## Riemann sums

Let $$f(x)$$ be defined for $$a\leq x\leq b$$ (Figure 1). Assume at pleasure $$n-1$$ points between $$a$$ and $$b$$ $a=x_{0}<x_{1}<x_{2}<\cdots<x_{n}=b,$ thus dividing the interval $$I=[a,b]$$ into $$n$$ subintervals: $[a,x_{1}],[x_{1},x_{2}],\cdots,[x_{n-1},b].$ In each of these subintervals, take a value of $$x=x_{k}^{*}$$ where $$x_{k-1}\leq x_{k}^{*}\leq x_{k}$$, and form the sum $S_{n}=f(x_{1}^{*})(x_{1}-x_{0})+f(x_{2}^{*})(x_{2}-x_{1})+\cdots+f(x_{n}^{*})(x_{n}-x_{n-1})\tag{8.1.1}$ This sum is called a Riemann sum for $$f$$ on the interval $$[a,b]$$, after German mathematician George Friedrich Bernhard Riemann (1826–1866). Now let’s see what the geometrical interpretation of $$S_{n}$$ is. First suppose $$f(x)\geq0$$ for all $$x\in[a,b]$$. In this case

$$f(x_{1}^{*})(x_{1}-x_{0})=$$ the area of the rectangle $$AP_{1}R_{1}M_{1},$$

$$f(x_{2}^{*})(x_{2}-x_{1})=$$ the area of the rectangle $$M_{1}P_{2}R_{2}M_{2},$$

$$\vdots$$

$$f(x_{n-1}^{*})(x_{n-1}-x_{n-2})=$$ the area of the rectangle $$M_{n-2}P_{n-1}R_{n-1}M_{n-1}$$,

$$f(x_{n}^{*})(x_{n}-x_{n-1})=$$ the area of the rectangle $$M_{n-1}P_{n}R_{n}B$$.

Thus $$S_{n}$$ represents the total area of all rectangles in Figure 1, and approximates the area of the region that lies between the curve $$y=f(x)$$ and the $$x$$-axis between $$x=a$$ and $$x=b$$.

If $$f(x)$$ takes both positive and negative values on $$[a,b]$$, then $$S_{n}$$ is an approximation for the net signed area; that is, the area below the curve $$y=f(x)$$ and above the $$x$$-axis minus the area above the curve $$y=f(x)$$ and below the $$x$$-axis from $$x=a$$ to $$x=b$$, as suggested in Figure 2. Figure 2. The Riemann sum is an approximation to the net area if $$f$$ takes on both positive and negative values.

The Riemann sum is often shortened as $S_{n}=\sum_{k=1}^{n}f(x_{k}^{*})(x_{k}-x_{k-1}),$ where $$\Sigma$$ (sigma), Greek form of the letter $$S$$, stands for the word “summation.” The whole expression indicates that the sum is to be taken of all terms obtained from $$f(x_{k}^{*})(x_{k}-x_{k-1})$$ by giving to $$k$$ in succession the values of $$1,2,3,\dots,n$$.

By introducing the symbol $\Delta_{k}x=x_{k}-x_{k-1},$ we can shorten it further and write: $S_{n}=\sum_{k=1}^{n}f(x_{k}^{*})\Delta_{k}x.\tag{8.1.2}$ Please note that here $$\Delta_{k}$$ is not a factor, but denotes the difference (here the length of the $$k$$-th subinterval).

• Some books, write $$\Delta x_{k}$$ instead of $$\Delta_{k}x$$.

## Some Specific Types of Riemann Sums

• If the sample point in each subinterval $$x_{k}^{*}$$ is the left endpoint of the subinterval; that is, if $$x_{k}^{*}=x_{k-1}$$ for all $$k$$, then $$S_{n}$$ is called a left Riemann sum (see Figure 3).
• If the sample point in each interval $$x_{k}^{*}$$ is the right endpoint of the subinterval; that is, if $$x_{k}^{*}=x_{k}$$ for all $$k$$, then $$S_{n}$$ is called a right Riemann sum (see Figure 4).
• If the sample point in each interval $$x_{k}^{*}$$ is the midpoint of the subinterval; that is, if $$x_{k}^{*}=(x_{k}+x_{k-1})/2$$ for all $$k$$, then $$S_{n}$$ is called a middle Riemann sum (see Figure 5).
• If the sample point in each interval $$x_{k}^{*}$$ is chosen such that $$f(x_{k}^{*})$$ is the maximum value of $$f$$ on the subinterval $$[x_{k-1},x_{k}]$$ for all $$k$$, then $$S_{n}$$ is called an upper Riemann sum (see Figure 6).
• If the sample point in each interval $$x_{k}^{*}$$ is chosen such that $$f(x_{k}^{*})$$ is the minimum value of $$f$$ on the subinterval $$[x_{k-1},x_{k}]$$ for all $$k$$, then $$S_{n}$$ is called a lower Riemann sum (see Figure 7).

## Limit of the Riemann Sums as $$n\to\infty$$

As shown in Figure 8, if the number of subintervals $$n$$ increases such that the subintervals lengths (or widths) get smaller and the rectangles become thinner, then the Riemann sum $$S_{n}$$ will be a better approximation to the net area between the curve $$y=f(x)$$ and the $$x$$-axis.

 n=11 n=20 n=40   Figure 8. As the number of subintervals $n$ increases, $S_n$ becomes a better approximation to the net area that lie between the curve $y = f(x)$ and the x-axis.

Now let the number of subintervals $$n$$ increase indefinitely $$(n\to\infty)$$, while the width of the widest subinterval approaches zero. If the sum (8.1.1) (or (8.1.2)) approaches a number which is independent of the choice of the $$x_{k}$$’s and of the $$x_{k}^{*}$$’s, its limit is called the definite integral of $$f(x)$$ between $$a$$ and $$b$$ and is denoted by $${\int_{a}^{b}f(x)}dx$$.

Definition 1. The limit of the sum (8.1.2) provided that the number of subintervals $$n$$ tends to infinity and at the same time the length of each subinterval approaches zero is called the definite integral of $$f(x)$$ from $$a$$ to $$b$$ and is denoted by $${\int_{a}^{b}f(x)}dx}$$ If this limit exists, the function is called integrable on the interval $$[a,b]$$. $\int_{a}^{b}f(x)dx=\lim_{\max\Delta_{k}x\to0}\sum_{k=1}^{n}f(x_{k}^{*})\Delta_{k}x.$

We note that

• The use of the word “integral” and of the symbol $$\int$$ suggests a connection with the indefinite integrals in the previous Chapter. This connection will be shown in Section on the Fundamental Theorem of Calculus.

• The symbol $$\int$$ is a modified form of $$S$$ (that stands for summation).

• We call $$a$$ the lower limit and $$b$$ the upper limit of integration (or of the internal), and the function $$f(x)$$ the integrand.

• If the integral $$\int_{a}^{b}f(x)dx$$ exists, $$f(x)$$ is called integrable between $$a$$ and $$b$$.

• If $$f$$ is continuous on the interval $$(a,b)$$ then $$\int_{a}^{b}f(x)dx$$ exists. The proof of this theorem is beyond the scope of an elementary course. #### The variable of integration is a dummy variable

The letter $$k$$ chosen for the summation index in the Riemann sum $$S_{n}=\sum_{k=1}^{n}f(x_{k}^{*})\Delta_{k}x$$ is a dummy variable; that is, it can be replaced by any other letter such as $$i$$ without affecting the value of the sum (provided, of course, it is replaced in each place where $$k$$ occurs) $S_{n}=\sum_{k=1}^{n}f(x_{k}^{*})\Delta_{k}x=\sum_{i=1}^{n}f(x_{i}^{*})\Delta_{i}x.$ Similarly, the letter $$x$$ chosen for the variable of integration in $$\int_{a}^{b}f(x)dx$$ is a dummy variable, which disappears in the final result. In other words, it can be replaced by any other letter that we wish without affecting the integral; thus $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(u)du.$ For example, if $\int_{1}^{2}f(x)dx=32,$ then $\int_{1}^{2}f(u)du=32.$ Therefore, instead of $$\int_{a}^{b}f(x)dx$$, sometimes we may simply write $\int_{a}^{b}f.$ Unlike the case of the definite integral, the variable of integration in an indefinite integral is not a dummy variable because it appears in the final result. For example, $\int xdx=\frac{1}{2}x^{2}+C_{1},\quad\int udu=\frac{1}{2}u^{2}+C_{2}$ and in this sense $\int xdx\neq\int udu.$

In the previous discussion, we assumed $$a<b$$. However, if we omit this condition and assume $$b<a$$, we can still retain our definition of integral; the only change is that we have to number the subdivision points from the right side, which make the differences $$\Delta_{k}x$$ negative. We are thus led to the relation $\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx.$ In conformity, we define $\int_{a}^{a}f(x)dx=0.$

Definition 2.

$\int_{\color{red} a}^{\color{blue} b}f(x)dx=-\int_{\color{blue} b}^{\color{red} a}f(x)dx. \tag{ a and b interchanged}$ and $\int_{a}^{a}f(x)dx=0. \tag{same upper and lower limits}$

$\displaystyle{\int_a^b f(x)dx} =$ area above the $x$-axis $-$ area below the $x$-axis   (when $a<b$)

## Examples

Now let’s solve some examples.

Example 1

Find $$\int_{-1}^{3}(2-x)dx$$ by the limit definition.

Solution 1

Let divide the interval $$[-1,3]$$ into $$n$$ subintervals of equal width $$\Delta x=(3-(-1))/n=4/n$$. In each subinterval $$[x_{k},x_{k+1}]$$, let use $$x_{k}^{*}=x_{k}$$ as the sample point \begin{aligned} x_{0} & =-1\\ x_{1} & =-1+\Delta x=-1+\frac{4}{n}\\ x_{2} & =-1+2\Delta x=-1+2\left(\frac{4}{n}\right)\\ & \vdots\\ x_{k} & =-1+k\Delta x=-1+k\left(\frac{4}{n}\right)\\ & \vdots\\ x_{n} & =-1+n\Delta x=-1+n\left(\frac{4}{n}\right)=-1+4=3.\end{aligned} Therefore, $$f(x_{k}^{*})=f(x_{k})=2-x_{k}=2-\left[-1+k\left(\frac{4}{n}\right)\right]=3-\frac{4k}{n}$$ and \begin{aligned} \sum_{k=1}^{n}f(x_{k}^{*})\Delta_{k}x & =\sum_{k=1}^{n}\left(3-\frac{4k}{n}\right)\frac{4}{n}\\ & =\sum_{k=1}^{n}\left(\frac{12}{n}-\frac{16k}{n^{2}}\right)\\ & =\frac{12}{n}\sum_{k=1}^{n}1-\frac{16}{n^{2}}\sum_{k=1}^{n}k\\ & =\frac{12}{n}(n)-\frac{16}{n^{2}}\frac{n(n+1)}{2}\\ & =12-8\left(1+\frac{1}{n}\right)\end{aligned} Notice that in the above summation, $$n$$ is a fixed number (say $$n=10$$) and not a function of the summation index $$k$$. That is why we could move $$n$$ outside the sigma sign. For these calculations, we have also used the formulas $$\sum_{k=1}^{n}1=n$$ and $$\sum_{k=1}^{n}k=n(n+1)/2$$ that we reviewed in the Section on Sigma Notation.

Because the width of each subinterval is the same $$\max\Delta_{k}x=4/n\to0$$ is equivalent to $$n\to\infty$$, and we can conclude \begin{aligned} \int_{-1}^{3}(2-x)dx & =\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k}^{*})\Delta x\\ & =\lim_{n\to\infty}\left[12-8\left(1+\frac{1}{n}\right)\right]\\ & =12-8(1+0)=4.\end{aligned}

You can easily show that if we use $$x_{k}^{*}=x_{k+1}$$ (instead of $$x_{k}^{*}=x_{k}$$) as the sample point in each subinterval $$[x_{k},x_{k+1}]$$, we will arrive at the same result.

Because the graph of $$y=2-x$$ is a straight line, we can easily check our answer. We should subtract the area of the part that lies below the $$x$$-axis from the area of the above that is above the $$x$$-axis (see Figure 9). It follows from the familiar triangle area formula $$A=\text{base}\times\text{height}/2$$ that $\int_{-1}^{3}(2-x)dx=\frac{3\times3}{2}-\frac{1\times1}{2}=4.$ Figure 9. The integral of $$2-x$$ from $$-1$$ to $$3$$ is $$\int_{-1}^{3}(2-x)dx=A_{1}-A_{2}=9/2-1/2=4.$$
Example 2

Find $$\int_{1}^{3}x^{2}dx$$ by the limit definition.

Solution 2

Let’s divide the interval $$[1,3]$$ into $$n$$ subintervals of equal width $$\Delta_{k}x=\Delta x=\frac{(3-1)}{n}=\frac{2}{n}$$: \begin{aligned} x_{0} & =1\\ x_{1} & =1+\Delta x=1+\frac{2}{n}\\ x_{2} & =1+2\Delta x=1+2\left(\frac{2}{n}\right)\\ & \vdots\\ x_{k} & =1+k\Delta x=1+k\left(\frac{2}{n}\right)\\ & \vdots\\ x_{n} & =1+n\Delta x=1+n\left(\frac{2}{n}\right)=3.\end{aligned} Assume $$f(x_{k}^{*})=f(x_{k})=\left(1+\frac{2k}{n}\right)^{2}=1+\frac{4k}{n}+\frac{4k^{2}}{n^{2}}$$. Then \begin{aligned} \sum_{k=1}^{n}f(x_{k}^{*})\Delta x & =\sum_{k=1}^{n}\left[\frac{2}{n}+\frac{8k}{n^{2}}+\frac{8k^{2}}{n^{3}}\right]\\ & =\frac{2}{n}\sum_{k=1}^{n}1+\frac{8}{n^{2}}\sum_{k=1}^{n}k+\frac{8}{n^{3}}\sum_{k=1}^{n}k^{2}\\ & =\frac{2}{n}n+\frac{8}{n^{2}}\frac{n(n+1)}{2}+\frac{8}{n^{3}}\frac{n(n+1)(2n+1)}{6}\\ & =2+4\frac{n+1}{n}+\frac{4}{3}\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)\\ & =2+4\left(1+\frac{1}{n}\right)+\frac{4}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\end{aligned} The width of every subinterval is the same and $$\max\Delta_{k}x\to0$$ is equivalent to $$n\to\infty$$. As $$n\to\infty$$, $$\frac{1}{n}\to0$$ and hence \begin{aligned} \int_{1}^{3}x^{2}dx & =\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k}^{*})\Delta x\\ & =\lim_{n\to\infty}\left[2+4\left(1+\frac{1}{n}\right)+\frac{8}{6}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\right]\\ & =2+4(1+0)+\frac{4}{3}(1+0)(2+0)\\ & =\frac{26}{3}.\end{aligned}

There are no formulas from elementary geometry that can help us verify our answer (see Figure 10), but we will soon learn a theorem, called the Fundamental Theorem of Calculus, that enables us to evaluate such definite integrals easily.

Example 3

Find $\int_{-1}^{3}\left(|x|-2\right)dx$ by interpreting the integral as the net area under the graph.

Solution 3

We know how the graph of $y=|x|$ looks like. To sketch the graph of $y=|x|-2$, we simply shift the graph of $y=|x|$ downward $2$ units. The net area of the region between the graph of $y=|x|-2$ and the $x$-axis over $[-1,3]$ can be easily computed. According to the following figure, we have

\begin{align*}
\int_{-1}^{3}(|x|-2)dx & =-A_{1}-A_{2}-A_{3}+A_{4}\\
& =-1-\frac{1}{2}-2+\frac{1}{2}\\
& =-3.
\end{align*}

Example 4

Find $\int_{-1}^{1}\sqrt{1-x^{2}}\ }d$ using a geometric formula.

Solution 4

We recognize that the curve $y=\sqrt{1-x^{2}}$ is the upper semi-circle of radius 1 centered at the origin (Figure 12). Because we can easily compute the area between the semi-circle and the $x$-axis, taking the limit of Riemann sums is not required.

\begin{align*}
\int_{-1}^{1}\sqrt{1-x^{2}}\ dx & =\text{area between the semi-circle of radius 1 and the $x$-axis}\\
& =\frac{1}{2}(\text{area of a circle of radius 1)}\\
& =\frac{1}{2}\pi(1)^{2}=\frac{\pi}{2}.
\end{align*}