Integration as the limit of a sum

A definite integral is the exact limit of a sum of thin rectangular strips. For a continuous function $f$ on $[a,b]$,

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

This formula connects finite sums, limits, areas and integration. It also explains the notation $\mathrm dx$: each term has the form

$$\text{height}\times\text{small width}.$$

Before you begin

You should be able to:

• use sigma notation;
• evaluate simple limits and manipulate algebraic expressions;
• use the standard sums $\sum r$ and $\sum r^2$;
• interpret a definite integral as signed area.

From rectangles to an integral

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

$$\Delta x=\frac{b-a}{n}.$$

The right endpoint of strip $r$ is

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

Using $f(x_r)$ as the height, the area of this rectangle is

$$f(x_r)\Delta x.$$

Adding all $n$ rectangles gives the right endpoint sum

$$S_n=\sum_{r=1}^{n}f(x_r)\Delta x =\sum_{r=1}^{n} f\left(a+\frac{r(b-a)}{n}\right)\frac{b-a}{n}.$$

For finite $n$, curved pieces remain between the rectangles and the graph. As $n$ increases, $\Delta x\to0$ and the approximation approaches a fixed value. The integral is defined to be that limit:

$$\int_a^b f(x)\,\mathrm dx=\lim_{n\to\infty}S_n.$$

The number of rectangles tends to infinity, but their widths tend to zero. It is the balance between these two changes that produces a finite limit.

Worked example 1: write a rectangle sum

Write the right endpoint sum with $n$ equal strips for

$$\int_2^5 (x^2+1)\,\mathrm dx.$$

First find the width:

$$\Delta x=\frac{5-2}{n}=\frac3n.$$

The right endpoint of strip $r$ is

$$x_r=2+\frac{3r}{n}.$$

Substitute this into the function and multiply by the width:

$$\boxed{S_n=\sum_{r=1}^{n} \left[\left(2+\frac{3r}{n}\right)^2+1\right]\frac3n.}$$

Therefore

$$\int_2^5(x^2+1)\,\mathrm dx=\lim_{n\to\infty}S_n.$$

Do not omit the factor $3/n$. The function value is a height, not an area.

Left, right and midpoint choices

The sample point can be chosen differently within each strip.

Method	Sample point in strip $r$
left endpoint	$a+(r-1)\Delta x$
right endpoint	$a+r\Delta x$
midpoint	$a+\left(r-\frac12\right)\Delta x$

For a continuous function, all three sums approach the same integral as $n\to\infty$. Their finite approximations need not be equal.

If $f$ is increasing and positive, left endpoint rectangles lie below the curve and right endpoint rectangles extend above it. Hence

$$L_n<\int_a^b f(x)\,\mathrm dx<R_n.$$

The inequalities reverse in the appropriate way for a decreasing function. Always reason from the graph rather than memorising one rule for every situation.

Worked example 2: compare finite sums

Estimate $\int_0^2 x^2\,\mathrm dx$ using four equal strips and left and right endpoints.

Here

$$\Delta x=\frac{2-0}{4}=\frac12.$$

The left endpoints are $0,\frac12,1,\frac32$, so

$$\begin{aligned} L_4 &=\frac12\left[0^2+\left(\frac12\right)^2+1^2+\left(\frac32\right)^2\right]\\ &=\frac12\left(0+\frac14+1+\frac94\right)\\ &=\frac74. \end{aligned}$$

The right endpoints are $\frac12,1,\frac32,2$, giving

$$\begin{aligned} R_4 &=\frac12\left[\left(\frac12\right)^2+1^2+\left(\frac32\right)^2+2^2\right]\\ &=\frac12\left(\frac14+1+\frac94+4\right)\\ &=\frac{15}{4}. \end{aligned}$$

Since $x^2$ is increasing on $[0,2]$,

$$\frac74<\int_0^2x^2\,\mathrm dx<\frac{15}{4}.$$

The exact value is $8/3$, which lies between these bounds.

Evaluating a limit of sums exactly

The standard formulae

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

turn many Riemann sums into algebraic expressions in $n$. Simplify before taking the limit.

Worked example 3: derive $\boldsymbol{\int_0^3 x\,\mathrm dx}$

Use $n$ right endpoint rectangles. Their width is

$$\Delta x=\frac3n,$$

and the $r$th right endpoint is $x_r=3r/n$. Therefore

$$\begin{aligned} S_n &=\sum_{r=1}^{n}\frac{3r}{n}\frac3n\\ &=\frac9{n^2}\sum_{r=1}^{n}r\\ &=\frac9{n^2}\cdot\frac{n(n+1)}2\\ &=\frac92\left(1+\frac1n\right). \end{aligned}$$

Now let $n\to\infty$. Since $1/n\to0$,

$$\boxed{\int_0^3x\,\mathrm dx =\lim_{n\to\infty}S_n =\frac92.}$$

This agrees with the triangular area $\frac12\times3\times3=9/2$.

Worked example 4: a quadratic on a shifted interval

Use a limit of sums to evaluate

$$\int_1^2 x^2\,\mathrm dx.$$

The strip width and right endpoints are

$$\Delta x=\frac1n, \qquad x_r=1+\frac rn.$$

Thus

$$\begin{aligned} S_n &=\sum_{r=1}^{n}\left(1+\frac rn\right)^2\frac1n\\ &=\frac1n\sum_{r=1}^{n} \left(1+\frac{2r}{n}+\frac{r^2}{n^2}\right)\\ &=\frac1n\sum_{r=1}^{n}1 +\frac2{n^2}\sum_{r=1}^{n}r +\frac1{n^3}\sum_{r=1}^{n}r^2. \end{aligned}$$

Insert the standard sums:

$$S_n =1+\frac{n+1}{n} +\frac{n(n+1)(2n+1)}{6n^3}.$$

Taking the limit term by term gives

$$\begin{aligned} \int_1^2x^2\,\mathrm dx &=1+1+\frac{1\cdot1\cdot2}{6}\\ &=\boxed{\frac73}. \end{aligned}$$

In the last fraction, divide numerator and denominator by $n^3$ before taking the limit. Substituting $n=\infty$ is not valid algebra.

Recognising an integral hidden in a limit

A sum may be presented without an integral. Compare it with

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

Look for three ingredients:

1. a width of the form $(b-a)/n$;
2. a sample point of the form $a+r(b-a)/n$;
3. a function evaluated at that sample point.

Worked example 5: convert a sum into an integral

Evaluate

$$\lim_{n\to\infty}\frac2n \sum_{r=1}^{n}\sqrt{1+\frac{2r}{n}}.$$

The width is $2/n$, so choose an interval of length $2$. The sample points are

$$1+\frac{2r}{n},$$

which are right endpoints on $[1,3]$. The function is $f(x)=\sqrt{x}$. Hence

$$\lim_{n\to\infty}\frac2n \sum_{r=1}^{n}\sqrt{1+\frac{2r}{n}} =\int_1^3\sqrt{x}\,\mathrm dx.$$

Now integrate:

$$\begin{aligned} \int_1^3x^{1/2}\,\mathrm dx &=\left[\frac23x^{3/2}\right]_1^3\\ &=\boxed{2\sqrt3-\frac23}. \end{aligned}$$

Signed area, not always geometric area

If $f(x)<0$, then $f(x_r)\Delta x$ is negative. The limit therefore measures signed area:

$$\int_a^b f(x)\,\mathrm dx =\text{area above the axis}-\text{area below the axis}.$$

For example,

$$\int_{-1}^{1}x\,\mathrm dx=0$$

because the negative and positive contributions cancel. The total geometric area is not zero. To find that, split the interval where the graph crosses the axis and make each area positive.

Common misconceptions

The integral is not one rectangle

The expression $f(x)\,\mathrm dx$ suggests a typical thin contribution, but $\mathrm dx$ is not an ordinary fixed width. The definite integral represents the limit of the whole sum of contributions.

• Forgetting the width: $\sum f(x_r)$ adds heights. A Riemann sum adds $f(x_r)\Delta x$.
• Using the wrong endpoint: with $r=1,\ldots,n$, $a+r\Delta x$ gives right endpoints. Left endpoints use $a+(r-1)\Delta x$.
• Treating a finite sum as exact: rectangles generally approximate the integral until the limit is taken.
• Assuming every integral is positive: values below the $x$ axis contribute negatively.
• Replacing $n$ by infinity: infinity is not a number. Simplify the expression, then use limit laws such as $1/n\to0$.

Self check

1. Write a right endpoint sum for $\int_1^4(2x-3)\,\mathrm dx$ using $n$ equal strips.
2. For an increasing function on $[a,b]$, which is larger: its left endpoint sum or its right endpoint sum?
3. Evaluate $\displaystyle \lim_{n\to\infty}\frac1{n^2}\sum_{r=1}^{n}r$.
4. Express $\displaystyle \lim_{n\to\infty}\frac4n\sum_{r=1}^{n}\left(2+\frac{4r}{n}\right)^2$ as a definite integral.
5. Why can $\int_{-2}^{2}x^3\,\mathrm dx$ equal zero even though the graph encloses regions with the axis?

Answers

1. Here $\Delta x=3/n$ and $x_r=1+3r/n$, so

   $$\sum_{r=1}^{n}\left[2\left(1+\frac{3r}{n}\right)-3\right]\frac3n.$$

2. The right endpoint sum is larger. On every strip, its rectangle uses the greater function value.

3. Using $\sum r=n(n+1)/2$,

   $$\lim_{n\to\infty}\frac{n(n+1)}{2n^2} =\frac12.$$

4. The width is $4/n$, the interval is $[2,6]$, and $f(x)=x^2$. Therefore the limit is

   $$\int_2^6x^2\,\mathrm dx.$$

5. The function $x^3$ is odd. The negative signed area on $[-2,0]$ cancels the equal positive signed area on $[0,2]$.

What to learn next

• Practise finding values and interpreting signs in definite integrals and areas.
• Review antiderivatives in integration basics.
• Use sigma notation confidently when manipulating finite sums.

The limit definition explains what a definite integral means. In routine calculations, the fundamental theorem of calculus lets you evaluate that limit efficiently by finding an antiderivative.