Introduction to Definite Integrals, calculus lesson

Riemann sums give approximations, and every approximation invites the question of what it approximates. When all Riemann sums of a function settle on one number as the pieces get thin, that number is the definite integral. It is the exact area under a curve, the exact distance from a speed record, the exact total from any rate.

Key terms

The Riemann integral

Definition Let $f$ be defined on $[a, b]$ with $a < b$ . We say $f$ is Riemann integrable on $[a, b]$ if there is a number $I$ with the following property. For every $\varepsilon > 0$ there is a $\delta > 0$ such that

\left\lvert \sum_{i=1}^{n} f(x_i^*) \, \Delta x_i - I \right\rvert < \varepsilon

for every partition $P$ of $[a, b]$ with $\lVert P \rVert < \delta$ and every choice of tags $x_i^* \in [x_{i-1}, x_i]$ . The number $I$ is unique when it exists. It is called the definite integral of $f$ from $a$ to $b$ and written

\int_a^b f(x) \, dx = \lim_{\lVert P \rVert \to 0} \sum_{i=1}^{n} f(x_i^*) \, \Delta x_i.

In plain words The integral is the single number that all Riemann sums approach once every piece of the partition is thin enough, no matter how the pieces are cut or where the sample points sit.

Example For the constant $f(x) = 3$ on $[0, 2]$ every Riemann sum equals $\sum 3 \, \Delta x_i = 3 \cdot 2 = 6$ , whatever the partition and tags. So

\int_0^2 3 \, dx = 6,

the area of a $2$ by $3$ rectangle.

Integrability

Definition A function is integrable on $[a, b]$ when the limit above exists. Every integrable function is bounded on $[a, b]$ . An equivalent test, due to Darboux, says that a bounded $f$ is integrable on $[a, b]$ if and only if for every $\varepsilon > 0$ there is a partition $P$ with

U(f, P) - L(f, P) < \varepsilon.

In that case $\int_a^b f(x) \, dx$ is the one number lying between all lower sums and all upper sums.

In plain words A function is integrable when the lower and upper sums can be squeezed as close together as you like. If they stay apart, Riemann sums keep jumping around and there is no single answer.

Example Let $f(x) = 1$ for rational $x$ and $f(x) = 0$ for irrational $x$ , on $[0, 1]$ . Every subinterval of every partition contains both kinds of numbers, so $m_i = 0$ and $M_i = 1$ . Then $L(f, P) = 0$ and $U(f, P) = 1$ for every $P$ , the gap never shrinks, and $f$ is not integrable.

Continuity implies integrability

Definition If $f$ is continuous on the closed interval $[a, b]$ , then $f$ is integrable on $[a, b]$ . More generally, a function that is bounded on $[a, b]$ and has only finitely many discontinuities there is integrable, and so is every monotonic function on $[a, b]$ .

In plain words For the functions you meet most often, the integral exists and you may compute it with whatever convenient Riemann sums you like, for example equal pieces with right endpoints.

Example The function $f(x) = x^2$ is continuous on $[0, 1]$ , so the integral exists and equals the limit of any particular family of Riemann sums with mesh tending to zero. Right endpoints with $n$ equal pieces give

R_n = \frac{(n+1)(2n+1)}{6n^2} \to \frac{1}{3},

so $\int_0^1 x^2 \, dx = \frac{1}{3}$ .

Integrand and bounds

Definition In $\int_a^b f(x) \, dx$ the function $f$ is the integrand. The numbers $a$ and $b$ are the lower and upper bounds, also called the limits of integration, and $[a, b]$ is the interval of integration. The value depends only on $f$ , $a$ and $b$ .

In plain words The integrand is what you add up, and the bounds say where you start and stop.

Example In $\int_1^4 \sqrt{x} \, dx$ the integrand is $\sqrt{x}$ , the lower bound is $1$ and the upper bound is $4$ . The value is a number, here $\frac{14}{3}$ , not a function of $x$ .

Dummy variable

Definition The letter used for the variable of integration is a dummy variable. Renaming it does not change the value,

\int_a^b f(x) \, dx = \int_a^b f(t) \, dt = \int_a^b f(u) \, du.

The dummy variable has no meaning outside the integral.

In plain words It works like the index $i$ in a sum. You could call it anything, and the result is the same number.

Example The integrals $\int_0^1 x^2 \, dx$ and $\int_0^1 t^2 \, dt$ are both $\frac{1}{3}$ . Writing $\int_0^x x^2 \, dx$ is confusing, because the letter $x$ plays two roles, so one writes $\int_0^x t^2 \, dt$ instead.

Differential notation

Definition The symbol $dx$ in $\int_a^b f(x) \, dx$ marks the variable of integration and comes from the width $\Delta x_i$ in the Riemann sum. The integral sign is a stretched S for sum. The notation records the passage

\sum_{i=1}^{n} f(x_i^*) \, \Delta x_i \quad \longrightarrow \quad \int_a^b f(x) \, dx.

In plain words Read $f(x) \, dx$ as a height times a very thin width. The integral adds up these thin products.

Example In $\int_0^2 a x \, dx$ the $dx$ tells you that $x$ varies and $a$ is a constant. The value is $2a$ . In $\int_0^2 a x \, da$ the roles swap, $a$ varies from $0$ to $2$ , and the value is $2x$ .

Signed net area

Definition For $f$ integrable on $[a, b]$ , the integral counts area above the $x$ -axis as positive and area below it as negative,

\int_a^b f(x) \, dx = A_{\text{above}} - A_{\text{below}},

where $A_{\text{above}}$ is the area between the graph and the axis where $f \ge 0$ and $A_{\text{below}}$ is the area where $f \le 0$ . If $f \ge 0$ on $[a, b]$ the integral is the area under the graph.

In plain words Parts of the graph below the axis subtract. The integral is a net amount, not a total amount.

Example On $[-1, 2]$ the graph of $f(x) = x$ is below the axis on $[-1, 0]$ , a triangle of area $\frac{1}{2}$ , and above it on $[0, 2]$ , a triangle of area $2$ . So

\int_{-1}^{2} x \, dx = 2 - \frac{1}{2} = \frac{3}{2}.

Midpoint sums

Definition For $n$ equal subintervals of $[a, b]$ with $\Delta x = \frac{b - a}{n}$ and midpoints $\bar{x}_i = \frac{x_{i-1} + x_i}{2}$ , the midpoint sum is

M_n = \sum_{i=1}^{n} f(\bar{x}_i) \, \Delta x.

If $f''$ is continuous on $[a, b]$ and $\lvert f''(x) \rvert \le K$ there, the error satisfies

\left\lvert \int_a^b f(x) \, dx - M_n \right\rvert \le \frac{K (b - a)^3}{24 n^2}.

In plain words Sampling at the middle of each piece usually beats both ends, because the overshoot on one half of the piece partly cancels the undershoot on the other.

Example For $\int_0^1 x^2 \, dx$ with $n = 4$ the midpoints are $0.125, 0.375, 0.625, 0.875$ .

M_4 = 0.25 \, (0.015625 + 0.140625 + 0.390625 + 0.765625) = 0.328125.

The error is $\frac{1}{3} - 0.328125 = \frac{1}{192}$ . With $K = 2$ , $b - a = 1$ and $n = 4$ the bound is

\frac{2 \cdot 1^3}{24 \cdot 4^2} = \frac{1}{192},

met exactly here. Compare $L_4 = 0.21875$ and $R_4 = 0.46875$ .

Check yourself

Use geometry to find $\int_{-2}^{2} \sqrt{4 - x^2} \, dx$ .

Show the answer

The graph is the upper half of the circle of radius $2$ , so the integral is half of $\pi \cdot 2^2$ , which is $2\pi$ .

Find $\int_0^3 (x - 1) \, dx$ as signed area.

Show the answer

Below the axis on $[0, 1]$ there is a triangle of area $\frac{1}{2}$ . Above on $[1, 3]$ there is a triangle of area $\frac{1}{2} \cdot 2 \cdot 2 = 2$ . The integral is $2 - \frac{1}{2} = \frac{3}{2}$ .

Estimate $\int_1^3 \frac{1}{x} \, dx$ with the midpoint sum $M_2$ .

Show the answer

Here $\Delta x = 1$ and the midpoints are $1.5$ and $2.5$ , so

M_2 = \frac{1}{1.5} + \frac{1}{2.5} = \frac{2}{3} + \frac{2}{5} \approx 1.067.

The exact value is $\ln 3 \approx 1.099$ .

Introduction to Definite Integrals

Theory, part 99

Practice, part 100

Key terms

The Riemann integral

Integrability

Continuity implies integrability

Integrand and bounds

Dummy variable

Differential notation

Signed net area

Midpoint sums

Check yourself

How this lesson connects

Builds on

Used later in

Resources for this lesson

Riemann sum explorer