Definite integrals | Integration | Achievable AP Calculus AB

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 provides a compact way to write the total of many terms added together. A right Riemann sum with $n$ subintervals can be written as

$k = 1 \sum n Δ x \cdot f (x_{k})$

where

$Σ$ stands for “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.

$Δ x$ is still the width of each rectangle, or $\frac{b - a}{n}$

The sample point $x_{k}$ is defined as

$x_{k} = a + k Δ x$

A left Riemann sum can be written instead as

$k = 0 \sum n - 1 Δ x \cdot f (x_{k})$

Where the index $k$ starts at $0$ and the 1st sample point is $x_{0} = a$ (the left endpoint).

From approximations to integrals

Riemann sums are merely approximations of the area. But the more rectangles there are that fit under the curve nicely, the more accurate the area becomes. As the number of rectangles $n$ approaches $\infty$ , then $Δ x$ approaches $0$ , and the approximation becomes exact. This leads to what is called the definite integral, which is defined to be the net sum of the areas of the infinitely many and narrow rectangles.

Definite integral

$\int_{a}^{b} f (x) d x = n \to \infty lim k = 1 \sum n Δ x \cdot f (x_{k})$

The left-hand side is read as the “definite integral $f (x)$ from $a$ to $b$ ,” where $d x$ represents the width $Δ x$ that is infinitesimally small.

$x = a$ and $x = b$ are the limits of integration where $a$ is the lower bound and $b$ is the upper bound.

$f (x)$ is the integrand (that is being integrated).

Evaluating the definite integral will produce a value that is the exact net signed area under the curve $f (x)$ from $x = a$ to $x = b$ .

In this half of calculus, integrals are just as important as derivatives were in the 1st half. The integral sign $\int$ is an elongated S that represents summation and tells us to add up small changes or pieces.

Just as the notation $f^{'} (x)$ represents the derivative of $f (x)$ , the integral sign is a key symbol you’ll see repeatedly when working with area problems and more.

Definite integral as a limit

You may be expected to either write or recognize the definite integral as a limit of a Riemann sum. Because

$Δ x = \frac{b - a}{n}$ and
$x_{k} = a + k Δ x$

the limit with $Δ x \cdot f (x_{k})$ can be rewritten as

$n \to \infty lim k = 1 \sum n Δ x (\frac{b - a}{n}) \cdot f (x_{k} a + \frac{b - a}{n} \cdot k)$

Examples

1. Express the definite integral as the limit of a Riemann sum

$\int_{2}^{4} sin (x) d x$

Solution

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

$Δ x = \frac{4 - 2}{n}$

$= \frac{2}{n}$

and the sample point is

$x_{k} = 2 + k \cdot \frac{2}{n}$

Then

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

The limit is then

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

Next is the other way around - writing a limit as a definite integral.

2. Express the limit of the Riemann sum as a definite integral

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

Solution

(spoiler)

The key is to recognize each component of the sum.

$\frac{1}{n}$ corresponds to $Δ x = \frac{b - a}{n}$ .

This means that $b - a = 1$

$(1 + \frac{k}{n})^{2}$ corresponds to $f (x_{k})$ .

Since $x_{k} = a + k Δ x$ , it suggests that suggests that $a = 1$ and the function $f (x) = x^{2}$ .

If $a = 1$ and $(b - a) = 1$ , then $b = 2$ , and one such definite integral that produces this sum could be

$\int_{1}^{2} x^{2} d x$

However, this is not the only definite integral that could work. If we let $a = 0, b = 1,$ and $f (x) = (1 + x)^{2}$ , then

$Δ x = \frac{1}{n}$

as well, and

$f (x_{k}) = (1 + (a + k Δ x))^{2}$

$= (1 + (0 + k \cdot \frac{1}{n}))^{2}$

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

which produces the same summation of

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

Therefore another answer (and not limited to just these two) could be

$\int_{0}^{1} (1 + x)^{2} d x$

Graphically, this should make sense - the 2nd function $(1 + x)^{2}$ is simply the 1st one $(x^{2})$ shifted left by $1$ unit. Since the bounds of integration, $a$ and $b$ , are also shifted left by $1$ unit, the entire region shifts and the area remains the same.

Evaluating definite integrals

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

Later on, techniques to evaluate definite integrals with more complicated functions in the integrand will be covered. For now, the next few examples involve simple geometric shapes to familiarize you with definite integral notation.

Examples

1. Evaluate

$\int_{- 2}^{1} 2 x d x$

Solution

Evaluating this integral will give the net signed area as shaded in the graph below. $A_{1}$ will be negative since it’s below the $x$ -axis and $A_{2}$ will be positive.

$A_{1}$ is a triangle with a base length of $2$ and a height of $4$ , so the signed area is

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

$= - 4$

$A_{2}$ is a triangle with a base length of $1$ and a height of $2$ , and the signed area is

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

$= 1$

Then the net signed area is the sum, or

$- 4 + 1 = - 3$

Therefore

$\int_{- 2}^{1} 2 x d x = - 3$

2. Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

Solution

(spoiler)

The region formed between $f (x) = ∣ x ∣$ and the $x$ -axis consists of 2 congruent triangles, both above the $x$ -axis - one from $x = - 1$ to $x = 0$ and the other from $x = 0$ to $x = 1$ . Each triangle has a base length of $1$ and a height of $1$ . Then the total area is

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$

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 provides a compact way to write the total of many terms added together. A right Riemann sum with $n$ subintervals can be written as

$k = 1 \sum n Δ x \cdot f (x_{k})$

where

$Σ$ stands for “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.

$Δ x$ is still the width of each rectangle, or $\frac{b - a}{n}$

The sample point $x_{k}$ is defined as

$x_{k} = a + k Δ x$

A left Riemann sum can be written instead as

$k = 0 \sum n - 1 Δ x \cdot f (x_{k})$

Where the index $k$ starts at $0$ and the 1st sample point is $x_{0} = a$ (the left endpoint).

From approximations to integrals

Definite integral

$\int_{a}^{b} f (x) d x = n \to \infty lim k = 1 \sum n Δ x \cdot f (x_{k})$

The left-hand side is read as the “definite integral $f (x)$ from $a$ to $b$ ,” where $d x$ represents the width $Δ x$ that is infinitesimally small.

$x = a$ and $x = b$ are the limits of integration where $a$ is the lower bound and $b$ is the upper bound.

$f (x)$ is the integrand (that is being integrated).

Evaluating the definite integral will produce a value that is the exact net signed area under the curve $f (x)$ from $x = a$ to $x = b$ .

Just as the notation $f^{'} (x)$ represents the derivative of $f (x)$ , the integral sign is a key symbol you’ll see repeatedly when working with area problems and more.

Definite integral as a limit

You may be expected to either write or recognize the definite integral as a limit of a Riemann sum. Because

$Δ x = \frac{b - a}{n}$ and
$x_{k} = a + k Δ x$

the limit with $Δ x \cdot f (x_{k})$ can be rewritten as

$n \to \infty lim k = 1 \sum n Δ x (\frac{b - a}{n}) \cdot f (x_{k} a + \frac{b - a}{n} \cdot k)$

Examples

1. Express the definite integral as the limit of a Riemann sum

$\int_{2}^{4} sin (x) d x$

Solution

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

$Δ x = \frac{4 - 2}{n}$

$= \frac{2}{n}$

and the sample point is

$x_{k} = 2 + k \cdot \frac{2}{n}$

Then

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

The limit is then

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

Next is the other way around - writing a limit as a definite integral.

2. Express the limit of the Riemann sum as a definite integral

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

Solution

(spoiler)

The key is to recognize each component of the sum.

$\frac{1}{n}$ corresponds to $Δ x = \frac{b - a}{n}$ .

This means that $b - a = 1$

$(1 + \frac{k}{n})^{2}$ corresponds to $f (x_{k})$ .

Since $x_{k} = a + k Δ x$ , it suggests that suggests that $a = 1$ and the function $f (x) = x^{2}$ .

If $a = 1$ and $(b - a) = 1$ , then $b = 2$ , and one such definite integral that produces this sum could be

$\int_{1}^{2} x^{2} d x$

However, this is not the only definite integral that could work. If we let $a = 0, b = 1,$ and $f (x) = (1 + x)^{2}$ , then

$Δ x = \frac{1}{n}$

as well, and

$f (x_{k}) = (1 + (a + k Δ x))^{2}$

$= (1 + (0 + k \cdot \frac{1}{n}))^{2}$

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

which produces the same summation of

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

Therefore another answer (and not limited to just these two) could be

$\int_{0}^{1} (1 + x)^{2} d x$

Evaluating definite integrals

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

Examples

1. Evaluate

$\int_{- 2}^{1} 2 x d x$

Solution

Evaluating this integral will give the net signed area as shaded in the graph below. $A_{1}$ will be negative since it’s below the $x$ -axis and $A_{2}$ will be positive.

$A_{1}$ is a triangle with a base length of $2$ and a height of $4$ , so the signed area is

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

$= - 4$

$A_{2}$ is a triangle with a base length of $1$ and a height of $2$ , and the signed area is

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

$= 1$

Then the net signed area is the sum, or

$- 4 + 1 = - 3$

Therefore

$\int_{- 2}^{1} 2 x d x = - 3$

2. Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

Solution

(spoiler)

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$