pi/2}\cos x,dx = [\sin x]_0^{\pi/2} = 1 - 0 = 1$$
Example 3 — Using FTC Part 1
$G(x) = \int_2^x (t^3+1),dt$. Find $G'(x)$.
By FTC Part 1: $G'(x) = x^3 + 1$ (the integrand evaluated at $x$).
Example 4 — Chain Rule with FTC Part 1
$H(x) = \int_1^{x^2}\sin t,dt$. Find $H'(x)$.
Let $u = x^2$. $H = \int_1^u\sin t,dt$, so $\dfrac{dH}{du} = \sin u$ (FTC Part 1).
By chain rule: $H'(x) = \sin(x^2) \cdot 2x$
Example 5 — FTC with Both Bounds Varying
$K(x) = \int_{x}^{x^2}e^t,dt$. Find $K'(x)$.
$K(x) = \int_c^{x^2}e^t,dt - \int_c^x e^t,dt$ for any constant $c$.
$K'(x) = e^{x^2} \cdot 2x - e^x$
Why Part 2 Follows from Part 1
Let $G(x) = \int_a^x f(t),dt$, so $G'(x) = f(x)$ (Part 1).
Any antiderivative $F$ satisfies $F(x) = G(x) + C$.
$\int_a^b f(x),dx = G(b) - G(a) = [F(b) - C] - [F(a) - C] = F(b) - F(a)$. ✓
Summary
FTC Part 1: $\dfrac{d}{dx}\int_a^x f(t),dt = f(x)$ — the derivative of an integral with variable upper limit is the integrand.