Definite Integral

The Definite Integral: A Concept-First Introduction

Earlier you met indefinite integrals — families of functions that differ by a constant. The definite integral instead has a single, unique numerical value; it is a number, not a family of functions.

The definite integral of a function $f(x)$ from $x = a$ to $x = b$ is written as:

Here, $a$ is the lower limit of integration and $b$ is the upper limit; the interval $[a, b]$ is the domain over which we integrate.

There are two fundamental ways to understand this symbol:

1. As a limit of a sum (the Riemann sum approach) — adding up infinitely many infinitely thin rectangles under a curve.
2. Using an antiderivative — if $F(x)$ is an antiderivative of $f(x)$ on $[a, b]$, then

This second result is the Fundamental Theorem of Calculus: to evaluate a definite integral we need not compute a limit of sums; we simply find an antiderivative and subtract.

---

Properties of the Definite Integral

Each property below is stated and then proved using the definition $\int_a^b f(x) \, dx = F(b) - F(a)$, where $F'(x) = f(x)$.

Property 1: Reversing the Limits

Proof. With $F$ an antiderivative of $f$:

Property 2: Zero Width Interval

Proof. $\int_a^a f(x) \, dx = F(a) - F(a) = 0$ — the area under a curve from a point to itself is zero.

Property 3: Constant Multiple

Proof. $kF$ is an antiderivative of $kf$ (since $\frac{d}{dx}[kF(x)] = kf(x)$), so $\int_a^b k f(x) \, dx = kF(b) - kF(a) = k[F(b) - F(a)] = k \int_a^b f(x) \, dx$.

Property 4: Sum/Difference

Proof. With $F$, $G$ antiderivatives of $f$, $g$, the function $F \pm G$ is an antiderivative of $f \pm g$:

Property 5: Splitting the Interval

Proof. $\int_a^c f(x) \, dx + \int_c^b f(x) \, dx = [F(c) - F(a)] + [F(b) - F(c)] = F(b) - F(a) = \int_a^b f(x) \, dx$.

Property 6: Inequality Preservation

If $f(x) \ge g(x)$ for all $x$ in $[a, b]$, then $\int_a^b f(x) \, dx \ge \int_a^b g(x) \, dx$.

Proof. Let $h(x) = f(x) - g(x) \ge 0$ on $[a, b]$, with antiderivative $H$. Since $h \ge 0$, $H$ is non-decreasing, so $\int_a^b h(x) \, dx = H(b) - H(a) \ge 0$; but this equals $\int_a^b f(x)\,dx - \int_a^b g(x)\,dx$, giving the result.

Property 7: Bounding the Integral …