6.3 Definite integrals

Achievable AP Calculus AB

6. Integration

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Definite integrals

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What you’ll learn:

Limit definitions: Convert between Riemann sums and definite integrals.
Geometric evaluation: Evaluate definite integrals as net signed area.

From approximations to integrals

A definite integral calculates the exact net signed area under a curve by taking the limit of a right Riemann sum as the number of rectangles ( $n$ ) approaches infinity.

As $n \to \infty$ , the width of each rectangle ( $Δ x$ ) approaches $0$ , turning an approximation into an exact value.

Limit definition of the definite integral:

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

where:

$Δ x = \frac{b - a}{n}$ is the width of each subinterval.
$x_{k} = a + k Δ x$ is the right-endpoint sample point.

Anatomy of an integral

The integral symbol $\int$ replaces the limit and summation, while $d x$ replaces the width $Δ x$ .

$\int_{a}^{b}$ (limits of integration): $a$ is the lower bound (start) and $b$ is the upper bound (end) on the $x$ -axis.
$f (x)$ (integrand): The function that determines the height of the region.
$d x$ (variable of integration): Identifies the variable we are integrating with respect to, representing $Δ x \to 0$ . Always include this at the end.

To convert between Riemann sums and definite integrals, map the structural components directly:

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

Examples

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)

To convert this to a definite integral, match the components of the limit form.

1. Identify $Δ x$ to find $[a, b]$ :

The term outside the function is the width:

$n \to \infty lim k = 1 \sum n Δ x (\frac{1}{n}) \cdot f (x_{k}) (1 + \frac{k}{n})^{2}$

Since $Δ x = \frac{1}{n} = \frac{b - a}{n}$ , we know:

$b - a = 1$

2. Identify $x_{k}$ and find the lower bound ( $a$ ):

Since $Δ x = \frac{1}{n}$ , the sample point $x_{k}$ is

$x_{k} = a + k Δ x = a + \frac{k}{n}$

This matches the expression in the limit $1 + \frac{k}{n}$ , meaning $a = 1$ .

3. Find the upper bound ( $b$ ):

Since $a = 1$ and the total interval width is $b - a = 1$ , then

$b - 1 b = 1 = 2$

4. Identify the function $f (x)$ :

Since $x_{k} = 1 + \frac{k}{n}$ is being squared, the outer matching function is:

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

Bringing it all together on the interval $[1, 2]$ , the definite integral represented by the limit is:

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

Sidenote

An alternative choice

You could also define the inside variable as $x_{k} = \frac{k}{n}$ , which means $a = 0$ and $b = 1$ . This changes the function to $f (x) = (1 + x)^{2}$ , resulting in $\int_{0}^{1} (1 + x)^{2} d x$ .

Both integrals represent the exact same geometric area, just shifted along the $x$ -axis.

Definite integral to limit

Next, the process is reversed.

Express the definite integral as the limit of a Riemann sum

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

Solution

(spoiler)

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

1. Find $Δ x$ :

$Δ x = \frac{b - a}{n} = \frac{4 - 2}{n} = \frac{2}{n}$

2. Write the sample point $x_{k}$ :

$x_{k} = a + k Δ x = 2 + \frac{2 k}{n}$

3. Substitute into $f (x_{k}) = sin (x_{k})$ :

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

4. Write in limit form:

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

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

Evaluate

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

Solution

(spoiler)

Evaluating this integral gives the net signed area between $f (x) = 2 x$ 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 $4$ with signed area:

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

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

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

The integral gives the net signed area, which is the sum of the areas:

$\int_{- 2}^{1} 2 x d x = - 4 + 1 = - 3$

Try one with an absolute value function:

Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

(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

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$

Net signed area vs. total area

A definite integral measures net signed area: regions below the $x$ -axis count as negative. For example, $\int_{- 2}^{1} 2 x d x = - 3$ because the negative region outweighs the positive one.

By contrast, $\int_{- 1}^{1} ∣ x ∣ d x = 1$ because $∣ x ∣ \geq 0$ everywhere - there’s no cancellation. If you were instead asked for $\int_{- 1}^{1} x d x$ , the answer would be $0$ : the two triangles are equal in size but opposite in sign, so they cancel completely. When $f$ changes sign on $[a, b]$ , the definite integral and the total (geometric) area between the curve and the $x$ -axis are different quantities.

Summation notation

Compact form for adding similar terms: $\sum_{k = 1}^{n} Δ x \cdot f (x_{k})$
$Δ x = \frac{b - a}{n}$ (width of each rectangle)
Right Riemann sum: $x_{k} = a + k Δ x$ ; left Riemann sum: $x_{k} = a + (k - 1) Δ x$ (or $k$ from $0$ to $n - 1$ )

From approximations to integrals

Riemann sums approximate area under a curve using rectangles
Definite integral: $\int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{k = 1}^{n} Δ x \cdot f (x_{k})$
$a$ , $b$ = limits of integration; $f (x)$ = integrand

Definite integral as a limit

$Δ x = \frac{b - a}{n}$ , $x_{k} = a + k Δ x$
Limit of Riemann sum: $lim_{n \to \infty} \sum_{k = 1}^{n} \frac{b - a}{n} f (a + \frac{b - a}{n} k)$
Definite integral equals exact net signed area under $f (x)$ from $a$ to $b$

Examples: Converting between forms

To write $\int_{2}^{4} sin (x) d x$ as a limit: $lim_{n \to \infty} \sum_{k = 1}^{n} \frac{2}{n} sin (2 + \frac{2 k}{n})$
To write $lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n} (1 + \frac{k}{n})^{2}$ as a definite integral:
- Possible answers: $\int_{1}^{2} x^{2} d x$ or $\int_{0}^{1} (1 + x)^{2} d x$

Evaluating definite integrals

Definite integral = net signed area under $f (x)$ from $a$ to $b$
For simple functions, use geometric area formulas

Examples: Evaluating definite integrals

$\int_{- 2}^{1} 2 x d x = - 3$
- Area below $x$ -axis is negative, above is positive
$\int_{- 1}^{1} ∣ x ∣ d x = 1$
- Total area is sum of two congruent triangles above $x$ -axis

Definite integrals

What you’ll learn:

Limit definitions: Convert between Riemann sums and definite integrals.
Geometric evaluation: Evaluate definite integrals as net signed area.

From approximations to integrals

A definite integral calculates the exact net signed area under a curve by taking the limit of a right Riemann sum as the number of rectangles ( $n$ ) approaches infinity.

As $n \to \infty$ , the width of each rectangle ( $Δ x$ ) approaches $0$ , turning an approximation into an exact value.

Limit definition of the definite integral:

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

where:

$Δ x = \frac{b - a}{n}$ is the width of each subinterval.
$x_{k} = a + k Δ x$ is the right-endpoint sample point.

Anatomy of an integral

The integral symbol $\int$ replaces the limit and summation, while $d x$ replaces the width $Δ x$ .

$\int_{a}^{b}$ (limits of integration): $a$ is the lower bound (start) and $b$ is the upper bound (end) on the $x$ -axis.
$f (x)$ (integrand): The function that determines the height of the region.
$d x$ (variable of integration): Identifies the variable we are integrating with respect to, representing $Δ x \to 0$ . Always include this at the end.

To convert between Riemann sums and definite integrals, map the structural components directly:

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

Examples

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)

To convert this to a definite integral, match the components of the limit form.

1. Identify $Δ x$ to find $[a, b]$ :

The term outside the function is the width:

$n \to \infty lim k = 1 \sum n Δ x (\frac{1}{n}) \cdot f (x_{k}) (1 + \frac{k}{n})^{2}$

Since $Δ x = \frac{1}{n} = \frac{b - a}{n}$ , we know:

$b - a = 1$

2. Identify $x_{k}$ and find the lower bound ( $a$ ):

Since $Δ x = \frac{1}{n}$ , the sample point $x_{k}$ is

$x_{k} = a + k Δ x = a + \frac{k}{n}$

This matches the expression in the limit $1 + \frac{k}{n}$ , meaning $a = 1$ .

3. Find the upper bound ( $b$ ):

Since $a = 1$ and the total interval width is $b - a = 1$ , then

$b - 1 b = 1 = 2$

4. Identify the function $f (x)$ :

Since $x_{k} = 1 + \frac{k}{n}$ is being squared, the outer matching function is:

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

Bringing it all together on the interval $[1, 2]$ , the definite integral represented by the limit is:

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

Sidenote

An alternative choice

Both integrals represent the exact same geometric area, just shifted along the $x$ -axis.

Definite integral to limit

Next, the process is reversed.

Express the definite integral as the limit of a Riemann sum

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

Solution

(spoiler)

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

1. Find $Δ x$ :

$Δ x = \frac{b - a}{n} = \frac{4 - 2}{n} = \frac{2}{n}$

2. Write the sample point $x_{k}$ :

$x_{k} = a + k Δ x = 2 + \frac{2 k}{n}$

3. Substitute into $f (x_{k}) = sin (x_{k})$ :

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

4. Write in limit form:

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

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

Evaluate

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

Solution

(spoiler)

Evaluating this integral gives the net signed area between $f (x) = 2 x$ 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 $4$ with signed area:

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

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

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

The integral gives the net signed area, which is the sum of the areas:

$\int_{- 2}^{1} 2 x d x = - 4 + 1 = - 3$

Try one with an absolute value function:

Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

(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

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$

Net signed area vs. total area

A definite integral measures net signed area: regions below the $x$ -axis count as negative. For example, $\int_{- 2}^{1} 2 x d x = - 3$ because the negative region outweighs the positive one.

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.