Chapter 5.5, Problem 19E

a.

To determine

To state: The approximate value of the integral ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$ with $n=4$ using Simpson’s Rule.

Answer to Problem 19E

The resultant answer is 35.

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$ .

Formula used: To approximate ${\displaystyle \underset{-1}{\overset{3}{\int }}f\left(x\right)dx}$ use $S=\frac{h}{3}\left(y_0+4y_1+4y_3+\cdots +2y_{n-2}+4y_{n-1}+y_n\right)$ where $\left[a,b\right]$ is partitioned into an even number n of subintervals of equal length $h=\frac{b-a}{n}$ .

Consider the given integral:

  ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$

Now apply the Simpson’s Rule with $n=4$ .

  $\begin{array}{c}f\left(x\right)=x^3-2x\\ \left[a,b\right]=\left[-1,3\right]\end{array}$

Now find $h=\frac{b-a}{n}$ .

  $\begin{array}{c}h=\frac{3-\left(-1\right)}{4}\\ =1\end{array}$

Now form a table as shown below:

x		$y=f\left(x\right)$
$x_0$	-1	$y_0$	1
$x_1$	0	$y_1$	0
$x_2$	1	$y_2$	-1
$x_3$	2	$y_3$	4
$x_4$	3	$y_4$	21

Now substitute the value from the above table in the equation:

  $\begin{array}{c}{\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+y_4\right)\\ =\frac{1}{3}\left(1+4\left(0\right)+2\left(-1\right)+4\left(4\right)+21\right)\\ =\frac{1}{3}\left(1+0-2+16+21\right)\\ =35\end{array}$

Therefore, the interval of convergence is $\left(-3,5\right)$ .

b.

To determine

To state: The exact value of the integral ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$ also state the error $\left|E_S\right|$ .

Answer to Problem 19E

The resultant exact value is 12 and the error $\left|E_S\right|=0$ .

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$ .

Consider the given integral:

  ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$

If $f$ is continuous at every point of $\left[a,b\right]$ , and if $F$ is any anti-derivative of $f$ on $\left[a,b\right]$ then ${\displaystyle \underset{-1}{\overset{3}{\int }}f\left(x\right)dx=F\left(b\right)-F\left(a\right)}$ .

This part of the fundamental theorem is called the integral evaluation theorem.

Now use the above formula to find the value of integral.

  $F\left(x\right)=\frac{x^4}{4}-x^2$

  $\begin{array}{c}{\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}=\left(\frac{3^4}{4}-3^2\right)-\left(\frac{{\left(-1\right)}^4}{4}-{\left(-1\right)}^2\right)\\ =\left(\frac{81}{4}-9\right)-\left(\frac{1}{4}-1\right)\\ =12\end{array}$

  $\begin{array}{c}\left|E_S\right|=\left|\text{value obtained from Simpson's rule}-\text{exact value of the integral}\right|\\ =\left|12-12\right|\\ =0\end{array}$

c.

To determine

To state: How a person predicted the value found in part (b) from knowing the error-bound formula.

Answer to Problem 19E

The resultant answer is $\left|E_s\right|=0$ .

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$ .

The error $E_S$ in Simpson’s rule depends on h and the fourth derivate. It satisfies the inequality:

  $\left|E_S\right|\le \frac{b-a}{180}\le h^4{M}_{f^{\left(4\right)}}$

Here $\left[a,b\right]$ is the interval of integration, h is the length of each subinterval, and $M_{f^{\left(4\right)}}$ is the maximum value of $\left|f^{\left(4\right)}\right|\text{ on }\left[a,b\right]$ , provided that $f\left(4\right)$ is continuous.

  $\begin{array}{c}\left|f^{\left(4\right)}\right|=\frac{d^4\left(x^3-2x\right)}{d{x}^4}\\ =\frac{d^4\left(x^3\right)}{d{x}^4}-\frac{d^4\left(2x\right)}{d{x}^4}\\ =0-0\\ =0\end{array}$

Hence $\left|E_s\right|=0$ .

d.

To determine

To state: The general statement about using Simpson’s Rule to approximate integrals of cubic polynomials.

Answer to Problem 19E

The fourth derivate of the cubic polynomial always vanishes so error is zero. Hence, approximation given by Simpson’s rule is always equal to exact value of the integral.

Explanation of Solution

Given information:

The given integral is ${\displaystyle \underset{-1}{\overset{3}{\int }}\left(x^3-2x\right)dx}$ .

The error $E_S$ in Simpson’s rule depends on h and the fourth derivate. It satisfies the inequality:

  $\left|E_S\right|\le \frac{b-a}{180}\le h^4{M}_{f^{\left(4\right)}}$

Here $\left[a,b\right]$ is the interval of integration, h is the length of each subinterval, and $M_{f^{\left(4\right)}}$ is the maximum value of $\left|f^{\left(4\right)}\right|\text{ on }\left[a,b\right]$ , provided that $f\left(4\right)$ is continuous.

A general cubic polynomial is: $y=a{x}^3+b{x}^2+cx+p$

Take fourth derivate both sides:

  $\begin{array}{c}\frac{d^{4}y}{d{x}^4}=\frac{d^4\left(a{x}^3\right)}{d{x}^4}+\frac{d^4\left(b{x}^2\right)}{d{x}^4}+\frac{d^4\left(cx\right)}{d{x}^4}+\frac{d^4\left(p\right)}{d{x}^4}\\ =0+0+0\\ =0\end{array}$

Since fourth derivate of the cubic polynomial always vanishes so error is zero. Hence, approximation given by Simpson’s rule is always equal to exact value of the integral.