6.3 Definite integrals

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Definite integrals

6 min read

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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

$k = 1 \sum n Δ x \cdot f (x_{k})$

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.

Here, $Δ x$ is the width of each rectangle:

$Δ x = \frac{b - a}{n}$

For a right Riemann sum, the sample point $x_{k}$ is

$x_{k} = a + k Δ x$

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

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

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

From approximations to integrals

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

As the number of rectangles $n$ approaches $\infty$ , the width $Δ x$ approaches $0$ , and the approximation becomes exact. This leads to the definite integral, defined as the net sum of the areas of infinitely many rectangles of infinitesimal width.

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 of $f (x)$ from $a$ to $b$ .” The $d x$ corresponds to the rectangle width $Δ x$ in the limit (an infinitesimally small width).

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).

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

The symbol $\int$ is an elongated S, reminding you 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

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

the expression $Δ x \cdot f (x_{k})$ can be rewritten in terms of $a$ , $b$ , $n$ , and $k$ :

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

Examples

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 $f (x) = sin (x)$ .

First find $Δ x$ :

$Δ x = \frac{4 - 2}{n}$

$= \frac{2}{n}$

Then write the right-endpoint sample point $x_{k}$ :

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

Now substitute into $f (x_{k})$ :

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

So the limit form is

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

Next is the other direction: writing a limit as a definite integral.

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 match each part of the sum to the Riemann-sum form $\sum Δ x f (x_{k})$ .

$\frac{1}{n}$ 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 $x_{k} = a + k Δ x$ , the expression $1 + \frac{k}{n}$ suggests $a = 1$ and $Δ x = \frac{1}{n}$ . That matches the function $f (x) = x^{2}$ .

If $a = 1$ and $b - a = 1$ , then $b = 2$ . One definite integral that produces this sum is

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

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

$Δ x = \frac{1}{n}$

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:

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

So another valid answer is

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

Graphically, this makes sense: $(1 + x)^{2}$ is $x^{2}$ shifted left by $1$ unit. Shifting the bounds left by $1$ unit shifts the entire region the same way, so the area stays the same.

Evaluating definite integrals

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 to build intuition for what the notation means.

Examples

Evaluate

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

Solution

Evaluating this integral gives the net signed area 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

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

Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

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

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$

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:

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

$k = 1 \sum n Δ x \cdot f (x_{k})$

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.

Here, $Δ x$ is the width of each rectangle:

$Δ x = \frac{b - a}{n}$

For a right Riemann sum, the sample point $x_{k}$ is

$x_{k} = a + k Δ x$

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

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

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

From approximations to integrals

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

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 of $f (x)$ from $a$ to $b$ .” The $d x$ corresponds to the rectangle width $Δ x$ in the limit (an infinitesimally small width).

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).

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

The symbol $\int$ is an elongated S, reminding you 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

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

the expression $Δ x \cdot f (x_{k})$ can be rewritten in terms of $a$ , $b$ , $n$ , and $k$ :

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

Examples

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 $f (x) = sin (x)$ .

First find $Δ x$ :

$Δ x = \frac{4 - 2}{n}$

$= \frac{2}{n}$

Then write the right-endpoint sample point $x_{k}$ :

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

Now substitute into $f (x_{k})$ :

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

So the limit form is

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

Next is the other direction: writing a limit as a definite integral.

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 match each part of the sum to the Riemann-sum form $\sum Δ x f (x_{k})$ .

$\frac{1}{n}$ 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 $x_{k} = a + k Δ x$ , the expression $1 + \frac{k}{n}$ suggests $a = 1$ and $Δ x = \frac{1}{n}$ . That matches the function $f (x) = x^{2}$ .

If $a = 1$ and $b - a = 1$ , then $b = 2$ . One definite integral that produces this sum is

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

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

$Δ x = \frac{1}{n}$

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:

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

So another valid answer is

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

Graphically, this makes sense: $(1 + x)^{2}$ is $x^{2}$ shifted left by $1$ unit. Shifting the bounds left by $1$ unit shifts the entire region the same way, so the area stays the same.

Evaluating definite integrals

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 $x = b$ .

Examples

Evaluate

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

Solution

$A_{1}$ 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

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

Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

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

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Definite integrals

6 min read

Font

Discuss

Feedback

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

$k = 1 \sum n Δ x \cdot f (x_{k})$

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.

Here, $Δ x$ is the width of each rectangle:

$Δ x = \frac{b - a}{n}$

For a right Riemann sum, the sample point $x_{k}$ is

$x_{k} = a + k Δ x$

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

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

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

From approximations to integrals

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

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 of $f (x)$ from $a$ to $b$ .” The $d x$ corresponds to the rectangle width $Δ x$ in the limit (an infinitesimally small width).

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).

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

The symbol $\int$ is an elongated S, reminding you 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

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

the expression $Δ x \cdot f (x_{k})$ can be rewritten in terms of $a$ , $b$ , $n$ , and $k$ :

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

Examples

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 $f (x) = sin (x)$ .

First find $Δ x$ :

$Δ x = \frac{4 - 2}{n}$

$= \frac{2}{n}$

Then write the right-endpoint sample point $x_{k}$ :

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

Now substitute into $f (x_{k})$ :

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

So the limit form is

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

Next is the other direction: writing a limit as a definite integral.

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 match each part of the sum to the Riemann-sum form $\sum Δ x f (x_{k})$ .

$\frac{1}{n}$ 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 $x_{k} = a + k Δ x$ , the expression $1 + \frac{k}{n}$ suggests $a = 1$ and $Δ x = \frac{1}{n}$ . That matches the function $f (x) = x^{2}$ .

If $a = 1$ and $b - a = 1$ , then $b = 2$ . One definite integral that produces this sum is

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

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

$Δ x = \frac{1}{n}$

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:

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

So another valid answer is

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

Graphically, this makes sense: $(1 + x)^{2}$ is $x^{2}$ shifted left by $1$ unit. Shifting the bounds left by $1$ unit shifts the entire region the same way, so the area stays the same.

Evaluating definite integrals

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 $x = b$ .

Examples

Evaluate

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

Solution

$A_{1}$ 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

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

Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

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

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$

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:

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

$k = 1 \sum n Δ x \cdot f (x_{k})$

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.

Here, $Δ x$ is the width of each rectangle:

$Δ x = \frac{b - a}{n}$

For a right Riemann sum, the sample point $x_{k}$ is

$x_{k} = a + k Δ x$

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

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

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

From approximations to integrals

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

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 of $f (x)$ from $a$ to $b$ .” The $d x$ corresponds to the rectangle width $Δ x$ in the limit (an infinitesimally small width).

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).

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

The symbol $\int$ is an elongated S, reminding you 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

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

the expression $Δ x \cdot f (x_{k})$ can be rewritten in terms of $a$ , $b$ , $n$ , and $k$ :

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

Examples

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 $f (x) = sin (x)$ .

First find $Δ x$ :

$Δ x = \frac{4 - 2}{n}$

$= \frac{2}{n}$

Then write the right-endpoint sample point $x_{k}$ :

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

Now substitute into $f (x_{k})$ :

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

So the limit form is

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

Next is the other direction: writing a limit as a definite integral.

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 match each part of the sum to the Riemann-sum form $\sum Δ x f (x_{k})$ .

$\frac{1}{n}$ 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 $x_{k} = a + k Δ x$ , the expression $1 + \frac{k}{n}$ suggests $a = 1$ and $Δ x = \frac{1}{n}$ . That matches the function $f (x) = x^{2}$ .

If $a = 1$ and $b - a = 1$ , then $b = 2$ . One definite integral that produces this sum is

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

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

$Δ x = \frac{1}{n}$

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:

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

So another valid answer is

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

Graphically, this makes sense: $(1 + x)^{2}$ is $x^{2}$ shifted left by $1$ unit. Shifting the bounds left by $1$ unit shifts the entire region the same way, so the area stays the same.

Evaluating definite integrals

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 $x = b$ .

Examples

Evaluate

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

Solution

$A_{1}$ 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

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

Evaluate

$\int_{- 1}^{1} ∣ x ∣ d x$

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

$2 \times Area of 1 triangle$

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

$= 1$

Therefore

$\int_{- 1}^{1} ∣ x ∣ d x = 1$