Math Core

A Riemann sum with four rectangles gives a rough estimate. With forty rectangles the estimate improves, and with four thousand it is nearly perfect. The definite integral is what you get when you push this all the way: the exact value that the Riemann sums approach as the rectangles become infinitely thin.

From sums to a limit

For the rest of this lesson, split $[a, b]$ into $n$ subintervals of equal width

\Delta x = \frac{b - a}{n}.

The right endpoints are then $x_k = a + k\,\Delta x$ for $k = 1, 2, \ldots, n$ , and the right Riemann sum can be written compactly in sigma notation:

R_n = \sum_{k=1}^{n} f(x_k)\,\Delta x = \sum_{k=1}^{n} f\!\left(a + \frac{k(b - a)}{n}\right) \frac{b - a}{n}.

As $n$ grows, each rectangle gets narrower and hugs the curve more closely. For a continuous function, the left, right and midpoint sums all approach the same number.

Definition

Definite integral

If $f$ is continuous on $[a, b]$ , the definite integral of $f$ from $a$ to $b$ is

\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\,\Delta x, \quad \text{where } \Delta x = \frac{b - a}{n} \text{ and } x_k = a + k\,\Delta x.

The numbers $a$ and $b$ are the lower and upper limits of integration, and $f(x)$ is the integrand.

The notation is a reminder of where it came from. The elongated S, $\int$ , stands for "sum," $f(x)$ is a rectangle's height, and $dx$ is its infinitely thin width.

Converting a limit of sums into an integral

A classic AP multiple-choice question hands you a limit of a Riemann sum and asks which integral it equals. Read off three pieces:

Find $\Delta x$ . It's the factor like $\dfrac{3}{n}$ that multiplies everything. That gives $b - a$ .
Find $x_k$ . It's the expression plugged into the function, usually of the form $a + \dfrac{(b - a)k}{n}$ . The constant part is $a$ .
Find $f$ . It's what the function does to $x_k$ .

Worked example: Reading a limit of sums

Write $\displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{1 + \frac{3k}{n}} \cdot \frac{3}{n}$ as a definite integral.

The width is $\Delta x = \dfrac{3}{n}$ , so $b - a = 3$ . The input to the square root is $x_k = 1 + \dfrac{3k}{n}$ , so $a = 1$ and $b = 1 + 3 = 4$ . The function is $f(x) = \sqrt{x}$ .

\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{1 + \frac{3k}{n}} \cdot \frac{3}{n} = \int_1^4 \sqrt{x}\,dx.

Another correct answer is $\displaystyle \int_0^3 \sqrt{1 + x}\,dx$ , which uses $x_k = \dfrac{3k}{n}$ starting from $a = 0$ and puts the 1 inside the function. Both integrals have the same value.

Signed area

When $f$ is positive, $\displaystyle\int_a^b f(x)\,dx$ is the area under the curve. When $f$ is negative, each rectangle has a negative height, so it contributes negative area. The integral is a net or signed area:

The integral is signed area

\int_a^b f(x)\,dx = (\text{area above the } x\text{-axis}) - (\text{area below the } x\text{-axis}).

If $f$ is a rate of change, this is the net change in the quantity from $x = a$ to $x = b$ .

When the graph is made of lines and circles, you can evaluate an integral exactly with geometry.

The graph of y = x − 1 on [0, 3]. The small triangle below the axis counts as negative area; the larger triangle above counts as positive.Open in grapher →

Worked example: Net area with triangles

Evaluate $\displaystyle\int_0^3 (x - 1)\,dx$ .

The line crosses the $x$ -axis at $x = 1$ .

From 0 to 1 there is a triangle below the axis with base 1 and height 1: area $\dfrac{1}{2}$ .
From 1 to 3 there is a triangle above the axis with base 2 and height 2: area $2$ .

\int_0^3 (x - 1)\,dx = 2 - \frac{1}{2} = \frac{3}{2}.

Worked example: A semicircle

Evaluate $\displaystyle\int_{-2}^{2} \sqrt{4 - x^2}\,dx$ .

Squaring $y = \sqrt{4 - x^2}$ gives $x^2 + y^2 = 4$ with $y \ge 0$ : the upper half of a circle of radius 2. The integral is the area of that half-circle:

\int_{-2}^{2} \sqrt{4 - x^2}\,dx = \frac{1}{2}\pi(2)^2 = 2\pi.

Common mistake

"Evaluate the integral" and "find the area" are different questions when the graph dips below the axis. The integral subtracts the area below the axis; the total area adds it. In the first example the integral is $\dfrac{3}{2}$ , but the total area between the line and the axis is $2 + \dfrac{1}{2} = \dfrac{5}{2}$ .

Integrals of rates and units

Since an integral is a limit of (height) × (width) sums, its units are the units of $f$ times the units of $x$ . If $v(t)$ is a velocity in meters per second, then $\displaystyle\int_0^{10} v(t)\,dt$ is measured in meters and gives the displacement (net change in position) over the first 10 seconds. If $R(t)$ is a rate in people per hour, $\displaystyle\int_2^5 R(t)\,dt$ counts people, the net number who arrived between hour 2 and hour 5.

Worked example: A piecewise graph

The graph of $f$ consists of a line segment from $(0, 2)$ to $(2, 2)$ and a line segment from $(2, 2)$ to $(5, -1)$ . Evaluate $\displaystyle\int_0^5 f(x)\,dx$ .

The second segment has slope $-1$ and crosses the axis at $x = 4$ .

From 0 to 2: a rectangle of area $2 \times 2 = 4$ .
From 2 to 4: a triangle above the axis with base 2 and height 2, area 2.
From 4 to 5: a triangle below the axis with base 1 and height 1, area $\dfrac{1}{2}$ .

\int_0^5 f(x)\,dx = 4 + 2 - \frac{1}{2} = \frac{11}{2}.

Practice

Practice 1

Use geometry to evaluate $\displaystyle\int_0^4 (2x + 1)\,dx$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Evaluate $\displaystyle\int_{-3}^{3} \sqrt{9 - x^2}\,dx$ . Give an exact answer in terms of $\pi$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Evaluate $\displaystyle\int_0^6 (3 - x)\,dx$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Evaluate $\displaystyle\int_{-1}^{3} |x|\,dx$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Which of the following is equal to $\displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} \left(1 + \frac{2k}{n}\right)^3 \frac{2}{n}$ ?

Practice 6

Which of the following is equal to $\displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n} \sin\!\left(\frac{\pi k}{n}\right)$ ?

Practice 7

On $[0, 4]$ , the graph of $f$ is the lower half of the circle of radius 2 centered at $(2, 0)$ . On $[4, 6]$ , $f(x) = x - 4$ . Evaluate $\displaystyle\int_0^6 f(x)\,dx$ . Give an exact answer.

The graph of f: a lower semicircle on [0, 4] followed by a line segment on [4, 6].Open in grapher →

Enter a number. Fractions like 3/4 and sqrt(2) are OK.