The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) is the ultimate bridge in calculus. It connects the two main operations you've learned: differentiation (finding the rate of change) and integration (finding the accumulated area). The theorem is split into two parts, each showing that derivatives and integrals are essentially inverse operations.

Part 1: The Derivative of an Integral

The First Fundamental Theorem of Calculus (FTC 1) tells us that if you integrate a continuous function and then take the derivative of that integral, you get the original function back.

Mathematically, if $f$ is a continuous function on $[a, b]$, and you define a new function $g(x)$ by an integral: $g(x) = \int_a^x f(t) \, dt$

Then the derivative of $g(x)$ is simply $f(x)$: $g'(x) = \frac{d}{dx} \int_a^x f(t) \, dt = f(x)$

Example 1

Problem: Find $\frac{d}{dx} \int_1^x (t^3 + 1) \, dt$.

Solution: Using FTC Part 1, we don't even need to evaluate the integral first. Because we are taking the derivative with respect to $x$ of an integral that goes from a constant to $x$, we just replace the dummy variable $t$ with $x$.

$\frac{d}{dx} \int_1^x (t^3 + 1) \, dt = x^3 + 1$

Part 2: Evaluating Definite Integrals

The Second Fundamental Theorem of Calculus (FTC 2) provides a highly practical way to evaluate definite integrals. Instead of calculating complex limits of Riemann sums, you can just use an antiderivative.

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ (meaning $F'(x) = f(x)$), then: $\int_a^b f(x) \, dx = F(b) - F(a)$

Example 2

Problem: Evaluate $\int_0^\pi \sin x \, dx$ using the Fundamental Theorem of Calculus.

Solution:

1. Find an antiderivative: We need a function $F(x)$ whose derivative is $\sin x$. We know that the derivative of $\cos x$ is $-\sin x$, so the antiderivative is $F(x) = -\cos x$.
2. Apply FTC Part 2: Evaluate $F(x)$ at the upper limit ($\pi$) and subtract $F(x)$ evaluated at the lower limit ($0$).

$\int_0^\pi \sin x \, dx = [-\cos x]_0^\pi$ $= (-\cos(\pi)) - (-\cos(0))$

3. Simplify: We know $\cos(\pi) = -1$ and $\cos(0) = 1$. $= (-(-1)) - (-1)$ $= 1 + 1 = 2$

The definite integral evaluates to $2$.

Summary

• FTC Part 1 proves that integration and differentiation reverse each other: $\frac{d}{dx} \int_a^x f(t) \, dt = f(x)$.
• FTC Part 2 gives us the shortcut to compute areas under curves: to evaluate $\int_a^b f(x) \, dx$, find an antiderivative $F(x)$ and calculate $F(b) - F(a)$.