Chapter 6.5, Problem 27E

(a)

To determine

The trapezoidal rule approximation for n=10, 100, 1000.

Answer to Problem 27E

The value of the integral is 1.98, 1.99. based on the n values.

Explanation of Solution

Given information:

Consider the integral ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$

Formula used:

  $h=\frac{b-a}{n}$

Calculation:

In the trapezoidal rule, if [a, b] is partitioned into n subintervals of equal length $h=\frac{b-a}{n}$ the graph of on [a, b] can be approximated by a straight line segment at each subinterval.

The region between the curve and the x-axis is then approximated by the trapezoids, the area of each trapezoid being the length of its horizontal “altitude" times the average of two vertical “bases". That is,

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=h\frac{y_0+y_1}{2}+h\frac{y_1+y_2}{2}+\mathrm{........}+h\frac{y_{n-1}+y_n}{2}\\ =h\left(\frac{y_0}{2}+y_1+y_2+\mathrm{.....}+y_{n-1}+\frac{y_n}{2}\right)\\ {\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\frac{h}{2}\left(y_0+2y_1+2y_2+\mathrm{.....}+2y_{n-1}+y_n\right)\end{array}$

Where,

  $\begin{array}{l}{y}_0=f\left(a\right)\\ y_1=f\left(x_1\right)\\ .\\ .\\ y_{n-1}=f\left(x_{n-1}\right)\\ y_n=f\left(b\right)\end{array}$

To find integral use your calculator program of Trapezoidal rule approximations with

  $\begin{array}{l}f\left(x\right)=\sin x\\ \left[a,b\right]=\left[0,\pi \right]\\ n=10,100\text{and}\text{1000}\end{array}$

  

Conclusion:

The value of the integral is 1.98, 1.99.

(b)

To determine

The decimal places based on recording the errors.

Answer to Problem 27E

The value of error is $1.645\times {10}^{-6}$

Explanation of Solution

Given information:

Consider the integral ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$

Formula used:

  $\left|E_r\right|\le \frac{b-a}{12}{h}^2{M}_{\stackrel{·}{f}}$

Calculation:

If T represents the approximation to ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ given by the trapezoidal rule, then the errors ET satisfy

  $\left|E_r\right|\le \frac{b-a}{12}{h}^2{M}_{\stackrel{·}{f}}$

  $M_{\stackrel{·}{f}}$ It is maximum value of $\left|\stackrel{·}{f}\right|$ on [a, b]

The error is determined as below,

  $\begin{array}{l}\text{error}\left(\left|E_r\right|\right)={\displaystyle \underset{0}{\overset{\pi }{\int }}\sin x-T_n}\\ ={\left[-\cos x\right]}_0^{\pi }-T_n\\ =-\cos \pi -\left[-\cos 0\right]-T_n\\ \text{error}\left(\left|E_r\right|\right)=2-T_n\end{array}$

Where Tn represents the value of integral by trapezoidal rule for different value of n

  

Conclusion:

The value of error is $1.645\times {10}^{-6}$

(c)

To determine

The number of patterns.

Answer to Problem 27E

The value is $\left|E_{T_{10n}}\right|={10}^{-2}\left|E_{T_n}\right|$

Explanation of Solution

Given information:

Consider the integral ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$

Formula used:

Multiplication is used.

Calculation:

Observe the last column or above table, error gets multiplied approximately each time by 10-2

So,

  $\left|E_{T_{10n}}\right|={10}^{-2}\left|E_{T_n}\right|$

Conclusion:

The value is $\left|E_{T_{10n}}\right|={10}^{-2}\left|E_{T_n}\right|$

(d)

To determine

The error bound for ET accounts for the pattern.

Answer to Problem 27E

The error bound that accounts for the pattern is $\left|E_{r_n}\right|\times {10}^{-2}$

Explanation of Solution

Given information:

Consider the integral ${\displaystyle \underset{0}{\overset{\pi }{\int }}\sin xdx}$

Formula used:

  $\left|E_r\right|\le \frac{b-a}{12}{h}^2{M}_{\stackrel{·}{f}}$

Calculation:

Now, use the formula for error bound

  $\left|E_r\right|\le \frac{b-a}{12}{h}^2{M}_{\stackrel{·}{f}}$

Now,

  $h=\frac{b-a}{n}$

So,

  $\begin{array}{l}\left|E_{r_n}\right|\le \frac{\pi -0}{12}\times {\left(\frac{\pi -0}{n}\right)}^2{M}_{\stackrel{·}{f}}\\ \le \frac{{\pi }^3}{12n^2}\times M_{\stackrel{·}{f}}\end{array}$

  $\begin{array}{l}\left|E_{r_{{10}_n}}\right|\le \frac{\pi -0}{12}\times {\left(\frac{\pi -0}{10n}\right)}^2{M}_{\stackrel{·}{f}}\\ \le \frac{{\pi }^3}{12n^2}\times {10}^{-2}\times M_{\stackrel{·}{f}}\\ \le \left|E_{r_n}\right|\times {10}^{-2}\end{array}$

Hence, error bound accounts for the pattern.

Conclusion:

The error bound that accounts for the pattern is $\left|E_{r_n}\right|\times {10}^{-2}$