Chapter 5.2, Problem 8E

To determine

To estimate: The integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ using three equal subintervals with right end points, left end points and midpoints and compare it with the exact value of integral.

Answer to Problem 8E

The value of the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ for the right end points is $\underset{\_}{4.2}$.

The value of the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ for the left end points is $-\underset{\_}{6.2}$.

The value of the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ for the mid points is $\underset{\_}{-0.8}$.

Explanation of Solution

Given information:

The function f is an increasing function.

The number of intervals, $n=3$.

Each interval value is $\Delta x=2$.

The Riemann sum $R_n$ provides the right end point, $L_5$ provides the left end points, and $M_n$ provides the mid points estimate for the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$.

The expression to find the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ with right end points is shown below:

$\begin{array}{c}{R}_n={\displaystyle \sum _{i=1}^{3}f\left(x_i\right)\Delta x}\\ =\Delta x\left[f\left(x_9\right)+f\left(x_7\right)+f\left(x_5\right)\right]\end{array}$ (1)

Substitute3 for n, 2 for $\Delta x$, and corresponding x values in Equation (1).

$R_3=2\left[f\left(9\right)+f\left(7\right)+f\left(5\right)\right]$ (2)

Substitute $1.8$ for $f\left(9\right)$, $0.9$ for $f\left(7\right)$, and $-0.6$ for $f\left(5\right)$ in Equation (2).

$\begin{array}{c}{R}_3=2\left[f\left(9\right)+f\left(7\right)+f\left(5\right)\right]\\ =2\left(1.8+0.9-0.6\right)\\ =4.2\end{array}$

Therefore, the estimation of the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ with right end points is $\underset{\_}{4.2}$.

For the increasing function of f,the right end points are greater than the exact value of the integral that is overestimate.

The expression to find the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ with left end points is shown below:

$\begin{array}{c}{R}_n={\displaystyle \sum _{i=1}^{3}f\left(x_{i-1}\right)\Delta x}\\ =\Delta x\left[f\left(x_3\right)+f\left(x_5\right)+f\left(x_7\right)\right]\end{array}$ (3)

Substitute 3 for n,2 for $\Delta x$, and corresponding x values in Equation (3).

$L_3=2\left[f\left(3\right)+f\left(5\right)+f\left(7\right)\right]$ (4)

Substitute $-3.4$ for $f\left(3\right)$, $-0.6$ for $f\left(5\right)$, and $0.9$ for $f\left(7\right)$ in Equation (4).

$\begin{array}{c}{L}_3=2\left[f\left(3\right)+f\left(5\right)+f\left(7\right)\right]\\ =2\left(-3.4-0.6+0.9\right)\\ =-6.2\end{array}$

Therefore, the estimation of the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ with left end points is $\underset{\_}{-6.2}$.

For the increasing function of f,the left end points are smaller than the exact value of the integral that is underestimating.

The expression to find the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ with mid points is shown below:

$\begin{array}{c}{M}_n={\displaystyle \sum _{i=1}^{3}f\left({\overline{x}}_i\right)\Delta x}\\ =\Delta x\left[f\left({\overline{x}}_4\right)+f\left({\overline{x}}_6\right)+f\left({\overline{x}}_8\right)\right]\end{array}$ (5)

Substitute 3 for n,2 for $\Delta x$, and corresponding x values in Equation (5).

$M_3=2\left[f\left(4\right)+f\left(6\right)+f\left(8\right)\right]$ (6)

Substitute $-2.1$ for $f\left(4\right)$, $0.3$ for $f\left(6\right)$, and $1.4$ for $f\left(8\right)$ in Equation (6).

$\begin{array}{c}{M}_3=2\left[f\left(4\right)+f\left(6\right)+f\left(8\right)\right]\\ =2\left(-2.1+0.3+1.4\right)\\ =-0.8\end{array}$

Therefore, the estimation of the integral ${\displaystyle {\int }_3^{9}f\left(x\right)dx}$ with mid points is $\underset{\_}{-0.8}$.

Due to increasing function of f, cannot be concluded about midpoint estimate when compared to the exact value of integral.