Area and Limits of Riemann Sums

In this article, I begin with sigma notation and working with the properties of finite sums. Once the reader is familiar with this, I motivate the definition of a Riemann sum. Then, estimating the area under a curve using Riemann sums is discussed. Finally, finding area using the limits of Riemann sum is detailed.

We discuss expressing finite sums using sigma notation and basic properties of these sums. The discussion then turns to approximating the area under a curve over a bounded closed interval using Riemann sums. We then consider partitions of closed bounded intervals and a limiting process which yields the exact area under the curve of a continuous function.

Sigma Notation

In order to understand Riemann sums and integration theory correctly it is important to understand summations using sigma notation.

Definition. If $a_m,$ $ a_{m+1}, \ldots, a_n$ are real numbers such that $m\leq n,$ then the summation of these numbers written in sigma notation is \begin{equation} \sum_{i=m}^n a_i=a_m+a{m+1}+\cdot \cdot \cdot +a_{n-1}+a_n \end{equation} and also using functional notation, \begin{equation} \sum_{i=m}^n f(i)=f(m)+f(m+1)+\cdot \cdot \cdot +f(n-1)+f(n) \end{equation} where $f(i)=a_i.$ The $i$ is called the index of summation, the $a_i$ are called the $i$th term of the sum, and the upper and lower bounds of the summation are $n$ and $m,$ respectively.

Example. Write the sum $$ \sum_{k=0}^4 \frac{2k-1}{2k+1} $$ in expanded form.

Solution. In expanded form the sum is \begin{align} \sum_{k=0}^4 \frac{2k-1}{2k+1} & =\frac{2(0)-1}{2(0)+1}+\frac{2(1)-1}{2(1)+1}+\frac{2(2)-1}{2(2)+1}+\frac{2(3)-1}{2(3)+1}+\frac{2(4)-1}{2(4)+1} \\ & =-1+\frac{1}{3}+\frac{3}{5}+\frac{5}{7}+\frac{7}{9} =\frac{449}{315}. \end{align} as desired.

Example. Write the sum $$ \left(\sum_{i=1}^6 \frac{i}{i+1}\right)-40 $$ in expanded form.

Solution. In expanded form, the sum is \begin{align} \left(\sum_{i=1}^6 \frac{i}{i+1}\right)-40 & =\frac{1}{1+1}+\frac{2}{2+1}+\frac{3}{3+1}+\frac{4}{4+1}+\frac{5}{5+1}+\frac{6}{6+1}-40 \\ & =\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}+\frac{6}{7}-40 =-\frac{4983}{140}. \end{align} as desired.

Example. Write the sum $$ \frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\cdot \cdot \cdot +\frac{1}{7225} $$ in sigma notation.

Solution. In sigma notation, the sum is \begin{equation} \frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\cdots+\frac{1}{7225}=\sum_{k=1}^{85} \frac{1}{k^2}. \end{equation} as desired.

Example. Write the sum $$ \frac{3}{7}+\frac{4}{8}+\frac{5}{9}+\frac{6}{10} +\cdots +\frac{23}{27} $$ in sigma notation.

Solution. In sigma notation the sum is \begin{equation} \sum_{k=7}^{27} \frac{k-4}{k}. \end{equation} as desired.

Example. Write the sum \begin{equation} \left[1-\left(\frac{1}{4}\right)^2\right]+\left[1-\left(\frac{2}{4}\right)^2\right]+\left[1-\left(\frac{3}{4}\right)^2\right]+ \cdots +\left[1-\left(\frac{4}{4}\right)^2\right] \end{equation} in sigma notation.

Solution. In sigma notation the sum is \begin{equation} \sum_{k=1}^4 \left[1-\left(\frac{k}{4}\right)^2\right] \end{equation} as desired.

Example. Write the sum \begin{equation} \left(\frac{1}{n}\right)\sqrt{1-\left(\frac{0}{n}\right)^2}+\left(\frac{1}{n}\right)\sqrt{1-\left(\frac{1}{n}\right)^2}+\cdots +\left(\frac{1}{n}\right)\sqrt{1-\left(\frac{n-1}{n}\right)^2} \end{equation} in sigma notation.

Solution. In sigma notation the sum is \begin{equation} \sum_{k=0}^{n-1} \left(\frac{1}{n}\right)\sqrt{1-\left(\frac{k}{n}\right)^2} \end{equation} as desired.

Properties of Finite Sums

The linearity rule, subtotal rule, and the dominance rule are the basic properties of summations illustrated below. We also give summation formulas for \begin{equation} \sum_{k=1}^n k^i \end{equation} when $i=1,2,\ldots,6$ followed by some examples on taking limits of sums.

Theorem. If $c$ and $d$ are real numbers that do not depend on integers $m$ and $n$, then

(1) $\sum_{k=1}^n \left(c a_k+d b_k\right)=c\left( \sum_{k=1}^n a_k \right)+d\left( \sum_{k=1}^n b_k \right)$

(2) If $\sum_{k=1}^n a_k=\sum_{k=1}^m a_k+\sum_{k=m+1}^n a_k $

(3) If $a_k\leq b_k $ for all $k,$ then $\sum_{k=1}^n a_k\leq \sum_{k=1}^n b_k.$

The following formulas will be necessary when determining limits of Riemann sums to find the area under a curve of a polynomial. The summation formulas for $\sum_{k=1}^n k^i$ when $i=1,2,\ldots,7.$

$(1) \quad \displaystyle \sum_{k=1}^n k=\frac{1}{2} n (n+1)$

$(2) \quad \displaystyle \sum_{k=1}^n k^2=\frac{1}{6} n (n+1) (2 n+1)$

$(3) \quad \displaystyle \sum_{k=1}^n k^3=\frac{1}{4} n^2 (n+1)^2$

$(4) \quad \displaystyle \sum_{k=1}^n k^4=\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)$

$(5) \quad \displaystyle \sum_{k=1}^n k^5=\frac{1}{12} n^2 (n+1)^2 \left(2 n^2+2 n-1\right)$

$(6) \quad \displaystyle \sum_{k=1}^n k^6=\frac{1}{42} n (n+1) (2 n+1) \left(3 n^4+6 n^3-3 n+1\right)$

Example. Find the value of the sum $$ \sum_{i=1}^{100} i \left(i^2+1\right). $$

Solution. The value of the sum is \begin{align} & \sum_{i=1}^{100} i \left(i^2+1\right) =\sum_{i=1}^{100} \left(i^3+i\right) =\sum_{i=1}^{100} i^3+\sum_{i=1}^{100} i \\ & \qquad =\frac{1}{4} 100^2 (100+1)^2+\frac{1}{2} 100 (100+1) \\ & \qquad =25502500+5050 =25507550. \end{align} as desired.

Example. Find the value of the sum $$ \sum_{i=1}^4 \left(2^i+i^2\right). $$

Solution. The value of the sum is \begin{align} \sum_{i=1}^4 \left(2^i+i^2\right) & =\left(2^1+1^2\right)+\left(2^2+2^2\right)+\left(2^3+3^2\right)+\left(2^4+4^2\right) \\ & =3+8+17+32 =60. \end{align} as desired.

Example. Find the value of the sum $$ \sum_{i=1}^n i^2 \left(i^2-i+1\right). $$

Solution. The value of the sum is \begin{align} & \sum_{i=1}^n i^2 \left(i^2-i+1\right) =\sum_{i=1}^n \left(i^4-i^3+i^2\right) =\sum_{i=1}^n i^4-\sum_{i=1}^n i^3+\sum_{i=1}^n i^2 \\ & \quad =\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)-\frac{1}{4} n^2 (n+1)^2+\frac{1}{6} n (n+1) (2 n+1) \\ & \quad =\frac{1}{60} n (n+1) \left(12 n^3+3 n^2+7 n+8\right). \end{align} as desired.

Definition of Riemann Sum

In this topic we illustrate how a Riemann sum can be used to approximate the area under a curve and in doing so, we anticipate the notion of definite integral. We will investigate the area under the curve $$y=2+6 x-5 x^2+x^3,$$ above the $x$-axis and between the vertical lines $x=0$ and $x=5.$

Definition. A Riemann sum for a function $f$ on the closed bounded interval $[a,b]$ is a sum of the form \begin{equation} \sum_{k=1}^n f\left( x_i^* \right) \Delta x_i \end{equation} where $a=x_0{i-1}\leq x_i^*\leq x_i$ for $i=1,\ldots,n$. Let $\Delta x_i=x_i-x_{i-1} $ denote the width of each subinterval. The set \begin{equation} \mathcal{P} = \{ x_0,x_1,\ldots,x_n\} \end{equation} is called a partition of $[a,b]$, the largest of the $\Delta x_i$ is called the norm of $\mathcal{P}$, and the $x_i^*$ are called the subinterval representatives for the Riemann sum.

Example. Given the function $f(x)=x^3,$ the closed bounded interval $[0,1],$ and the partition $\left\{0,\frac{1}{2},\frac{3}{4},\frac{5}{6},1\right\}$, compute a Riemann sum.

Solution. Organizing into a table we determine the values, \begin{equation} \begin{array}{c|c|c|c|c} i & \text{subinterval} & x_i^* & f\left(x_i^*\right) & \Delta x_i \\ \hline 1 & \left[0,\frac{1}{2}\right] & \frac{1}{3} & \left(\frac{1}{3}\right)^3=\frac{1}{27} & \frac{1}{2}-0=\frac{1}{2} \\ 2 & \left[\frac{1}{2},\frac{3}{4}\right] & \frac{2}{3} & \left(\frac{2}{3}\right)^3=\frac{8}{27} & \frac{3}{4}-\frac{1}{2}=\frac{1}{4} \\ 3 & \left[\frac{3}{4},\frac{5}{6}\right] & \frac{8}{10} & \left(\frac{8}{10}\right)^3=\frac{64}{125} & \frac{5}{6}-\frac{3}{4}=\frac{1}{12} \\ 4 & \left[\frac{5}{6},1\right] & \frac{9}{10} & \left(\frac{9}{10} \right)^3 =\frac{729}{1000} & 1-\frac{5}{6}=\frac{1}{6} \end{array} \end{equation} So the Riemann sum for these $x_i^*$ is \begin{align*} & \sum_{i=1}^4 f\left(x_i^*\right)\Delta x_i
=f\left(x_1^*\right)\Delta x_1+f\left(x_2^*\right)\Delta x_2+f\left(x_3^*\right)\Delta x_3+f\left(x_4^*\right)\Delta x_4 \\ & =\left(\frac{1}{27}\right)\left(\frac{1}{2}\right)+\left(\frac{8}{27}\right)\left(\frac{1}{4}\right)+\left(\frac{64}{125}\right)\left(\frac{1}{12}\right)+
\left(\frac{729}{1000}\right)\left(\frac{1}{6}\right) \\ & =\frac{2773}{10800}. \end{align*} as desired.

Estimating the Area Under a Curve

Example. Given the function $f(x)=2+6 x-5 x^2+x^3,$ the closed bounded interval $[0,5],$ and the partition ${0,1,2,3,4,5},$ compute a Riemann sum to approximate the area.

Solution. Organizing into a table we determine the values, \begin{equation} \begin{array}{c|c|c|c|c} i & \text{subinterval} & x_i^* & f\left(x_i^*\right) & \Delta x_i \\ \hline 1 & [0,1] & 0 & 2+6 (0)-5 (0)^2+(0)^3=2 & 1 \\ 2 & [1,2] & 1 & 2+6 (1)-5 (1)^2+(1)^3=4 & 1 \\ 3 & [2,3] & 2 & 2+6 (2)-5 (2)^2+(2)^3=2 & 1 \\ 4 & [3,4] & 3 & 2+6 (3)-5 (3)^2+(3)^3=2 & 1 \\ 5 & [4,5] & 4 & 2+6 (4)-5 (4)^2+(4)^3=10 & 1 \end{array} \end{equation} So the Riemann sum for these $x_i^*$ is \begin{equation} \sum_{i=1}^5 f\left(x_i^*\right)\Delta x_i =\text{(2)(1)+(4)(1)+(2)(1)+(2)(1)+(10)(1)} =20. \end{equation} the figure shows the Riemann sum as the approximate area under the given curve.

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Refining Partitions

For our next example we will use a finer partition, say a partition using 14 subintervals between $x=0$ and $x=5$ (also with uniform width). We will still use left-endpoints for our subinterval representatives. Our second estimate for the area is $28.1481$.

Example. Given the function $f(x)=2+6 x-5 x^2+x^3,$ the closed bounded interval $[0,5],$ and the partition \begin{equation} \left\{0,\frac{1}{3},\frac{2}{3},1,\frac{4}{3},\frac{5}{3},2,\frac{7}{3},\frac{8}{3},3,\frac{10}{3},\frac{11}{3},4,\frac{13}{3},\frac{14}{3},5\right\}, \end{equation} compute a Riemann sum to approximate the area.

Solution. Organizing into a table we determine the values, \begin{equation} \begin{array}{c|c|c|c|c} i & \text{subinterval} & x_i^* & f\left(x_i^*\right) & \Delta x_i \\ \hline 1 & \left[0,\frac{1}{3}\right] & 0 & 2+6 (0)-5 (0)^2+(0)^3=2 & \frac{1}{3} \\ 2 & \left[\frac{1}{3},\frac{2}{3}\right] & \frac{1}{3} & 2+6 \left(\frac{1}{3}\right)-5 \left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3=\frac{94}{27} & \frac{1}{3} \\ 3 & \left[\frac{2}{3},1\right] & \frac{2}{3} & 2+6 \left(\frac{2}{3}\right)-5 \left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3=\frac{110}{27} & \frac{1}{3} \\ 4 & \left[1,\frac{4}{3}\right] & 1 & 2+6 (1)-5 (1)^2+(1)^3=4 & \frac{1}{3} \\ 5 & \left[\frac{4}{3},\frac{5}{3}\right] & \frac{4}{3} & 2+6 \left(\frac{4}{3}\right)-5 \left(\frac{4}{3}\right)^2+\left(\frac{4}{3}\right)^3=\frac{94}{27} & \frac{1}{3} \\ 6 & \left[\frac{5}{3},2\right] & \frac{5}{3} & 2+6 \left(\frac{5}{3}\right)-5 \left(\frac{5}{3}\right)^2+\left(\frac{5}{3}\right)^3=\frac{74}{27} & \frac{1}{3} \\ 7 & \left[2,\frac{7}{3}\right] & 2 & 2+6 (2)-5 (2)^2+(2)^3=2 & \frac{1}{3} \\ 8 & \left[\frac{7}{3},\frac{8}{3}\right] & \frac{7}{3} & 2+6 \left(\frac{7}{3}\right)-5 \left(\frac{7}{3}\right)^2+\left(\frac{7}{3}\right)^3=\frac{40}{27} & \frac{1}{3} \\ 9 & \left[\frac{8}{3},3\right] & \frac{8}{3} & 2+6 \left(\frac{8}{3}\right)-5 \left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^3=\frac{38}{27} & \frac{1}{3} \\ 10 & \left[3,\frac{10}{3}\right] & 3 & 2+6 (3)-5 (3)^2+(3)^3=2 & \frac{1}{3} \\ 11 & \left[\frac{10}{3},\frac{11}{3}\right] & \frac{10}{3} & 2+6 \left(\frac{10}{3}\right)-5 \left(\frac{10}{3}\right)^2+\left(\frac{10}{3}\right)^3=\frac{94}{27} & \frac{1}{3} \\ 12 & \left[\frac{11}{3},4\right] & \frac{11}{3} & 2+6 \left(\frac{11}{3}\right)-5 \left(\frac{11}{3}\right)^2+\left(\frac{11}{3}\right)^3=\frac{164}{27} & \frac{1}{3} \\ 13 & \left[4,\frac{13}{3}\right] & 4 & 2+6 (4)-5 (4)^2+(4)^3=10 & \frac{1}{3} \\ 14 & \left[\frac{13}{3},\frac{14}{3}\right] & \frac{13}{3} & 2+6 \left(\frac{13}{3}\right)-5 \left(\frac{13}{3}\right)^2+\left(\frac{13}{3}\right)^3=\frac{418}{27} & \frac{1}{3} \\ 15 & \left[\frac{14}{3},5\right] & \frac{14}{3} & 2+6 \left(\frac{14}{3}\right)-5 \left(\frac{14}{3}\right)^2+\left(\frac{14}{3}\right)^3=\frac{614}{27} & \frac{1}{3} \end{array} \end{equation} So the Riemann sum $\sum_{i=1}^{15} f\left(x_i^*\right)\Delta x_i $ for these $x_i^*$ is $\frac{760}{27}$.

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Area and Limits of Riemann Sums

Example. Use a Riemann sum to approximate the area under the graph of $f(x)=6x^2+2x+4$ on $[1,3]$ with 8 subintervals.

Solution. As a partition we choose $\mathcal{P}=\left\{1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2,\frac{9}{4},\frac{5}{2},\frac{11}{4},3\right\}.$ Organizing our computations and choices for our subinterval representations we have \begin{equation} \begin{array}{c|c|c|c|c|c} i & \text{subinterval} & x_i^* & f\left(x_i^*\right) & \Delta x_i & f\left(x_i^*\right)\Delta x_i \\ \hline 1 & \left[1,\frac{5}{4}\right] & 1 & 6(1)^2+2(1)+4=12 & \frac{1}{4} & 3 \\ 2 & \left[\frac{5}{4},\frac{3}{2}\right] & \frac{5}{4} & 6\left(\frac{5}{4}\right)^2+2\left(\frac{5}{4}\right)+4=\frac{127}{8} & \frac{1}{4} & \frac{127}{32} \\ 3 & \left[\frac{3}{2},\frac{7}{4}\right] & \frac{3}{2} & 6\left(\frac{3}{2}\right)^2+2\left(\frac{3}{2}\right)+4=\frac{41}{2} & \frac{1}{4} & \frac{41}{8} \\ 4 & \left[\frac{7}{4},2\right] & \frac{7}{4} & 6\left(\frac{7}{4}\right)^2+2\left(\frac{7}{4}\right)+4=\frac{207}{8} & \frac{1}{4} & \frac{207}{32} \\ 5 & \left[2,\frac{9}{4}\right] & 2 & 6(2)^2+2(2)+4=32 & \frac{1}{4} & 8 \\ 6 & \left[\frac{9}{4},\frac{5}{2}\right] & \frac{9}{4} & 6\left(\frac{9}{4}\right)^2+2\left(\frac{9}{4}\right)+4 =\frac{311}{8} & \frac{1}{4} & \frac{311}{32} \\ 7 & \left[\frac{5}{2},\frac{11}{4}\right] & \frac{5}{2} & 6\left(\frac{5}{2}\right)^2+2\left(\frac{5}{2}\right)+4=\frac{93}{2} & \frac{1}{4} & \frac{93}{8} \\ 8 & \left[\frac{11}{4},3\right] & \frac{11}{4} & 6\left(\frac{11}{4}\right)^2 +2\left(\frac{11}{4}\right) +4=\frac{439}{8} & \frac{1}{4} & \frac{439}{32} \end{array}
\end{equation} So the Riemann sum for this partition $\mathcal{P}$ and these $x_i^*$ is $$ \sum_{i=1}^8 f\left(x_i^*\right)\Delta x_i =493/8. $$ Thus, the approximate the area under the graph of $f(x)=6x^2+2x+4$ on $[1,3]$ with 8 subintervals is $\frac{493}{8}$ as shown in the figure.

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Example. Use a Riemann sum to approximate the area under the graph of $f(x)=\sqrt{1+x^2} $ on $[0,1]$ with 10 subintervals.

Solution. As a partition we choose \begin{equation} \mathcal{P}=\left\{0,\frac{1}{10},\frac{1}{5},\frac{3}{10},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{7}{10},\frac{4}{5},\frac{9}{10},1\right\}. \end{equation} Organizing our computations and choices for our subinterval representations we have \begin{equation} \begin{array}{c|c|c|c|c|c} i & \text{subinterval} & x_i^* & f\left(x_i^*\right) & \Delta x_i & f\left(x_i^*\right)\Delta x_i \\ \hline
1 & \left[0,\frac{1}{10}\right] & \frac{1}{10} & \sqrt{1+\left(\frac{1}{10}\right)^2} = \frac{\sqrt{101}}{10} & \frac{1}{10} & \frac{\sqrt{101}}{100}
\\ 2 & \left[\frac{1}{10},\frac{1}{5}\right] & \frac{1}{5} & \sqrt{1+\left(\frac{1}{5}\right)^2} =\frac{\sqrt{26}}{5} & \frac{1}{10} & \frac{\sqrt{\frac{13}{2}}}{25} \\ 3 & \left[\frac{1}{5},\frac{3}{10}\right] & \frac{3}{10} & \sqrt{1+\left(\frac{3}{10}\right)^2} =\frac{\sqrt{109}}{10} & \frac{1}{10} & \frac{\sqrt{109}}{100} \\ 4 & \left[\frac{3}{10},\frac{2}{5}\right] & \frac{2}{5} & \sqrt{1+\left(\frac{2}{5}\right)^2}=\frac{\sqrt{29}}{5} & \frac{1}{10} & \frac{\sqrt{29}}{50} \\ 5 & \left[\frac{2}{5},\frac{1}{2}\right] & \frac{1}{2} & \sqrt{1+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2} & \frac{1}{10} & \frac{1}{4 \sqrt{5}} \\ 6 & \left[\frac{1}{2},\frac{3}{5}\right] & \frac{3}{5} & \sqrt{1+\left(\frac{3}{5}\right)^2}=\frac{\sqrt{34}}{5} & \frac{1}{10} & \frac{\sqrt{\frac{17}{2}}}{25} \\ 7 & \left[\frac{3}{5},\frac{7}{10}\right] & \frac{7}{10} & \sqrt{1+\left(\frac{7}{10}\right)^2}=\frac{\sqrt{149}}{10} & \frac{1}{10} & \frac{\sqrt{149}}{100} \\ 8 & \left[\frac{7}{10},\frac{4}{5}\right] & \frac{4}{5} & \sqrt{1+\left(\frac{4}{5}\right)^2}=\frac{\sqrt{41}}{5} & \frac{1}{10} & \frac{\sqrt{41}}{50} \\ 9 & \left[\frac{4}{5},\frac{9}{10}\right] & \frac{9}{10} & \sqrt{1+\left(\frac{9}{10}\right)^2}=\frac{\sqrt{181}}{10} & \frac{1}{10} & \frac{\sqrt{181}}{100} \\ 10 & \left[\frac{9}{10},1\right] & 1 & \sqrt{1+(1)^2}=\sqrt{2} & \frac{1}{10} & \frac{1}{5 \sqrt{2}} \\ \hline & & & & \text{Total}\approx & 1.16909 \end{array} \end{equation} So the approximate the area under the graph of $f(x)=\sqrt{1+x^2}$ on $[0,1]$ using 10 subintervals is 1.16909 as shown in the following figure.

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Area using Limits of Riemann Sums

It is important to be able to evaluate limits of sums so here are a couple of examples.

Theorem. Suppose $f$ is continuous and $f(x)\geq 0$ throughout the interval $[a,b].$ Then the area of the region under the curve $y=f(x)$ over this interval is \begin{equation} A=\lim_{\Delta x\to 0}\sum_{k=1}^n f(a+k \Delta x)\Delta x \end{equation} where $\Delta x=(b-a)/n.$

Example. Evaluate $$ \lim_{n\to +\infty }\sum_{k=1}^n \frac{k}{n^2}. $$

Solution. The value of the limit is \begin{align} \lim_{n\to +\infty }\sum_{k=1}^n \frac{k}{n^2}
& =\lim_{n\to +\infty }\frac{1}{n^2} \left(\sum_{k=1}^n k\right) \\
& =\lim_{n\to +\infty }\frac{1}{n^2} \left(\frac{1}{2} n (n+1)\right) \\ & =\lim_{n\to +\infty }\frac{n+1}{2 n} \\ & =\frac{1}{2}. \end{align} as desired.

Example. Evaluate the limit $$\lim_{n\to +\infty }\sum_{k=1}^n \left(1+\frac{2k}{n}\right)^2\left(\frac{2}{n}\right). $$

Solution. The value of the limit is \begin{align} \lim_{n\to +\infty }\sum_{k=1}^n \left(1+\frac{2k}{n}\right)^2\left(\frac{2}{n}\right) & =\lim_{n\to +\infty }\left(\frac{2}{n}\right)\sum_{k=1}^n \left(1+\frac{2k}{n}\right)^2 \\ & =\lim_{n\to +\infty }\left(\frac{2}{n}\right)\sum_{k=1}^n \left(\frac{4 k^2}{n^2}+\frac{4 k}{n}+1\right) \\ & =\lim_{n\to +\infty }\left(\frac{2}{n}\right)\left(\frac{13 n^2+12 n+2}{3 n}\right) \\ & =\lim_{n\to +\infty }\frac{2 \left(13 n^2+12 n+2\right)}{3 n^2} \\ & =\frac{26}{3}. \end{align} as desired.

Example. Find the exact area under the curve $$ y=x^2+4 $$ on $[2,10].$

Solution. We will use the formula \begin{equation}
A=\lim_{\Delta x\to 0}\sum_{k=1}^n f(a+k \Delta x)\Delta x \end{equation} with $b=10$ and $a=2.$ We see $\Delta x=\frac{10-2}{n}=\frac{8}{ n};$ and we notice that $n\to +\infty $ as $\Delta x\to 0$ and so the area $A$ is given by \begin{align} A&=\lim_{\Delta x\to 0}\sum_{k=1}^n f(a+k \Delta x)\Delta x \\ & =\lim_{n\to \infty }\sum_{k=1}^n f\left(2+k \frac{8}{ n}\right)\frac{8}{ n} \\
& =\lim_{n\to \infty }\frac{8}{ n}\sum_{k=1}^n \left[\left(2+\frac{8 k}{ n}\right)^2+4\right] \\ & =\lim_{n\to \infty }\frac{8}{ n}\sum_{k=1}^n \left(8+4\left(\frac{8 k}{n}\right)+\frac{64 k^2}{n^2}\right) \\ & =\lim_{n\to \infty }\frac{8}{ n}\sum_{k=1}^n \left(8+4\left(\frac{8 k}{n}\right)+\frac{64 k^2}{n^2}\right) \\ & =\lim_{n\to \infty }\frac{8}{ n}\left[\sum_{k=1}^n 8+\frac{32}{n}\sum_{k=1}^n k+\frac{64 }{n^2}\sum_{k=1}^n k^2\right] \\ & =\lim_{n\to \infty }\frac{8}{ n}\left[8n+\frac{32}{n}\left(\frac{1}{2} n (n+1)\right)+\frac{64 }{n^2}\left(\frac{1}{6} n (n+1) (2 n+1)\right)\right] \\ & =\lim_{n\to \infty }\frac{64 \left(4+18 n+17 n^2\right)}{3 n^2} \qquad \text{(after algebraic simplification)} \\ & =\frac{64(17)}{3} \\ & =\frac{1088}{3}. \end{align} Therefore, the exact area under the curve $y=x^2+4$ bounded by the lines $y=0,$ $x=2,$ and $x=10$ is $\frac{1088}{3}.$

Example. Find the exact area under the curve $$ y=4x^3+3x^2 $$ on $[0,1].$

Solution. We will use the formula \begin{equation} A=\lim_{\Delta x\to 0}\sum_{k=1}^n f(a+k \Delta x)\Delta x \end{equation} with $b=1$ and $a=0.$ We see $\Delta x=\frac{1}{n};$ and we notice that $n\to +\infty $ as $\Delta x\to 0$ and so the area $A$ is given by \begin{align} A & =\lim_{\Delta x\to 0}\sum_{k=1}^n f(a+k \Delta x)\Delta x \\& =\lim_{n\to \infty }\sum_{k=1}^n f\left(\frac{k}{n}\right)\frac{1}{n} \\& =\lim_{n\to \infty }\frac{1}{ n}\sum_{k=1}^n \left[4\left(\frac{k}{n}\right)^3+3\left(\frac{k}{n}\right)^2\right] \\ & =\lim_{n\to \infty }\frac{1}{ n}\left[\frac{4}{n^3}\sum_{k=1}^n k^3+\frac{3}{n^2}\sum_{k=1}^n k^2\right] \\ & =\lim_{n\to \infty }\frac{1}{ n}\left[\frac{4}{n^3}\left(\frac{1}{4} n^2 (n+1)^2\right)+\frac{3}{n^2}\left(\frac{1}{6} n (n+1) (2 n+1)\right)\right] \\ & =\lim_{n\to \infty }\frac{3+7 n+4 n^2}{2 n^2} \qquad \text{(after algebraic simplification)} \\ & =2. \end{align} Therefore, the exact area under the curve $y=4x^3+3x^2$ bounded by the lines $y=0,$ $x=0,$ and $x=1$ is $2.$

Exercises on Area and Limits of Riemann Sums

Exercise. Use finite approximations to estimate the area under the graph of $f(x)=x^3$ between $x=0$ and $x=1.$

Exercise. Use finite approximations to estimate the area under the graph of $f(x)=4-x^2$ between $x=-2$ and $x=2.$

Exercise. Using rectangles whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule) estimate the area under the graph of the function $f(x)=x^2$ between $x=0$ and $x=1,$ using first two and then four rectangles.

Exercise. Using rectangles whose height is given by the value of the function at the midpoint of the rectangle’s base (the midpoint rule) estimate the area under the graph of the function $f(x)=4-x^2$ between $x=-2$ and $x=2,$ using first two and then four rectangles.

Exercise. An object is shot straight upward from sea level with an initial velocity of $400 \text{ft}/\sec .$ (a) Assuming that gravity is the only force acting on the object, give an upper estimate for its velocity after 5 sec have elapsed. Use $g=-32 \text{ft}\left/\sec ^2\right.$ for the gravitational acceleration. (b) Find a lower estimate for the height attained after 5 sec.

Exercise. Use a finite sum to estimate the average of $$ f(x)=\left(\frac{1}{2}\right)+\sin ^2\pi t $$ on the closed bounded interval $[0,2]$ by partitioning the interval into four subintervals of equal length and evaluating $f$ at the subinterval midpoints.

Exercise. Write the sum $$ \sum_{k=1}^2 \frac{6k}{k+1} $$ without sigma notation and then evaluate the sum.

Exercise. Write the sum $$ \sum_{k=1}^4 (-1)^k\cos k \pi $$ without sigma notation and then evaluate the sum.

Exercise. Rewrite the expression $1-2+4-8+16-32$ in sigma notation.

Exercise. Which formula is not equivalent to the other two?

$(1) \displaystyle \sum_{k=2}^4 \frac{(-1)^{k-1}}{k-1}$

$(2) \displaystyle\sum_{k=0}^2 \frac{(-1)^k}{k+1}$

$(3) \displaystyle \sum_{k=-1}^1 \frac{(-1)^k}{k+2}$

Exercise. Rewrite the sum using sigma notation $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}.$

Exercise. Rewrite the sum using sigma notation $\frac{-1}{5}+\frac{2}{5}-\frac{3}{5}+\frac{4}{5}-\frac{5}{5}.$

Exercise. Suppose that $\sum_{k=1}^n a_k=0$ and $\sum_{k=1}^n b_k=1.$ Find the values of

$(1) \quad \displaystyle \sum_{k=1}^n 8 a_k$

$(2) \quad \displaystyle\sum_{k=1}^n 250 b_k$

$(3) \quad \displaystyle\sum_{k=1}^n \left( a_k+1\right)$, and

$(4) \quad \displaystyle\sum_{k=1}^n \left(b_k-1\right)$.

Exercise. Evaluate the following sums.

$(1) \quad \displaystyle\sum_{k=1}^7 (-2k)$

$(2) \quad \displaystyle\sum_{k=1}^6 \left(k^2-5\right)$

$(3) \quad \displaystyle\sum_{k=1}^5 \frac{k^3}{225}+\left( \sum_{k=1}^5 k \right)^3$

Exercise. Graph the function $f(x)=-x^2$ over the interval $[0,2].$ Partition the interval into four subintervals of equal length. Then add to your sketch the rectangles associated with the Riemann sum $$\sum_{k=1}^n f\left(c_k\right)\Delta x_k $$ given that $c_k$ is the (a) left-hand endpoint, (b) right-hand endpoint, (c) the midpoint of the $k$-th subinterval. Make a separate sketch for each set of rectangles.

Exercise. Graph the function $f(x)=\sin x+1$ over the interval $[-\pi ,\pi ].$ Partition the interval into four subintervals of equal length.Then add to your sketch the rectangles associated with the Riemann sum $$\sum_{k=1}^n f\left(c_k\right)\Delta x_k $$ given that $c_k$ is the (a) left-hand endpoint, (b) right-hand endpoint, (c) the midpoint of the $k$-th subinterval. Make a separate sketch for each set of rectangles.

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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