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

What you'll learn

How rectangles can approximate the area under a curve.
How to write these approximations using sigma notation.
Why taking a limit gives the exact value of a definite integral.
How to evaluate and recognise limits of sums in exam questions.

Prerequisites

Definite integrals

A definite integral gives the signed area between a graph and the horizontal axis over a specified interval.

If $f(x)≥0f(x)\geq 0$ between $x = a$ and $x = b$ , then

dx\int_a^b f(x)\,dx

is the area under the curve $y = f (x)$ from $x = a$ to $x = b$ .

You already know that this can be evaluated using an antiderivative:

dx=[F(x)]ab=F(b)−F(a),\int_a^b f(x)\,dx=[F(x)]_a^b=F(b)-F(a),

where $F^{'} (x) = f (x)$ .

In this topic, you will see where this integral comes from.

Sigma notation

The Greek capital letter $∑\sum$ , read as “sum”, is used to add a sequence of terms.

For example,

∑r=1nr=1+2+3+⋯+n.\sum_{r=1}^{n}r=1+2+3+\cdots+n.

Here, $r$ is the index: it changes from its starting value to its finishing value.

You should know these standard results:

∑r=1n1=n,\sum_{r=1}^{n}1=n,

∑r=1nr=n(n+1)2,\sum_{r=1}^{n}r=\frac{n(n+1)}{2},

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

and, when required,

∑r=1nr3=(n(n+1)2)2.\sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}{2}\right)^2.

Example

Evaluating a sum of squares

Evaluate $∑r=1n(3r2+2r)\displaystyle \sum_{r=1}^{n}(3r^2+2r)$ .

Split the sum using the fact that sums are linear:
$∑r=1n(3r2+2r)=3∑r=1nr2+2∑r=1nr.\sum_{r=1}^{n}(3r^2+2r) =3\sum_{r=1}^{n}r^2+2\sum_{r=1}^{n}r.$
Substitute the standard formulae:
$=3(n(n+1)(2n+1)6)+2(n(n+1)2).=3\left(\frac{n(n+1)(2n+1)}{6}\right) +2\left(\frac{n(n+1)}{2}\right).$
Simplify:
$=n(n+1)(2n+1)2+n(n+1)=n(n+1)(2n+3)2.\begin{aligned} &=\frac{n(n+1)(2n+1)}{2}+n(n+1)\\ &=\frac{n(n+1)(2n+3)}{2}. \end{aligned}$

Approximating an area with rectangles

Suppose you want the area under $y = f (x)$ from $x = a$ to $x = b$ .

Divide the interval into $n$ equal strips. The width of each strip is

Δx=b−an.\Delta x=\frac{b-a}{n}.

The symbol $Δx\Delta x$ means “change in $x$ ”.

If the height of each rectangle is taken at the right-hand end of its strip, the right 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 area of rectangle $r$ is therefore approximately

f(xr)Δx.f(x_r)\Delta x.

Adding all the rectangles gives the approximation

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

Definition

Riemann sum

A Riemann sum is a sum of rectangular areas used to approximate a definite integral, such as $∑r=1nf(xr)Δx\displaystyle \sum_{r=1}^{n}f(x_r)\Delta x$ .

Right-endpoint rectangles approximating the area under the curve y equals x squared

For an increasing function, right-endpoint rectangles usually give an overestimate. Left-endpoint rectangles usually give an underestimate. Both become more accurate as the strips become thinner.

Key Idea

More strips, better approximation

As $n$ increases, the strip width $Δx\Delta x$ approaches zero and the tops of the rectangles fit the curve more closely.

Taking the limit

The notation $n→∞n\to\infty$ means that the number of strips increases without bound.

As this happens,

Δx=b−an→0.\Delta x=\frac{b-a}{n}\to 0.

The limiting value of the Riemann sum is the exact definite integral:

dx=lim⁡n→∞∑r=1nf(a+r(b−a)n)b−an\boxed{ \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} }

Definition

Integration as the limit of a sum

Integration as the limit of a sum means defining a definite integral as the limiting value of a Riemann sum when the number of strips tends to infinity.

Common Mistake

Forgetting the strip width

The summand must include the rectangle width $Δx\Delta x$ . Summing only the heights does not approximate an area.

Evaluating an integral from its limiting sum

To evaluate a limit of a sum, you usually:

write the rectangle width and endpoint;
expand the summand;
apply standard sigma formulae;
simplify before taking the limit.

Example

Finding the area under a quadratic

Use the limit of a sum to evaluate $dx\displaystyle \int_0^2 x^2\,dx$ .

Divide the interval from zero to two into $n$ equal strips. Then
$Δx=2n,xr=2rn.\Delta x=\frac{2}{n}, \qquad x_r=\frac{2r}{n}.$
Form the right-endpoint Riemann sum:
$dx=lim⁡n→∞∑r=1n(2rn)22n.\int_0^2x^2\,dx = \lim_{n\to\infty} \sum_{r=1}^{n} \left(\frac{2r}{n}\right)^2\frac{2}{n}.$
Simplify the summand and take constants outside the sum:
$=lim⁡n→∞8n3∑r=1nr2.=\lim_{n\to\infty} \frac{8}{n^3}\sum_{r=1}^{n}r^2.$
Use $∑r=1nr2=n(n+1)(2n+1)6\displaystyle \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}$ :
$=lim⁡n→∞8n3⋅n(n+1)(2n+1)6.=\lim_{n\to\infty} \frac{8}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}.$
Divide numerator and denominator by $n^3$ :
$=lim⁡n→∞4n(n+1)(2n+1)3n3=lim⁡n→∞43(1+1n)(2+1n).\begin{aligned} &=\lim_{n\to\infty} \frac{4n(n+1)(2n+1)}{3n^3}\\ &=\lim_{n\to\infty} \frac{4}{3} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right). \end{aligned}$
As $n→∞n\to\infty$ , both $1n\frac{1}{n}$ terms tend to zero:
$dx=43×1×2=83.\int_0^2x^2\,dx =\frac{4}{3}\times 1\times 2 =\frac{8}{3}.$

Tip

Taking the limit efficiently

Rewrite expressions using powers of $1n\frac{1}{n}$ . Every positive power of $1n\frac{1}{n}$ tends to zero as $n→∞n\to\infty$ .

Recognising an integral from a sum

You may also need to work backwards and identify the integral represented by a limiting sum.

Compare the sum with

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

Look for a factor behaving like the strip width and an expression behaving like the sample point.

Example

Converting a limit into an integral

Express the following limit as a definite integral:

lim⁡n→∞∑r=1n(1+3rn)23n.\lim_{n\to\infty} \sum_{r=1}^{n} \left(1+\frac{3r}{n}\right)^2\frac{3}{n}.

Identify the strip width:
$Δx=3n.\Delta x=\frac{3}{n}.$
Therefore, the total interval length is $nΔx=3n\Delta x=3$ .
Identify the right endpoint:
$xr=1+3rn.x_r=1+\frac{3r}{n}.$
This has the form $a+rΔxa+r\Delta x$ , so the lower limit is $a = 1$ .
Add the interval length to the lower limit to obtain the upper limit:
$b = 1 + 3 = 4.$
The rectangle height is $f(x_r)=x_r^2$ , so $f(x)=x^2$ . Hence,
$dx.\lim_{n\to\infty} \sum_{r=1}^{n} \left(1+\frac{3r}{n}\right)^2\frac{3}{n} = \int_1^4x^2\,dx.$

Common Mistake

Misreading the interval length

In a term such as $3n\frac{3}{n}$ , the numerator gives $b - a$ , not necessarily the upper limit $b$ . You must also identify the starting value $a$ .

Why the method matters

Integration as the limit of a sum connects two viewpoints:

Geometry: adding the areas of increasingly thin rectangles.
Algebra: evaluating a limit involving sigma notation.

The rectangles only approximate the curved region for finite $n$ . The exact area appears only after taking the limit as $n→∞n\to\infty$ .

Key Idea

The central result

A definite integral is the exact limiting value of a sum of increasingly thin rectangular areas.

Exam technique

In the exam

Write $Δx=b−an\Delta x=\frac{b-a}{n}$ and $xr=a+rΔxx_r=a+r\Delta x$ before forming the sum.
Keep the factor $Δx\Delta x$ visible until you have written the complete Riemann sum.
Use the standard formulae for $∑r\sum r$ , $∑r2\sum r^2$ and $∑r3\sum r^3$ , then simplify fully before taking the limit.
When working backwards, identify $Δx\Delta x$ , the starting value $a$ , the interval length $b - a$ , and the function $f (x)$ separately.

Self review

Check yourself

How would you form a right-endpoint Riemann sum for $dx\displaystyle \int_1^3 f(x)\,dx$ ?
Can you use a limit of a sum to evaluate $dx\displaystyle \int_0^1 x^2\,dx$ ?
Which definite integral is represented by $lim⁡n→∞∑r=1n(2+4rn)4n\displaystyle \lim_{n\to\infty}\sum_{r=1}^{n}\left(2+\frac{4r}{n}\right)\frac{4}{n}$ ?

What you'll learn

How rectangles can approximate the area under a curve.
How to write these approximations using sigma notation.
Why taking a limit gives the exact value of a definite integral.
How to evaluate and recognise limits of sums in exam questions.

Prerequisites

Definite integrals

A definite integral gives the signed area between a graph and the horizontal axis over a specified interval.

If $f(x)≥0f(x)\geq 0$ between $x = a$ and $x = b$ , then

dx\int_a^b f(x)\,dx

is the area under the curve $y = f (x)$ from $x = a$ to $x = b$ .

You already know that this can be evaluated using an antiderivative:

dx=[F(x)]ab=F(b)−F(a),\int_a^b f(x)\,dx=[F(x)]_a^b=F(b)-F(a),

where $F^{'} (x) = f (x)$ .

In this topic, you will see where this integral comes from.

Sigma notation

The Greek capital letter $∑\sum$ , read as “sum”, is used to add a sequence of terms.

For example,

∑r=1nr=1+2+3+⋯+n.\sum_{r=1}^{n}r=1+2+3+\cdots+n.

Here, $r$ is the index: it changes from its starting value to its finishing value.

You should know these standard results:

∑r=1n1=n,\sum_{r=1}^{n}1=n,

∑r=1nr=n(n+1)2,\sum_{r=1}^{n}r=\frac{n(n+1)}{2},

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

and, when required,

∑r=1nr3=(n(n+1)2)2.\sum_{r=1}^{n}r^3=\left(\frac{n(n+1)}{2}\right)^2.

Example

Evaluating a sum of squares

Evaluate $∑r=1n(3r2+2r)\displaystyle \sum_{r=1}^{n}(3r^2+2r)$ .

Split the sum using the fact that sums are linear:
$∑r=1n(3r2+2r)=3∑r=1nr2+2∑r=1nr.\sum_{r=1}^{n}(3r^2+2r) =3\sum_{r=1}^{n}r^2+2\sum_{r=1}^{n}r.$
Substitute the standard formulae:
$=3(n(n+1)(2n+1)6)+2(n(n+1)2).=3\left(\frac{n(n+1)(2n+1)}{6}\right) +2\left(\frac{n(n+1)}{2}\right).$
Simplify:
$=n(n+1)(2n+1)2+n(n+1)=n(n+1)(2n+3)2.\begin{aligned} &=\frac{n(n+1)(2n+1)}{2}+n(n+1)\\ &=\frac{n(n+1)(2n+3)}{2}. \end{aligned}$

Approximating an area with rectangles

Suppose you want the area under $y = f (x)$ from $x = a$ to $x = b$ .

Divide the interval into $n$ equal strips. The width of each strip is

Δx=b−an.\Delta x=\frac{b-a}{n}.

The symbol $Δx\Delta x$ means “change in $x$ ”.

If the height of each rectangle is taken at the right-hand end of its strip, the right 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 area of rectangle $r$ is therefore approximately

f(xr)Δx.f(x_r)\Delta x.

Adding all the rectangles gives the approximation

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

Definition

Riemann sum

A Riemann sum is a sum of rectangular areas used to approximate a definite integral, such as $∑r=1nf(xr)Δx\displaystyle \sum_{r=1}^{n}f(x_r)\Delta x$ .

Right-endpoint rectangles approximating the area under the curve y equals x squared

For an increasing function, right-endpoint rectangles usually give an overestimate. Left-endpoint rectangles usually give an underestimate. Both become more accurate as the strips become thinner.

Key Idea

More strips, better approximation

As $n$ increases, the strip width $Δx\Delta x$ approaches zero and the tops of the rectangles fit the curve more closely.

Taking the limit

The notation $n→∞n\to\infty$ means that the number of strips increases without bound.

As this happens,

Δx=b−an→0.\Delta x=\frac{b-a}{n}\to 0.

The limiting value of the Riemann sum is the exact definite integral:

dx=lim⁡n→∞∑r=1nf(a+r(b−a)n)b−an\boxed{ \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} }

Definition

Integration as the limit of a sum

Integration as the limit of a sum means defining a definite integral as the limiting value of a Riemann sum when the number of strips tends to infinity.

Common Mistake

Forgetting the strip width

The summand must include the rectangle width $Δx\Delta x$ . Summing only the heights does not approximate an area.

Evaluating an integral from its limiting sum

To evaluate a limit of a sum, you usually:

write the rectangle width and endpoint;
expand the summand;
apply standard sigma formulae;
simplify before taking the limit.

Example

Finding the area under a quadratic

Use the limit of a sum to evaluate $dx\displaystyle \int_0^2 x^2\,dx$ .

Divide the interval from zero to two into $n$ equal strips. Then
$Δx=2n,xr=2rn.\Delta x=\frac{2}{n}, \qquad x_r=\frac{2r}{n}.$
Form the right-endpoint Riemann sum:
$dx=lim⁡n→∞∑r=1n(2rn)22n.\int_0^2x^2\,dx = \lim_{n\to\infty} \sum_{r=1}^{n} \left(\frac{2r}{n}\right)^2\frac{2}{n}.$
Simplify the summand and take constants outside the sum:
$=lim⁡n→∞8n3∑r=1nr2.=\lim_{n\to\infty} \frac{8}{n^3}\sum_{r=1}^{n}r^2.$
Use $∑r=1nr2=n(n+1)(2n+1)6\displaystyle \sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}$ :
$=lim⁡n→∞8n3⋅n(n+1)(2n+1)6.=\lim_{n\to\infty} \frac{8}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}.$
Divide numerator and denominator by $n^3$ :
$=lim⁡n→∞4n(n+1)(2n+1)3n3=lim⁡n→∞43(1+1n)(2+1n).\begin{aligned} &=\lim_{n\to\infty} \frac{4n(n+1)(2n+1)}{3n^3}\\ &=\lim_{n\to\infty} \frac{4}{3} \left(1+\frac{1}{n}\right) \left(2+\frac{1}{n}\right). \end{aligned}$
As $n→∞n\to\infty$ , both $1n\frac{1}{n}$ terms tend to zero:
$dx=43×1×2=83.\int_0^2x^2\,dx =\frac{4}{3}\times 1\times 2 =\frac{8}{3}.$

Tip

Taking the limit efficiently

Rewrite expressions using powers of $1n\frac{1}{n}$ . Every positive power of $1n\frac{1}{n}$ tends to zero as $n→∞n\to\infty$ .

Recognising an integral from a sum

You may also need to work backwards and identify the integral represented by a limiting sum.

Compare the sum with

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

Look for a factor behaving like the strip width and an expression behaving like the sample point.

Example

Converting a limit into an integral

Express the following limit as a definite integral:

lim⁡n→∞∑r=1n(1+3rn)23n.\lim_{n\to\infty} \sum_{r=1}^{n} \left(1+\frac{3r}{n}\right)^2\frac{3}{n}.

Identify the strip width:
$Δx=3n.\Delta x=\frac{3}{n}.$
Therefore, the total interval length is $nΔx=3n\Delta x=3$ .
Identify the right endpoint:
$xr=1+3rn.x_r=1+\frac{3r}{n}.$
This has the form $a+rΔxa+r\Delta x$ , so the lower limit is $a = 1$ .
Add the interval length to the lower limit to obtain the upper limit:
$b = 1 + 3 = 4.$
The rectangle height is $f(x_r)=x_r^2$ , so $f(x)=x^2$ . Hence,
$dx.\lim_{n\to\infty} \sum_{r=1}^{n} \left(1+\frac{3r}{n}\right)^2\frac{3}{n} = \int_1^4x^2\,dx.$

Common Mistake

Misreading the interval length

In a term such as $3n\frac{3}{n}$ , the numerator gives $b - a$ , not necessarily the upper limit $b$ . You must also identify the starting value $a$ .

Why the method matters

Integration as the limit of a sum connects two viewpoints:

Geometry: adding the areas of increasingly thin rectangles.
Algebra: evaluating a limit involving sigma notation.

The rectangles only approximate the curved region for finite $n$ . The exact area appears only after taking the limit as $n→∞n\to\infty$ .

Key Idea

The central result

A definite integral is the exact limiting value of a sum of increasingly thin rectangular areas.

Exam technique

In the exam

Write $Δx=b−an\Delta x=\frac{b-a}{n}$ and $xr=a+rΔxx_r=a+r\Delta x$ before forming the sum.
Keep the factor $Δx\Delta x$ visible until you have written the complete Riemann sum.
Use the standard formulae for $∑r\sum r$ , $∑r2\sum r^2$ and $∑r3\sum r^3$ , then simplify fully before taking the limit.
When working backwards, identify $Δx\Delta x$ , the starting value $a$ , the interval length $b - a$ , and the function $f (x)$ separately.

Self review

Check yourself

How would you form a right-endpoint Riemann sum for $dx\displaystyle \int_1^3 f(x)\,dx$ ?
Can you use a limit of a sum to evaluate $dx\displaystyle \int_0^1 x^2\,dx$ ?
Which definite integral is represented by $lim⁡n→∞∑r=1n(2+4rn)4n\displaystyle \lim_{n\to\infty}\sum_{r=1}^{n}\left(2+\frac{4r}{n}\right)\frac{4}{n}$ ?

What you'll learn

Prerequisites

Definite integrals

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from its limiting sum

Recognising an integral from a sum

Why the method matters

Review 1.11.4 Integration as the limit of a sum (A-level only) by teaching Genie

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

What you'll learn

Prerequisites

Definite integrals

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from its limiting sum

Recognising an integral from a sum

Why the method matters

Review 1.11.4 Integration as the limit of a sum (A-level only) by teaching Genie

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

What you'll learn

Prerequisites

Definite integrals

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from its limiting sum

Recognising an integral from a sum

Why the method matters

Review 1.11.4 Integration as the limit of a sum (A-level only) by teaching Genie

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

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

What you'll learn

Prerequisites

Definite integrals

Sigma notation

Approximating an area with rectangles

Taking the limit

Evaluating an integral from its limiting sum

Recognising an integral from a sum

Why the method matters

Review 1.11.4 Integration as the limit of a sum (A-level only) by teaching Genie

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