1.8.7 Integration as the limit of a sum (A-level only)

What you'll learn

How rectangles can approximate the area under a graph.
How sigma notation represents a sum of rectangle areas.
Why taking a limit turns an approximation into a definite integral.
How to evaluate and recognise integrals written as limits of sums.

Prerequisites

Before starting, you should be comfortable with:

definite integrals such as $dx\int_a^b f(x)\,dx$ ;
the area of a rectangle: width multiplied by height;
sequences and limits, including the meaning of $n→∞n\to\infty$ ;
sigma notation, written using $∑\sum$ .

Sigma notation

The Greek capital letter sigma, $∑\sum$ , means “add together”.

For example,

∑r=1nr2=12+22+32+⋯+n2.\sum_{r=1}^{n} r^2=1^2+2^2+3^2+\cdots+n^2.

Here:

$r$ is the index: it changes as the terms are added;
$r = 1$ is the starting value;
$n$ is the finishing value;
$r^2$ gives the form of each term.

A useful standard result is

∑r=1nr2=n(n+1)(2n+1)6.\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}.

You may need this when evaluating a limit of a sum.

Example

Expanding sigma notation

Write out the first four terms of

∑r=14(2r+1).\sum_{r=1}^{4}\left(2r+1\right).

Substitute $r = 1, 2, 3, 4$ into the expression $2 r + 1$ :
$9.3,\ 5,\ 7,\ 9.$
Add the terms:
$∑r=14(2r+1)=3+5+7+9=24.\sum_{r=1}^{4}\left(2r+1\right)=3+5+7+9=24.$

Approximating an area with rectangles

Suppose $y = f (x)$ is above the horizontal axis between $x = a$ and $x = b$ . We want the area between the curve, the horizontal axis, and the lines $x = a$ and $x = b$ .

Divide the interval $[a, b]$ into $n$ strips of equal width.

The total interval has width $b - a$ , so each strip has width

Δx=b−an.\Delta x=\frac{b-a}{n}.

The symbol $Δx\Delta x$ means a change, or small increment, in $x$ .

A curve from x equals a to x equals b divided into right-endpoint rectangles, showing that more narrow rectangles give a closer area approximation

If we use the right-hand endpoint of each strip to choose its rectangle height, the right-hand endpoint of strip $r$ is

xr=a+rΔx=a+r(b−a)n.x_r=a+r\Delta x =a+\frac{r(b-a)}{n}.

The rectangle therefore has:

width $Δx\Delta x$ ;
height $f(x_r)$ ;
area $f(xr)Δxf(x_r)\Delta x$ .

Adding the areas of all $n$ rectangles gives

∑r=1nf(xr)Δx.\sum_{r=1}^{n}f(x_r)\Delta x.

Definition

Riemann sum

A Riemann sum is a sum of rectangle areas used to approximate the area under a curve. A right-endpoint Riemann sum is

∑r=1nf(a+r(b−a)n)b−an.\sum_{r=1}^{n}f\left(a+\frac{r(b-a)}{n}\right)\frac{b-a}{n}.

Example

Forming a rectangle sum

Form a right-endpoint rectangle sum for the area under $y=x^2$ between $x = 0$ and $x = 2$ , using $n$ equal strips.

The strip width is
$Δx=2−0n=2n.\Delta x=\frac{2-0}{n}=\frac{2}{n}.$
The right-hand endpoint of strip $r$ is
$xr=0+rΔx=2rn.x_r=0+r\Delta x=\frac{2r}{n}.$
Its height is therefore
$f(xr)=(2rn)2.f(x_r)=\left(\frac{2r}{n}\right)^2.$
Multiply each height by the strip width and sum:
$∑r=1n(2rn)22n=8n3∑r=1nr2.\sum_{r=1}^{n} \left(\frac{2r}{n}\right)^2\frac{2}{n} = \frac{8}{n^3}\sum_{r=1}^{n}r^2.$

Taking the limit

With only a few rectangles, there are visible gaps or overlaps between the rectangles and the curve. As $n$ increases:

the number of rectangles increases;
each width $Δx\Delta x$ decreases;
the sum of their areas becomes a better approximation.

As $n→∞n\to\infty$ , the rectangle widths approach zero. The limit of the rectangle sum gives the exact definite integral.

Key Idea

Integration as a limit of a sum

For a continuous function $f$ ,

dx=lim⁡n→∞∑r=1nf(a+r(b−a)n)b−an.\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{r=1}^{n} f\left(a+\frac{r(b-a)}{n}\right) \frac{b-a}{n}.

The definite integral is therefore the limit of a sum of rectangle areas.

The symbol $d x$ can be connected to the rectangle width $Δx\Delta x$ . In the limiting process, $Δx\Delta x$ tends to zero, and the sum notation becomes integral notation:

dx.\sum f(x_r)\Delta x \quad\longrightarrow\quad \int_a^b f(x)\,dx.

Common Mistake

Forgetting the rectangle width

The expression $∑f(xr)\sum f(x_r)$ adds rectangle heights, not rectangle areas. Every term must be multiplied by the strip width $Δx\Delta x$ .

Evaluating an integral from first principles

Returning to the area under $y=x^2$ from $x = 0$ to $x = 2$ , we found the rectangle sum

8n3∑r=1nr2.\frac{8}{n^3}\sum_{r=1}^{n}r^2.

Taking its limit gives the integral.

Example

Finding an area from a limiting sum

Evaluate $dx\int_0^2x^2\,dx$ by taking the limit of the rectangle sum.

Write the integral as the limit obtained from the rectangles:
$dx=lim⁡n→∞8n3∑r=1nr2.\int_0^2x^2\,dx = \lim_{n\to\infty} \frac{8}{n^3}\sum_{r=1}^{n}r^2.$
Substitute the standard formula for the sum of squares:
$dx=lim⁡n→∞8n3⋅n(n+1)(2n+1)6=lim⁡n→∞4n(n+1)(2n+1)3n3.\begin{aligned} \int_0^2x^2\,dx &= \lim_{n\to\infty} \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}\\ &= \lim_{n\to\infty} \frac{4n(n+1)(2n+1)}{3n^3}. \end{aligned}$
Divide the numerator and denominator by $n^3$ :
$4n(n+1)(2n+1)3n3=43(1+1n)(2+1n).\frac{4n(n+1)(2n+1)}{3n^3} = \frac{4}{3} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right).$
As $n→∞n\to\infty$ , both $1n\frac{1}{n}$ terms approach zero:
$dx=43(1)(2)=83.\int_0^2x^2\,dx = \frac{4}{3}(1)(2) = \frac{8}{3}.$

This agrees with ordinary integration:

dx=[x33]02=83.\int_0^2x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}.

Recognising a limit as an integral

You may also be given a limit of a sum and asked to express it as a definite integral.

The key form to recognise is

dx.\lim_{n\to\infty} \frac{b-a}{n} \sum_{r=1}^{n} f\left(a+\frac{r(b-a)}{n}\right) = \int_a^b f(x)\,dx.

For the special interval $[0, 1]$ , the strip width is $1n\frac{1}{n}$ and the right-hand endpoint is $rn\frac{r}{n}$ . Therefore,

dx.\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1f(x)\,dx.

Example

Converting a limit into an integral

Express the following limit as an integral, and hence evaluate it:

lim⁡n→∞1n∑r=1n(1+rn)3.\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \left(1+\frac{r}{n}\right)^3.

Compare the sum with
$1n∑r=1nf(rn).\frac{1}{n}\sum_{r=1}^{n}f\left(\frac{r}{n}\right).$
Here, $f(x)=(1+x)^3$ and the interval is $[0, 1]$ .
Replace the limiting sum with the corresponding integral:
$dx.\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \left(1+\frac{r}{n}\right)^3 = \int_0^1(1+x)^3\,dx.$
Integrate and apply the limits:
$dx=[(1+x)44]01=244−144=154.\begin{aligned} \int_0^1(1+x)^3\,dx &= \left[\frac{(1+x)^4}{4}\right]_0^1\\ &= \frac{2^4}{4}-\frac{1^4}{4}\\ &= \frac{15}{4}. \end{aligned}$

Tip

Matching the pieces

First identify the factor outside the sum as $Δx\Delta x$ . Then identify the sampled value of $x$ inside the function. Together, these reveal the interval and the integrand.

Signed area

A definite integral represents signed area. Regions above the horizontal axis contribute positively, while regions below it contribute negatively.

This also follows from rectangle sums: if $f(x_r)<0$ , then the rectangle’s signed height and hence its contribution $f(xr)Δxf(x_r)\Delta x$ are negative.

Therefore, the limit-of-a-sum definition still works when the graph crosses or lies below the horizontal axis, but the result may not equal the total geometrical area.

Common Mistake

Assuming every integral is positive

The limit of a rectangle sum gives a signed integral. To find total geometrical area, split the interval at any roots and make the area of each region positive.

Exam technique

In the exam

Identify the strip width $Δx=b−an\Delta x=\frac{b-a}{n}$ before finding any rectangle heights.
Write the sampled point carefully, usually $xr=a+rΔxx_r=a+r\Delta x$ for right endpoints.
Check that each summand has the form “height multiplied by width”.
When recognising a limiting sum, match $Δx\Delta x$ , the interval and the function before integrating.
Keep the limit notation until you have simplified to an expression whose limit you can evaluate.

Self review

Check yourself

How would you form a right-endpoint rectangle sum for the area under $y=xy=\sqrt{x}$ from $x = 1$ to $x = 5$ ?
Which definite integral is represented by $lim⁡n→∞2n∑r=1n(2rn)4\displaystyle \lim_{n\to\infty}\frac{2}{n}\sum_{r=1}^{n}\left(\frac{2r}{n}\right)^4$ ?
Why does increasing $n$ make a rectangle sum a better approximation to the area under a continuous curve?

What you'll learn

How rectangles can approximate the area under a graph.
How sigma notation represents a sum of rectangle areas.
Why taking a limit turns an approximation into a definite integral.
How to evaluate and recognise integrals written as limits of sums.

Prerequisites

Before starting, you should be comfortable with:

definite integrals such as $dx\int_a^b f(x)\,dx$ ;
the area of a rectangle: width multiplied by height;
sequences and limits, including the meaning of $n→∞n\to\infty$ ;
sigma notation, written using $∑\sum$ .

Sigma notation

The Greek capital letter sigma, $∑\sum$ , means “add together”.

For example,

∑r=1nr2=12+22+32+⋯+n2.\sum_{r=1}^{n} r^2=1^2+2^2+3^2+\cdots+n^2.

Here:

$r$ is the index: it changes as the terms are added;
$r = 1$ is the starting value;
$n$ is the finishing value;
$r^2$ gives the form of each term.

A useful standard result is

∑r=1nr2=n(n+1)(2n+1)6.\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}.

You may need this when evaluating a limit of a sum.

Example

Expanding sigma notation

Write out the first four terms of

∑r=14(2r+1).\sum_{r=1}^{4}\left(2r+1\right).

Substitute $r = 1, 2, 3, 4$ into the expression $2 r + 1$ :
$9.3,\ 5,\ 7,\ 9.$
Add the terms:
$∑r=14(2r+1)=3+5+7+9=24.\sum_{r=1}^{4}\left(2r+1\right)=3+5+7+9=24.$

Approximating an area with rectangles

Suppose $y = f (x)$ is above the horizontal axis between $x = a$ and $x = b$ . We want the area between the curve, the horizontal axis, and the lines $x = a$ and $x = b$ .

Divide the interval $[a, b]$ into $n$ strips of equal width.

The total interval has width $b - a$ , so each strip has width

Δx=b−an.\Delta x=\frac{b-a}{n}.

The symbol $Δx\Delta x$ means a change, or small increment, in $x$ .

A curve from x equals a to x equals b divided into right-endpoint rectangles, showing that more narrow rectangles give a closer area approximation

If we use the right-hand endpoint of each strip to choose its rectangle height, the right-hand endpoint of strip $r$ is

xr=a+rΔx=a+r(b−a)n.x_r=a+r\Delta x =a+\frac{r(b-a)}{n}.

The rectangle therefore has:

width $Δx\Delta x$ ;
height $f(x_r)$ ;
area $f(xr)Δxf(x_r)\Delta x$ .

Adding the areas of all $n$ rectangles gives

∑r=1nf(xr)Δx.\sum_{r=1}^{n}f(x_r)\Delta x.

Definition

Riemann sum

A Riemann sum is a sum of rectangle areas used to approximate the area under a curve. A right-endpoint Riemann sum is

∑r=1nf(a+r(b−a)n)b−an.\sum_{r=1}^{n}f\left(a+\frac{r(b-a)}{n}\right)\frac{b-a}{n}.

Example

Forming a rectangle sum

Form a right-endpoint rectangle sum for the area under $y=x^2$ between $x = 0$ and $x = 2$ , using $n$ equal strips.

The strip width is
$Δx=2−0n=2n.\Delta x=\frac{2-0}{n}=\frac{2}{n}.$
The right-hand endpoint of strip $r$ is
$xr=0+rΔx=2rn.x_r=0+r\Delta x=\frac{2r}{n}.$
Its height is therefore
$f(xr)=(2rn)2.f(x_r)=\left(\frac{2r}{n}\right)^2.$
Multiply each height by the strip width and sum:
$∑r=1n(2rn)22n=8n3∑r=1nr2.\sum_{r=1}^{n} \left(\frac{2r}{n}\right)^2\frac{2}{n} = \frac{8}{n^3}\sum_{r=1}^{n}r^2.$

Taking the limit

With only a few rectangles, there are visible gaps or overlaps between the rectangles and the curve. As $n$ increases:

the number of rectangles increases;
each width $Δx\Delta x$ decreases;
the sum of their areas becomes a better approximation.

As $n→∞n\to\infty$ , the rectangle widths approach zero. The limit of the rectangle sum gives the exact definite integral.

Key Idea

Integration as a limit of a sum

For a continuous function $f$ ,

dx=lim⁡n→∞∑r=1nf(a+r(b−a)n)b−an.\int_a^b f(x)\,dx = \lim_{n\to\infty} \sum_{r=1}^{n} f\left(a+\frac{r(b-a)}{n}\right) \frac{b-a}{n}.

The definite integral is therefore the limit of a sum of rectangle areas.

The symbol $d x$ can be connected to the rectangle width $Δx\Delta x$ . In the limiting process, $Δx\Delta x$ tends to zero, and the sum notation becomes integral notation:

dx.\sum f(x_r)\Delta x \quad\longrightarrow\quad \int_a^b f(x)\,dx.

Common Mistake

Forgetting the rectangle width

The expression $∑f(xr)\sum f(x_r)$ adds rectangle heights, not rectangle areas. Every term must be multiplied by the strip width $Δx\Delta x$ .

Evaluating an integral from first principles

Returning to the area under $y=x^2$ from $x = 0$ to $x = 2$ , we found the rectangle sum

8n3∑r=1nr2.\frac{8}{n^3}\sum_{r=1}^{n}r^2.

Taking its limit gives the integral.

Example

Finding an area from a limiting sum

Evaluate $dx\int_0^2x^2\,dx$ by taking the limit of the rectangle sum.

Write the integral as the limit obtained from the rectangles:
$dx=lim⁡n→∞8n3∑r=1nr2.\int_0^2x^2\,dx = \lim_{n\to\infty} \frac{8}{n^3}\sum_{r=1}^{n}r^2.$
Substitute the standard formula for the sum of squares:
$dx=lim⁡n→∞8n3⋅n(n+1)(2n+1)6=lim⁡n→∞4n(n+1)(2n+1)3n3.\begin{aligned} \int_0^2x^2\,dx &= \lim_{n\to\infty} \frac{8}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}\\ &= \lim_{n\to\infty} \frac{4n(n+1)(2n+1)}{3n^3}. \end{aligned}$
Divide the numerator and denominator by $n^3$ :
$4n(n+1)(2n+1)3n3=43(1+1n)(2+1n).\frac{4n(n+1)(2n+1)}{3n^3} = \frac{4}{3} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right).$
As $n→∞n\to\infty$ , both $1n\frac{1}{n}$ terms approach zero:
$dx=43(1)(2)=83.\int_0^2x^2\,dx = \frac{4}{3}(1)(2) = \frac{8}{3}.$

This agrees with ordinary integration:

dx=[x33]02=83.\int_0^2x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}.

Recognising a limit as an integral

You may also be given a limit of a sum and asked to express it as a definite integral.

The key form to recognise is

dx.\lim_{n\to\infty} \frac{b-a}{n} \sum_{r=1}^{n} f\left(a+\frac{r(b-a)}{n}\right) = \int_a^b f(x)\,dx.

For the special interval $[0, 1]$ , the strip width is $1n\frac{1}{n}$ and the right-hand endpoint is $rn\frac{r}{n}$ . Therefore,

dx.\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1f(x)\,dx.

Example

Converting a limit into an integral

Express the following limit as an integral, and hence evaluate it:

lim⁡n→∞1n∑r=1n(1+rn)3.\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \left(1+\frac{r}{n}\right)^3.

Compare the sum with
$1n∑r=1nf(rn).\frac{1}{n}\sum_{r=1}^{n}f\left(\frac{r}{n}\right).$
Here, $f(x)=(1+x)^3$ and the interval is $[0, 1]$ .
Replace the limiting sum with the corresponding integral:
$dx.\lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{n} \left(1+\frac{r}{n}\right)^3 = \int_0^1(1+x)^3\,dx.$
Integrate and apply the limits:
$dx=[(1+x)44]01=244−144=154.\begin{aligned} \int_0^1(1+x)^3\,dx &= \left[\frac{(1+x)^4}{4}\right]_0^1\\ &= \frac{2^4}{4}-\frac{1^4}{4}\\ &= \frac{15}{4}. \end{aligned}$

Tip

Matching the pieces

First identify the factor outside the sum as $Δx\Delta x$ . Then identify the sampled value of $x$ inside the function. Together, these reveal the interval and the integrand.

Signed area

A definite integral represents signed area. Regions above the horizontal axis contribute positively, while regions below it contribute negatively.

This also follows from rectangle sums: if $f(x_r)<0$ , then the rectangle’s signed height and hence its contribution $f(xr)Δxf(x_r)\Delta x$ are negative.

Therefore, the limit-of-a-sum definition still works when the graph crosses or lies below the horizontal axis, but the result may not equal the total geometrical area.

Common Mistake

Assuming every integral is positive

The limit of a rectangle sum gives a signed integral. To find total geometrical area, split the interval at any roots and make the area of each region positive.

Exam technique

In the exam

Identify the strip width $Δx=b−an\Delta x=\frac{b-a}{n}$ before finding any rectangle heights.
Write the sampled point carefully, usually $xr=a+rΔxx_r=a+r\Delta x$ for right endpoints.
Check that each summand has the form “height multiplied by width”.
When recognising a limiting sum, match $Δx\Delta x$ , the interval and the function before integrating.
Keep the limit notation until you have simplified to an expression whose limit you can evaluate.

Self review

Check yourself

How would you form a right-endpoint rectangle sum for the area under $y=xy=\sqrt{x}$ from $x = 1$ to $x = 5$ ?
Which definite integral is represented by $lim⁡n→∞2n∑r=1n(2rn)4\displaystyle \lim_{n\to\infty}\frac{2}{n}\sum_{r=1}^{n}\left(\frac{2r}{n}\right)^4$ ?
Why does increasing $n$ make a rectangle sum a better approximation to the area under a continuous curve?

What you'll learn

Prerequisites

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from first principles

Recognising a limit as an integral

Signed area

Review 1.8.7 Integration as the limit of a sum (A-level only) by teaching Genie

1.8.7 Integration as the limit of a sum (A-level only)

What you'll learn

Prerequisites

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from first principles

Recognising a limit as an integral

Signed area

Review 1.8.7 Integration as the limit of a sum (A-level only) by teaching Genie

1.8.7 Integration as the limit of a sum (A-level only)

What you'll learn

Prerequisites

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from first principles

Recognising a limit as an integral

Signed area

Review 1.8.7 Integration as the limit of a sum (A-level only) by teaching Genie

1.8.7 Integration as the limit of a sum (A-level only) Revision Guide

1.8.7 Integration as the limit of a sum (A-level only)

What you'll learn

Prerequisites

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from first principles

Recognising a limit as an integral

Signed area

Review 1.8.7 Integration as the limit of a sum (A-level only) by teaching Genie

1.8.7 Integration as the limit of a sum (A-level only) Revision Guide