Chapter 5.2, Problem 12E

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places.

${\displaystyle {\int }_0^{\pi }x{\sin }^{2}xdx},n=4$

To determine

To evaluate: The integral function ${\displaystyle {\int }_0^{\pi }x}{\sin }^{2}xdx$ using midpoint Rule.

Answer to Problem 12E

The approximate value of the integral function ${\displaystyle {\int }_0^{\pi }x}{\sin }^{2}xdx$ is2.4674 by using midpoint rule.

Explanation of Solution

Given information:

The integral function is ${\displaystyle {\int }_0^{\pi }x}{\sin }^{2}xdx$ and $n=4$.

The Expression for midpointrule is shown below:

$\begin{array}{c}{\displaystyle {\int }_a^{b}f\left(x\right)dx}\approx {\displaystyle \sum _{i=1}^{n}f\left({\overline{x}}_i\right)\Delta x}\\ =\Delta x\left[f\left({\overline{x}}_1\right)+\mathrm{...}+f\left({\overline{x}}_n\right)\right]\end{array}$ (1)

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

$\Delta x=\frac{b-a}{n}$ (2)

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

Substitute $\pi$ for b, 0 for a, and 4 for $n$ in Equation (2).

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

Calculate the value of right end points $\left(x_i\right)$ using the relation:

$x_i=a+i\Delta x$ (3)

Calculate the value of left end points $\left(x_{i-1}\right)$ by rearranging Equation (3).

Substitute $\left(i-1\right)$ for $i$ in Equation (3)

$x_{i-1}=a+\left(i-1\right)\Delta x$ (4)

Calculate the value of mid points $\left({\overline{x}}_i\right)$ using the relation:

${\overline{x}}_i=\frac{1}{2}\left(x_{i-1}+x_i\right)$ (5)

Substitute $a+i\Delta x$ for $x_i$ and $a+\left(i-1\right)\Delta x$ for $x_{i-1}$ in Equation (5).

$\begin{array}{c}{\overline{x}}_i=\frac{1}{2}\left[a+\left(i-1\right)\Delta x+a+i\Delta x\right]\\ =\frac{1}{2}\left(a+i\Delta x-\Delta x+a+i\Delta x\right)\\ =\frac{1}{2}\times \left[2\left(a+i\Delta x\right)-\Delta x\right]\\ =\left(a+i\Delta x\right)-\frac{\Delta x}{2}\end{array}$ (6)

Calculate the value of ${\overline{x}}_1$ using Equation (6).

Substitute 0 for $a$, 1 for $i$,and $\frac{\pi }{4}$ for $\Delta x$ in Equation (6).

$\begin{array}{c}{\overline{x}}_1=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =(0+1\times \frac{\pi }{4})-\frac{1}{2}\times \frac{\pi }{4}\\ =\frac{\pi }{4}-\frac{\pi }{8}\\ =\frac{\pi }{8}\end{array}$

Calculate the value of ${\overline{x}}_2$ using Equation (6).

Substitute 0 for $a$, 2 for $i$,and $\frac{\pi }{4}$ for $\Delta x$ in Equation (6).

$\begin{array}{c}{\overline{x}}_2=\left(0+2\times \frac{\pi }{4}\right)-\frac{1}{2}\times \frac{\pi }{4}\\ =\frac{2\pi }{4}-\frac{\pi }{8}\\ =\frac{3\pi }{8}\end{array}$

Calculate the value of ${\overline{x}}_3$ using Equation (6).

Substitute 0 for $a$, 3 for $i$,and $\frac{\pi }{4}$ for $\Delta x$ in Equation (6).

$\begin{array}{c}{\overline{x}}_3=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =0+\left(3\times \frac{\pi }{4}\right)-\left(\frac{1}{2}\times \frac{\pi }{4}\right)\\ =\frac{3\pi }{4}-\frac{\pi }{8}\\ =\frac{5\pi }{8}\end{array}$

Calculate the value of ${\overline{x}}_4$ using Equation (6).

Substitute 0 for $a$, 3 for $i$,and $\frac{\pi }{4}$ for $\Delta x$ in Equation (6).

$\begin{array}{c}{\overline{x}}_4=\left(a+i\Delta x\right)-\frac{\Delta x}{2}\\ =(0+4\times \frac{\pi }{4})-\frac{1}{2}\times \frac{\pi }{4}\\ =\pi -\frac{\pi }{8}\\ =\frac{7\pi }{8}\end{array}$

Consider $f\left(x\right)=x{\sin }^{2}x$.

Rearrange the expression $f\left(x\right)=x{\sin }^{2}x$ in terms of $f\left({\overline{x}}_i\right)$ as shown below.

Substitute ${\overline{x}}_i$ for $x$.

$f\left({\overline{x}}_i\right)={\overline{x}}_i{\sin }^2{\overline{x}}_i$ (7)

Calculate the value of $f\left({\overline{x}}_1\right)$ using Equation (7).

Substitute $\frac{\pi }{8}$ for ${\overline{x}}_1$ in Equation (7).

$\begin{array}{c}f\left({\overline{x}}_1\right)={\overline{x}}_1{\sin }^2{\overline{x}}_1\\ =\frac{\pi }{8}\times {\sin }^2\left(\frac{\pi }{8}\right)\end{array}$

Calculate the value of $f\left({\overline{x}}_2\right)$ using Equation (7).

Substitute $\frac{3\pi }{8}$ for ${\overline{x}}_2$ in Equation (7).

$\begin{array}{c}f\left({\overline{x}}_2\right)={\overline{x}}_2{\sin }^2{\overline{x}}_2\\ =\frac{3\pi }{8}\times {\sin }^2\left(\frac{3\pi }{8}\right)\end{array}$

Calculate the value of $f\left({\overline{x}}_3\right)$ using Equation (7).

Substitute $\frac{5\pi }{8}$ for ${\overline{x}}_3$ in Equation (7).

$\begin{array}{c}f\left({\overline{x}}_3\right)={\overline{x}}_3{\sin }^2{\overline{x}}_3\\ =\frac{5\pi }{8}\times {\sin }^2\left(\frac{5\pi }{8}\right)\end{array}$

Calculate the value of $f\left({\overline{x}}_4\right)$ using Equation (7).

Substitute $\frac{7\pi }{8}$ for ${\overline{x}}_4$ in Equation (7).

$\begin{array}{c}f\left({\overline{x}}_4\right)={\overline{x}}_4{\sin }^2{\overline{x}}_4\\ =\frac{7\pi }{8}\times {\sin }^2\left(\frac{7\pi }{8}\right)\end{array}$

Substitute $x{\sin }^{2}x$ for $f\left(x\right)$,$\frac{\pi }{4}$ for $\Delta x$,$\frac{\pi }{8}\times {\sin }^2\left(\frac{\pi }{8}\right)$ for $f\left({\overline{x}}_1\right)$, $\frac{3\pi }{8}\times {\sin }^2\left(\frac{3\pi }{8}\right)$ for $f\left({\overline{x}}_2\right)$, $\frac{5\pi }{8}\times {\sin }^2\left(\frac{5\pi }{8}\right)$ for $f\left({\overline{x}}_3\right)$,$\frac{7\pi }{8}\times {\sin }^2\left(\frac{7\pi }{8}\right)$ for $f\left({\overline{x}}_4\right)$, $0$ for $a$, and $\pi$ for $b$ in Equation (1).

$\begin{array}{c}{\displaystyle {\int }_0^{\pi }x}{\sin }^{2}xdx=\Delta x\left[f\left({\overline{x}}_1\right)+f\left({\overline{x}}_2\right)+f\left({\overline{x}}_3\right)+f\left({\overline{x}}_4\right)\right]\\ =\frac{\pi }{4}\times \left[\frac{\pi }{8}\times {\sin }^2\left(\frac{\pi }{8}\right)+\frac{3\pi }{8}\times {\sin }^2\left(\frac{3\pi }{8}\right)+\frac{5\pi }{8}\times {\sin }^2\left(\frac{5\pi }{8}\right)+\frac{7\pi }{8}\times {\sin }^2\left(\frac{7\pi }{8}\right)\right]\\ =\frac{\pi }{4}\times \left[\frac{\pi }{8}\times {\sin }^2\left(\frac{\pi }{8}\right)+\frac{3\pi }{8}\times {\sin }^2\left(\frac{3\pi }{8}\right)+\frac{5\pi }{8}\times {\sin }^2\left(\frac{5\pi }{8}\right)+\frac{7\pi }{8}\times {\sin }^2\left(\frac{7\pi }{8}\right)\right]\\ =\frac{\pi }{4}\times \left[\frac{\pi }{8}\times \left(0.1464\right)+\frac{3\pi }{8}\times \left(0.8536\right)+\frac{5\pi }{8}\times \left(0.8536\right)+\frac{7\pi }{8}\times \left(0.1464\right)\right]\end{array}$

$\begin{array}{c}=\frac{{\pi }^2}{32}\times \left[\left(0.1464\right)+3\left(0.8536\right)+5\left(0.8536\right)+7\left(0.1464\right)\right]\\ =\frac{{\pi }^2}{32}\times \left[0.1464+2.5607+4.2678+1.0251\right]\\ =0.3084\times 8\\ =2.4674\end{array}$

Thus, the value of integral function ${\displaystyle {\int }_0^{\pi }x}{\sin }^{2}xdx$ is 2.4674.