Chapter 5.2, Problem 26E

a)

To determine

To find: an approximation to the integral ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ using a Riemann sum with right endpoints.

Answer to Problem 26E

The approximation to the integral ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ using a Riemann sum with right endpoints is $\underset{\_}{-1.5}$.

Explanation of Solution

Given:

The integral function is, ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$.

The number of sub intervals as $n=8$.

The upper limit is 4 and lower limit is 0.

Theorem 4: If f is integrable on [a, b], then

${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$

Where $\Delta x=\frac{b-a}{n}$ and $x_i=a+i\Delta x$

Calculation:

The integral function is,

${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$

The Riemann sum with right endpointsformula is,

${\displaystyle {\int }_a^{b}f\left(x\right)dx}={\displaystyle \sum _{i=1}^{n}f\left(x_i^{*}\right)\Delta x}$ (1)

Calculate the subinterval $\left(\Delta x\right)$ using the formula:

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

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

Substitute 4 for b, 0 for a, and 8 for n.

$\begin{array}{c}\Delta x=\frac{4-0}{8}\\ =0.5\end{array}$

Hence, the subinterval length is 0.5 for the limits $\left[0,4\right]$.

The subinterval values are 0.5, 1.0, 1.5, 2.0, 2.5,3.0, 3.5,and 4.0for $x_1$, $x_2$, $x_3$, $x_4$, $x_5$, $x_6$, $x_7$, and $x_8$

Calculate the approximation to the integral using a Riemann sum with right endpoints:

Substitute 0.5 for $\Delta x$ and 8 for n in Equation (1).

$\begin{array}{c}{\displaystyle {\int }_a^{b}f\left(x\right)dx}={\displaystyle \sum _{i=1}^8\left[f\left(x_i\right)\right]0.5}\\ =\left[\begin{array}{l}f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)+f\left(x_4\right)+\\ f\left(x_5\right)+f\left(x_6\right)+f\left(x_7\right)+f\left(x_8\right)\end{array}\right]0.5\end{array}$ (2)

Apply the subinterval values in Equation (3).

${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\left[\begin{array}{l}f\left(0.5\right)+f\left(1.0\right)+f\left(1.5\right)+f\left(2.0\right)+\\ f\left(2.5\right)+f\left(3.0\right)+f\left(3.5\right)+f\left(4.0\right)\end{array}\right]0.5$ (3)

Consider $f\left(x\right)=x^2-3x$ (4)

Calculate $f\left(0.5\right)$

Substitute 0.5 for x in Equation (4).

$\begin{array}{c}f\left(0.5\right)={\left(0.5\right)}^2-3\left(0.5\right)\\ =-0.25\end{array}$

Calculate $f\left(1.0\right)$:

Substitute 1.0 for x in Equation (4).

$\begin{array}{c}f\left(1.0\right)={\left(1.0\right)}^2-3\left(1.0\right)\\ =-2.00\end{array}$

Calculate $f\left(1.5\right)$

Substitute 1.5 for x in Equation (4).

$\begin{array}{c}f\left(1.5\right)={\left(1.5\right)}^2-3\left(1.5\right)\\ =-2.25\end{array}$

Calculate $f\left(2.0\right)$:

Substitute 2.0 for x in Equation (4).

$\begin{array}{c}f\left(2.0\right)={\left(2.0\right)}^2-3\left(2.0\right)\\ =-2.00\end{array}$

Calculate $f\left(2.5\right)$

Substitute 2.5 for x in Equation (4).

$\begin{array}{c}f\left(2.5\right)={\left(2.5\right)}^2-3\left(2.5\right)\\ =-1.25\end{array}$

Calculate $f\left(3.0\right)$:

Substitute 3.0 for x in Equation (4).

$\begin{array}{c}f\left(3.0\right)={\left(3.0\right)}^2-3\left(3.0\right)\\ =0.00\end{array}$

Calculate $f\left(3.5\right)$

Substitute 3.5 for x in Equation (4).

$\begin{array}{c}f\left(3.5\right)={\left(3.5\right)}^2-3\left(3.5\right)\\ =-1.75\end{array}$

Calculate $f\left(4.0\right)$:

Substitute 4.0 for x in Equation (4).

$\begin{array}{c}f\left(4.0\right)={\left(4.0\right)}^2-3\left(4.0\right)\\ =4.00\end{array}$

Substitute -1.25, -2.00, -2.25, -2.00, -1.25, 0.00, 1.75, and 4.00 for $f\left(0.5\right)$,$f\left(1.0\right)$,$f\left(1.5\right)$, $f\left(2.0\right)$, $f\left(2.5\right)$, $f\left(3.0\right)$, $f\left(3.5\right)$, and $f\left(4.0\right)$ respectively in Equation (3).

$\begin{array}{c}{\displaystyle {\int }_a^{b}f\left(x\right)dx}=\left[-1.25-2.00-2.25-2.00-1.25+0.00+1.75+4.00\right]0.5\\ =-1.5\end{array}$

Hence, the approximation to the integral ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ using a Riemann sum with right endpoints is $\underset{\_}{-1.5}$.

b)

To determine

To draw: a diagram similar to figure 3 to illustrate the approximation in part (a).

Answer to Problem 26E

A diagram similar to figure 3 to illustrate the approximation in part (a) is drawn.

Explanation of Solution

Calculation:

Show the equation as below:

$y=x^2-3x$ (5)

Plot a graph for the equation $y=x^2-3x$ using the calculation as follows:

Calculate y value using Equation (5).

Substitute 0 for x in Equation (5).

$\begin{array}{c}y={\left(0\right)}^2-3\left(0\right)\\ =0\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(0,0\right)$.

Calculate y value using Equation (5).

Substitute 1 for x in Equation (5).

$\begin{array}{c}y={\left(1\right)}^2-3\left(1\right)\\ =-2\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(1,-2\right)$.

Calculate y value using Equation (5).

Substitute 2 for x in Equation (5).

$\begin{array}{c}y={\left(2\right)}^2-3\left(2\right)\\ =-2\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(2,-2\right)$.

Calculate y value using Equation (5).

Substitute 3 for x in Equation (5).

$\begin{array}{c}y={\left(3\right)}^2-3\left(3\right)\\ =0\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(3,0\right)$.

Calculate y value using Equation (5).

Substitute 4 for x in Equation (5).

$\begin{array}{c}y={\left(4\right)}^2-3\left(4\right)\\ =4\end{array}$

Hence, the co-ordinate of $\left(x,y\right)$ is $\left(4,4\right)$.

Draw the region as shown in Figure 1.

c)

To determine

To evaluate: the integral ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ using Theorem 4.

Answer to Problem 26E

The evaluation of the integral ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ using Theorem 4 is $\underset{\_}{-\frac{8}{3}}$.

Explanation of Solution

Calculation:

The Theorem 4 is,

${\displaystyle {\int }_a^{b}f\left(x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^{n}f\left(x_i\right)\Delta x}$ (6)

Calculate the subinterval $\left(\Delta x\right)$ using the formula:

$\Delta x=\frac{b-a}{n}$ (7)

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

Substitute 4 for b, and 0 for a.

$\begin{array}{c}\Delta x=\frac{4-0}{n}\\ =\frac{4}{n}\end{array}$

Calculate the $x_i$ using the formula:

$x_i=a+i\Delta x$ (8)

Substitute $\frac{4}{n}$ for $\Delta x$ and 0 for a in Equation (7)

$\begin{array}{c}{x}_i=0+i\frac{4}{n}\\ =\frac{4i}{n}\end{array}$

The integral function is,

${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ (9)

Apply the Theorem 4 in Equation (5).

${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left(x_i^2-3x_i\right)\Delta x}$ (10)

Substitute $\frac{4i}{n}$ for $x_i$ and $\frac{4}{n}$ for $\Delta x$ in Equation (10).

$\begin{array}{c}{\displaystyle {\int }_0^4\left(x^2-3x\right)dx}=\underset{n\to \infty }{\lim }{\displaystyle \sum _{i=1}^n\left[{\left(\frac{4i}{n}\right)}^2-3\left(\frac{4i}{n}\right)\right]\left(\frac{4}{n}\right)}\\ =\underset{n\to \infty }{\lim }\frac{4}{n}\left[\frac{16}{n^2}{\displaystyle \sum _{i=1}^n{i}^2}-\frac{12}{n}{\displaystyle \sum _{i=1}^{n}i}\right]\\ =\underset{n\to \infty }{\lim }\left[\frac{64}{n^3}\times \frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{48}{n^2}\times \frac{n\left(n+1\right)}{2}\right]\\ =\underset{n\to \infty }{\lim }\left[\frac{32}{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)-24\left(1+\frac{1}{n}\right)\right]\end{array}$

$\begin{array}{c}=\left(\frac{32}{3}\times 2\right)-24\\ =-\frac{8}{3}\end{array}$

Hence, the evaluation of the integral ${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}$ using Theorem 4 is $\underset{\_}{-\frac{8}{3}}$.

d)

To determine

To draw: a diagram similar to figure 4 to illustrate the integral in part (c) as a difference of areas.

Answer to Problem 26E

A diagram similar to figure 4 to illustrate the integral in part (c) as a difference of areas is drawn.

Explanation of Solution

Calculation:

Draw the region as shown in Figure 2.

Refer Figure 2

Positive sign $(+)$ refers area $A_1$ and negative sign $(-)$ refers area $A_2$.

The integral function is as follows:

${\displaystyle {\int }_0^4\left(x^2-3x\right)dx}=A_1-A_2$.