What you’ll learn:

• Converting between expressions of Riemann sums in summation notation and definite integrals
• Evaluating simple definite integrals as net signed area

Summation notation

Summation notation gives you a compact way to write a long addition of many similar terms. For example, a right Riemann sum with $n$ subintervals can be written as

$$\sum _{k=1}^n\Delta x\cdot f(x_k)$$

where

• $\Sigma$ means “sum.”
• The index variable $k$ starts at $1$ and increases by $1$ until it reaches $n$.
• $n$ terms are being added, one for each rectangle.

Here, $\Delta x$ is the width of each rectangle:

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

For a right Riemann sum, the sample point $x_k$ is

$$x_k=a+k\Delta x$$

A left Riemann sum uses left endpoints instead, and it can be written as

$$\sum _{k=0}^{n-1}\Delta x\cdot f(x_k)$$

Here the index $k$ starts at $0$, so the first sample point is $x_0=a$ (the left endpoint).

From approximation to definite integral

Riemann sums approximate area using rectangles. As the number of rectangles is increased, the approximation typically improves because the rectangles fit the curve more closely.

As the number of rectangles $n$ approaches $\infty$, the width $\Delta x$ approaches $0$, and the approximation of the area becomes exact.

The exact area is calculated by a definite integral, which is defined as the net sum of the areas of infinitely many rectangles of infinitesimal width. Like the derivative, the integral is defined using a limit.

Evaluating a definite integral gives the exact net signed area under the curve $f(x)$ from $x=a$ to $x=b$.

Integrals are a key concept in the second half of AP Calculus. While we’ll eventually use them to calculate exact areas and other accumulated quantities for different functions, for now let’s focus on becoming familiar with the notation.

Limit definition of the definite integral

$${\int }_a^{b}f(x)dx=\underset{n\to \infty }{\lim }\sum _{k=1}^n\Delta x\cdot f(x_k)$$

The left-hand side is read as “the definite integral of $f(x)$ from $a$ to $b$.” The $dx$ corresponds to the rectangle width $\Delta x$ in the limit (an infinitesimally small width). This variable of integration must always be written at the end of an integral.

In this notation:

• $a$ and $b$ are the limits of integration - $a$ is the lower bound and $b$ is the upper bound.
• $f(x)$ is the integrand - the function being integrated.

The integral symbol $\int$ is an elongated S, a reminder that integration is a kind of summation: it adds up many small pieces.

Definite integral as a limit

You may be asked to write (or recognize) a definite integral as the limit of a Riemann sum.

Because

• $\Delta x=\frac{b-a}{n}$ and

• $x_k=a+k\Delta x$

the expression $\Delta x\cdot f(x_k)$ can be rewritten in terms of $a$, $b$, $n$, and $k$ as:

$$\underset{n\to \infty }{\lim }\sum _{k=1}^n\underset{\Delta x}{\underbrace{\left(\frac{b-a}{n}\right)}}\cdot \underset{f(x_k)}{\underbrace{f(a+\frac{b-a}{n}\cdot k)}}$$

Examples

Express the definite integral as the limit of a Riemann sum

$${\int }_2^4\sin (x)dx$$

Solution

In this integral, $a=2$, $b=4$, and $f(x)=\sin (x)$.

First find $\Delta x$:

$$\begin{array}{rl}\Delta x& =\frac{b-a}{n}\\ & =\frac{4-2}{n}\\ & =\frac{2}{n}\end{array}$$

Then write the sample point $x_k$:

$$\begin{array}{rl}{x}_k& =a+k\Delta x\\ & =2+\frac{2k}{n}\end{array}$$

Now substitute into $f(x_k)=\sin (x_k)$:

$$f(x_k)=\sin (2+\frac{2k}{n})$$

So the integral as a limit is

$$\boxed{\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{2}{n}\cdot \sin (2+\frac{2k}{n})}$$

Limit as a definite integral

Next, we’ll move in the opposite direction and convert a limit to a definite integral.

Express the limit of the Riemann sum as a definite integral

$$\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{1}{n}(1+\frac{k}{n}{)}^2$$

Solution

The key is to match the expression in the sum to the form $\Delta x\cdot f(x_k)$ to identify the interval $[a,b]$ and the function $f(x)$ to write the integral ${\int }_a^{b}f(x)dx$.

First, identify $\Delta x$ and$f(x_k)$ in the limit:

$$\underset{n\to \infty }{\lim }\sum _{k=1}^n\underset{\Delta x}{\underbrace{\left(\frac{1}{n}\right)}}\cdot \underset{f(x_k)}{\underbrace{(1+\frac{k}{n}{)}^2}}$$

Since $\Delta x=\frac{1}{n}$, the sample point $x_k$ is

$$\begin{array}{rl}{x}_k& =a+k\Delta x\\ & =a+\frac{k}{n}\end{array}$$

and any expression in the limit that involves $\frac{k}{n}$ corresponds to $x_k$.

Next,

$$f(x_k)=(1+\frac{k}{n}{)}^2$$

suggests that $a=1$ and $f(x_k)=(x_k{)}^2$. So the function is

$$f(x)=x^2$$

Finally, since $\Delta x=\frac{1}{n}$, then $b-a=1$. We have $a=1$ so then $b=2$ and the integral represented by the limit is

$$\boxed{{\int }_1^2{x}^{2}dx}$$

However, this is not the only possible choice for the definite integral. For example, we could have also assumed $a=0$, leading to

$$x_k=\frac{k}{n}$$

Then the expression in the limit

$$f(x_k)=(1+\frac{k}{n}{)}^2$$

corresponds to

$$f(x_k)=(1+x_k{)}^2$$

so the function would be

$$f(x)=(1+x{)}^2$$

Since $a=0$, then $b=1$, and the limit could also be represented by the definite integral

$$\boxed{{\int }_0^1(1+x{)}^{2}dx}$$

Consider the two functions graphically: $(1+x{)}^2$ is simply $x^2$ shifted $1$ unit left. Shifting the bounds of integration left by $1$ unit as well shows that the area under $(1+x{)}^2$ from $0$ to $1$ is exactly the same as the area under $x^2$ from $1$ to $2$, so their definite integrals are equal.

Evaluating a definite integral

When a problem asks you to evaluate a definite integral, it’s asking for the net signed area of $f(x)$ from $x=a$ to $x=b$.

Later, you’ll learn techniques for evaluating definite integrals when the integrand is more complicated. For now, the next examples use simple geometric areas.

Examples

1. Evaluate

$${\int }_{-2}^{1}2xdx$$

Solution

Evaluating this integral gives the net signed area between $f(x)=2x$ and the $x$-axis from $x=-2$ to. $x=1$, as shown in the graph below.

The region labeled $A_1$ is below the $x$-axis (so it contributes negative area), and $A_2$ is above the $x$-axis (so it contributes positive area).

$A_1$ is a triangle with base length $2$ and height $4$, so its signed area is

$$A_1=-\frac{1}{2}(2)(4)$$

$$=-4$$

$A_2$ is a triangle with base length $1$ and height $2$, so its signed area is

$$A_2=\frac{1}{2}(1)(2)$$

$$=1$$

The net signed area is

$$-4+1=-3$$

Therefore

$${\int }_{-2}^{1}2xdx=\boxed{-3}$$

2. Evaluate

$${\int }_{-1}^1|x|dx$$

Solution

(spoiler)

The region between $f(x)=|x|$ and the $x$-axis from $x=-1$ to $x=1$ consists of two congruent triangles, both above the $x$-axis:

• one from $x=-1$ to $x=0$
• one from $x=0$ to $x=1$

Each triangle has base length $1$ and height $1$. So the total area is

$$2\times \text{Area of 1 triangle}$$

$$=2\times \frac{1}{2}(1)(1)$$

$$=1$$

Therefore

$${\int }_{-1}^1|x|dx=\boxed{1}$$