The definite integral

In areas and Riemann sums, we found areas and distances as limits of sums. The same kind of limit appears in many problems. So it gets its own name and symbol: the definite integral.

Definition

Let $f$ be a function on $[a,b]$. Cut $[a,b]$ into $n$ pieces of width $\Delta x=\dfrac{b-a}{n}$, and choose a sample point $x_i^*$ in each piece. The definite integral of $f$ from $a$ to $b$ is

$\displaystyle\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^{n}f(x_i^*)\,\Delta x$,

if this limit exists and is the same for every choice of sample points. Then we say $f$ is integrable on $[a,b]$.

The parts of the symbol have names:

$\displaystyle\int$ is the integral sign. It is a tall S, for "sum."
$a$ and $b$ are the limits of integration. The number $a$ is the lower limit, and $b$ is the upper limit.
$f(x)$ is the integrand, the function being integrated.
$dx$ shows that the variable is $x$. It comes from the width $\Delta x$ in the sum.

Every continuous function on $[a,b]$ is integrable. So is a function with a finite number of jumps.

The definite integral is a number, not a function. The letter $x$ could be replaced by any other letter, and the number stays the same:

$\displaystyle\int_a^b f(x)\,dx=\int_a^b f(t)\,dt$

Area and net area

If $f(x)\ge 0$, the definite integral is the area under the curve.

If $f$ is negative on part of the interval, the terms $f(x_i^*)\Delta x$ are negative there. So the integral counts the area below the axis as negative:

$\begin{align*}&\int_a^b f(x)\,dx\\&=(\text{area above the axis})\\&\qquad-(\text{area below the axis})\end{align*}$

This is called the net area.

For $\sin x$ on $[0,2\pi]$, the area above the axis (teal) and the area below (red) are equal. So $\displaystyle\int_0^{2\pi}\sin x\,dx=0$.

Example 1: Find each integral by thinking of it as an area.

(a) $\displaystyle\int_0^3(x+1)\,dx$

(b) $\displaystyle\int_{-2}^{2}\sqrt{4-x^2}\,dx$

(a) A trapezoid

(b) A half circle

Solution:

(a) The region is a trapezoid with parallel sides $f(0)=1$ and $f(3)=4$ and width $3$:

$\displaystyle\int_0^3(x+1)\,dx=\dfrac{1+4}{2}\cdot 3=7.5$

(b) The graph of $y=\sqrt{4-x^2}$ is the top half of the circle $x^2+y^2=4$, with radius $2$. So the integral is half the area of the circle:

$\displaystyle\int_{-2}^{2}\sqrt{4-x^2}\,dx=\dfrac{1}{2}\pi(2)^2=2\pi$

(c) From $0$ to $1$, the line is below the axis. That triangle has area $\dfrac{1}{2}(1)(1)=\dfrac{1}{2}$. From $1$ to $3$, it is above the axis. That triangle has area $\dfrac{1}{2}(2)(2)=2$. So

$\displaystyle\int_0^3(x-1)\,dx=2-\dfrac{1}{2}=\dfrac{3}{2}$

Most integrals cannot be found from simple shapes. In areas and Riemann sums, we needed a long limit to show that

$\displaystyle\int_0^1 x^2\,dx=\dfrac{1}{3}$.

In the next section, the Fundamental Theorem of Calculus gives a much faster way.

Properties of the definite integral

These properties come from the properties of sums and limits. Here $c$ is a constant, and $f$ and $g$ are integrable:

1. A constant gives a rectangle:

$\displaystyle\int_a^b c\,dx=c(b-a)$

2. Integrate a sum or difference one term at a time:

$\begin{align*}&\int_a^b[f(x)\pm g(x)]\,dx\\&=\int_a^b f(x)\,dx\pm\int_a^b g(x)\,dx\end{align*}$

3. A constant factor moves out:

$\displaystyle\int_a^b c\,f(x)\,dx=c\int_a^b f(x)\,dx$

4. No width, no area:

$\displaystyle\int_a^a f(x)\,dx=0$

5. Switching the limits changes the sign:

$\displaystyle\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx$

6. Two pieces add up to the whole:

$\begin{align*}&\int_a^c f(x)\,dx+\int_c^b f(x)\,dx\\&=\int_a^b f(x)\,dx\end{align*}$

Example 2: Use the properties and

$\displaystyle\int_0^1 x^2\,dx=\dfrac{1}{3}$

to find

$\displaystyle\int_0^1(4+3x^2)\,dx$.

Solution:

$\begin{align*}&\int_0^1(4+3x^2)\,dx\\&=\int_0^1 4\,dx+3\int_0^1 x^2\,dx\\&=4(1-0)+3\cdot\dfrac{1}{3}\\&=5\end{align*}$

Example 3: Suppose that

$\displaystyle\int_0^{10}f(x)\,dx=17$ and

$\displaystyle\int_0^{8}f(x)\,dx=12$.

Find $\displaystyle\int_8^{10}f(x)\,dx$.

Solution:

The two pieces add up to the whole:

$\begin{align*}&\int_0^{8}f(x)\,dx+\int_8^{10}f(x)\,dx\\&=\int_0^{10}f(x)\,dx\end{align*}$

$\displaystyle\int_8^{10}f(x)\,dx=17-12=5$.

Comparing integrals

Bigger functions have bigger integrals:

1. If $f(x)\ge 0$ on $[a,b]$, then

$\displaystyle\int_a^b f(x)\,dx\ge 0$.

2. If $f(x)\ge g(x)$ on $[a,b]$, then

$\displaystyle\int_a^b f(x)\,dx\ge\int_a^b g(x)\,dx$.

3. If $m\le f(x)\le M$ on $[a,b]$, then

$\displaystyle m(b-a)\le\int_a^b f(x)\,dx\le M(b-a)$.

The last one says: the area is between the area of a short rectangle (height $m$) and a tall rectangle (height $M$).

Example 4: Estimate

$\displaystyle\int_0^1 e^{-x^2}\,dx$.

Solution:

On $[0,1]$, the function $e^{-x^2}$ is decreasing. So its largest value is $e^0=1$ (at $x=0$), and its smallest value is $e^{-1}\approx 0.368$ (at $x=1$). So

$\displaystyle 0.368\le\int_0^1 e^{-x^2}\,dx\le 1$.

(The true value is about $0.747$. This integral has no formula made of simple functions, so estimates like this are useful.)

Summary

The definite integral $\displaystyle\int_a^b f(x)\,dx$ is the limit of Riemann sums. It is a number.
If $f\ge 0$, it is the area under the curve. In general, it is the net area: above the axis minus below.
Integrals add over pieces, and constants move out.
Switching the limits changes the sign.

← Areas and Riemann sums The Fundamental Theorem of Calculus →