Chapter 5.2, Problem 16E

Use a calculator or computer to make a table of values of left and right Riemann sums L_n and R_n for the integral ${\displaystyle {\int }_0^2{e}^{-x^2}dx}$ with n = 5, 10, 50, and 100. Between what two numbers must the value of the integral lie? Can you make a similar statement for the integral ${\displaystyle {\int }_{-1}^2{e}^{-x^2}dx}$? Explain.

To determine

To Calculate: The left Riemann sum $L_n$ and right Riemann sums $R_n$ for the integral ${\displaystyle {\int }_0^2{e}^{-x^2}}dx$ for $n=5,10,50,and100$ using the calculator.

To Find: The two numbers which the integral ${\displaystyle {\int }_0^2{e}^{-x^2}}dx$ lies between that two numbers.

To Explain: the statement for the integral ${\displaystyle {\int }_{-1}^2{e}^{-x^2}}dx$.

Answer to Problem 16E

The value of right Riemann sums $R_n$ for the integral ${\displaystyle {\int }_0^2{e}^{-x^2}}dx$ for $n=5,10,50,and100$ is tabulated in table 1.

The value of left Riemann sums $L_n$ for the integral ${\displaystyle {\int }_0^2{e}^{-x^2}}dx$ for $n=5,10,50,and100$ is tabulated in table 2.

The value of the integral lies between $0.09976$ and $0.09956$.

No, similar statement cannot be made for the integral ${\displaystyle {\int }_{-1}^2{e}^{-x^2}}dx$, since it is aincreasing function on limit $\left[-1,0\right]$ and decreasing function onlimit $\left[0,2\right]$.

Explanation of Solution

Given:

The integral function as ${\displaystyle {\int }_0^2{e}^{-x^2}}dx$.

Number of rectangles $n=5,10,50\text{and}100$.

Calculation:

Show the Equation of the integral as follows:

${\displaystyle {\int }_0^2{e}^{-x^2}}dx$ (1)

Consider the value of the function $f\left(x\right)=e^{-x^2}$ (2)

Calculate the value of the function $f\left(x\right)=e^{-x^2}$ within limit $\left[0,2\right]$.

Substitute 0 for x in Equation (6).

$\begin{array}{c}f\left(-1\right)=e^{-{\left(0\right)}^2}\\ =1\end{array}$

Substitute 2 for x in Equation (6).

$\begin{array}{c}f\left(2\right)=e^{-{\left(2\right)}^2}\\ =e^{-4}\\ =0.183\end{array}$

The function $f\left(x\right)=e^{-x^2}$ is a decreasing function within limits $\left[0,2\right]$.

For $n=5$

Find the width $\left(\Delta x\right)$ using the relation:

$\Delta x=\frac{b-a}{n}$ (3)

Here, the upper limit is b, the lower limit is a, and the number of rectangles is n.

The limits of the integral ${\displaystyle {\int }_0^2{e}^{-x^2}}dx$, $a=0$ and $b=2$.

Substitute $2$ for b, 5 for n, and 0 for a in Equation (3).

$\begin{array}{c}\Delta x=\frac{2-0}{5}\\ =\frac{2}{5}\end{array}$

Calculate the right Riemann Sum for $n=5,10,50\text{and}100$.

The right endpoints are $x_1=\frac{2}{5}$, $x_2=\frac{4}{5}$, $x_3=\frac{6}{5}$, $x_4=\frac{8}{5}$ and $x_5=2$.

The expression to find the right Riemann sum $R_n$ as shown below:

$\begin{array}{c}{R}_n={\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x\\ =f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{...}+f\left(x_n\right)\Delta x\end{array}$ (4)

Here, the right endpoint height of first rectangle is $f\left(x_1\right)$, the width is $\Delta x$, height of right endpoint of second rectangle is $f\left(x_2\right)$, and left endpoint height of n^th rectangle is $f\left(x_n\right)$.

Calculate the value of $f\left(x_1\right)$, $f\left(x_2\right)$, $f\left(x_3\right)$, $f\left(x_4\right)$, and $f\left(x_5\right)$ using the Equation (2).

Substitute $\frac{2}{5}$ for $x_1$ in Equation (2).

$\begin{array}{c}f\left(\frac{2}{5}\right)=e^{-{\left(\frac{2}{5}\right)}^2}\\ =e^{-\frac{4}{25}}\\ =0.8521\end{array}$

Substitute $\frac{4}{5}$ for $x_2$ in Equation (2).

$\begin{array}{c}f\left(\frac{4}{5}\right)=e^{-{\left(\frac{4}{5}\right)}^2}\\ =e^{-\frac{16}{25}}\\ =0.5272\end{array}$

Substitute $\frac{6}{5}$ for $x_3$ in Equation (2).

$\begin{array}{c}f\left(\frac{6}{5}\right)=e^{-{\left(\frac{6}{5}\right)}^2}\\ =e^{-\frac{36}{25}}\\ =0.2369\end{array}$

Substitute $\frac{8}{5}$ for $x_4$ in Equation (2).

$\begin{array}{c}f\left(\frac{8}{5}\right)=e^{-{\left(\frac{8}{5}\right)}^2}\\ =e^{-\frac{64}{25}}\\ =0.0773\end{array}$

Substitute $\frac{10}{5}$ for $x_5$ in Equation (2).

$\begin{array}{c}f\left(2\right)=e^{-{\left(\frac{10}{5}\right)}^2}\\ =e^{-\frac{100}{25}}\\ =0.0183\end{array}$

Calculate the right Riemann sum using calculator.

Substitute $\frac{2}{5}$ for $\Delta x$, 5 for n, and values of right end points in Equation (4).

$\begin{array}{c}{R}_5={\displaystyle \sum _{i=1}^{5}f\left(x_i\right)}\Delta x\\ =f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\mathrm{...}+f\left(x_5\right)\Delta x\\ =f\left(\frac{2}{5}\right)\times \frac{2}{5}+f\left(\frac{4}{5}\right)\times \frac{2}{5}+\mathrm{...}+f\left(2\right)\times \frac{2}{5}\\ =\left[\begin{array}{l}\left(0.8521\times 0.4\right)+\left(0.5272\times 0.4\right)+\left(0.2369\times 0.4\right)\\ +\left(0.0773\times 0.4\right)+\left(0.0183\times 0.4\right)\end{array}\right]\end{array}$

$=0.68479$

The value of right Riemann Sum for $n=5$ is $\underset{\_}{0.68479}$.

Similarly calculate the right Riemann sum for $n=10,50,100$.

Tabulate the values of right Riemann sum for $n=5,10,50\text{and}100$ as shown in table 1.

n	$R_n$
5	0.68479
10	0.68115
50	0.19652
100	0.09956

Table 1

Calculate the left Riemann Sum for $n=5,10,50\text{and}100$.

The left endpoints are $x_0=0$, $x_1=\frac{2}{5}$, $x_2=\frac{4}{5}$, $x_3=\frac{6}{5}$ and $x_4=\frac{8}{5}$.

The expression to find the left Riemann sum $L_n$ as shown below:

$\begin{array}{c}{L}_n={\displaystyle \sum _{i=1}^{n}f\left(x_i\right)}\Delta x\\ =f\left(x_0\right)\Delta x+f\left(x_1\right)\Delta x+\mathrm{...}+f\left(x_{n-1}\right)\Delta x\end{array}$ (5)

Here, the left endpoint height of first rectangle is $f\left(x_0\right)$, the width is $\Delta x$, height of left endpoint of second rectangle is $f\left(x_1\right)$, and left endpoint height of n^th rectangle is $f\left(x_{n-1}\right)$.

Calculate the value of $f\left(x_0\right)$, $f\left(x_1\right)$, $f\left(x_2\right)$, $f\left(x_3\right)$, and $f\left(x_4\right)$ using the Equation (2).

Substitute $0$ for $x_0$ in Equation (2).

$\begin{array}{c}f\left(0\right)=e^{-{\left(0\right)}^2}\\ =1\end{array}$

Substitute $\frac{2}{5}$ for $x_1$ in Equation (2).

$\begin{array}{c}f\left(\frac{2}{5}\right)=e^{-{\left(\frac{2}{5}\right)}^2}\\ =e^{-\frac{4}{25}}\\ =0.8521\end{array}$

Substitute $\frac{4}{5}$ for $x_2$ in Equation (2).

$\begin{array}{c}f\left(\frac{4}{5}\right)=e^{-{\left(\frac{4}{5}\right)}^2}\\ =e^{-\frac{16}{25}}\\ =0.5272\end{array}$

Substitute $\frac{6}{5}$ for $x_3$ in Equation (2).

$\begin{array}{c}f\left(\frac{6}{5}\right)=e^{-{\left(\frac{6}{5}\right)}^2}\\ =e^{-\frac{36}{25}}\\ =0.2369\end{array}$

Substitute $\frac{8}{5}$ for $x_4$ in Equation (2).

$\begin{array}{c}f\left(\frac{8}{5}\right)=e^{-{\left(\frac{8}{5}\right)}^2}\\ =e^{-\frac{64}{25}}\\ =0.0773\end{array}$

Calculate the left Riemann sum using calculator.

Substitute $\frac{2}{5}$ for $\Delta x$, 5 for n, and values of left end points in Equation (5).

$\begin{array}{c}{L}_5={\displaystyle \sum _{i=0}^{5-1}f\left(x_i\right)}\Delta x\\ =f\left(x_0\right)\Delta x+f\left(x_1\right)\Delta x+\mathrm{...}+f\left(x_4\right)\Delta x\\ =f\left(0\right)\times \frac{2}{5}+f\left(\frac{2}{5}\right)\times \frac{2}{5}+f\left(\frac{4}{5}\right)\times \frac{2}{5}+\mathrm{...}+f\left(\frac{8}{5}\right)\times \frac{2}{5}\\ =\left[\begin{array}{l}\left(1\times 0.4\right)+\left(0.8521\times 0.4\right)+\left(0.5272\times 0.4\right)\\ +\left(0.2369\times 0.4\right)+\left(0.0773\times 0.4\right)\end{array}\right]\end{array}$

$L_5=1.07746$

The value of left Riemann Sum for $n=5$ is $\underset{\_}{1.07746}$.

Similarly calculate the left Riemann sum for $n=10,50,100$.

Tabulate the values of left Riemann sum for $n=5,10,50\text{and}100$ as shown in table 2.

n	$L_n$
5	1.07746
10	0.80758
50	0.19809
100	0.09976

Table 2

Refer to Table 1 and Table 2.

The value of the integral lies as shown below:

$\begin{array}{l}{L}_n>{\displaystyle {\int }_0^2{e}^{-x^2}}dx>R_n\\ 0.09976>{\displaystyle {\int }_0^2{e}^{-x^2}}dx>0.09956\end{array}$

The value of the integral lies between $0.09976$ and $0.09956$.

whether similar statement can be made for the integral ${\displaystyle {\int }_{-1}^2{e}^{-x^2}}dx$.

Consider the value of the function $f\left(x\right)=e^{-x^2}$ (6)

Calculate the value of the function $f\left(x\right)=e^{-x^2}$ within limit $\left[-1,2\right]$.

Substitute $-1$ for x in Equation (6).

$\begin{array}{c}f\left(-1\right)=e^{-{\left(-1\right)}^2}\\ =e^{-1}\\ =0.3678\end{array}$

Substitute 0 for x in Equation (6).

$\begin{array}{c}f\left(0\right)=e^{-{\left(0\right)}^2}\\ =1\end{array}$

Substitute 2 for x in Equation (6).

$\begin{array}{c}f\left(2\right)=e^{-{\left(2\right)}^2}\\ =e^{-4}\\ =0.183\end{array}$

No, similar statement cannot be made for the integral ${\displaystyle {\int }_{-1}^2{e}^{-x^2}}dx$, since the function $f\left(x\right)=e^{-x^2}$ is a increasing function on limit $\left[-1,0\right]$ and decreasing function on limit $\left[0,2\right]$.