Chapter 6.5, Problem 19E

(a)

To determine

Approximate the value of integral using Simpson’s Rule with $n=4$ .

Answer to Problem 19E

With $n=4$:

The approximate value is 12.

Explanation of Solution

Given information:

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

Simpson’s rule formula:

  $S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+\mathrm{...}+2y_{n-2}+4y_{n-1}+y_n\right)$

Where,

  $h=\frac{b-a}{n}$

Now,

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

We have

  $a=1\text{and}b=3$

With $n=4$ :

  $h=\frac{3-\left(-1\right)}{4}=\frac{4}{4}=1$

Since

  $y=x^3-2x$

Then

For $x=-1$ :

  $y_0=y\left(-1\right)={\left(-1\right)}^3-2\left(-1\right)=-1+2=1$

For $x=0$ :

  $y_1=y\left(0\right)={\left(0\right)}^3-2\left(0\right)=0-0=0$

For $x=1$ :

  $y_2=y\left(1\right)={\left(1\right)}^3-2\left(1\right)=1-2=-1$

For $x=2$ :

  $y_3=y\left(2\right)={\left(2\right)}^3-2\left(2\right)=8-4=4$

For $x=3$ :

  $y_4=y\left(3\right)={\left(3\right)}^3-2\left(3\right)=27-6=21$

Substitute the above values:

  $\begin{array}{l}S=\frac{h}{3}\left(y_0+4y_1+2y_2+4y_3+y_4\right)\\ =\frac{1}{3}\left(1+4\left(0\right)+2\left(-1\right)+4\left(4\right)+21\right)\\ =\frac{1}{3}\left(1-2+16+21\right)\\ =\frac{1}{3}\left(36\right)\\ =12\end{array}$

(b)

To determine

Exact value of integral and error $\left|E_S\right|$ .

Answer to Problem 19E

Exact value of integral is also 12.

Error,

  $\left|E_s\right|=0$

Explanation of Solution

Given information:

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

Evaluate the integral:

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

Thus,

The actual value of integral is 12.

From Part (a),

The estimated value was 12.

Then

Error,

  $\left|E_s\right|=\left|\text{Actual}\text{value}-\text{Estimate}\right|=\left|12-12\right|=0$

(c)

To determine

Prediction of error $\left|E_S\right|$ according to error bound formula.

Answer to Problem 19E

From error bound formula,

  $\left|E_S\right|\le \frac{b-a}{180}{h}^4\left(0\right)=0$

Explanation of Solution

Given information:

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

According to error − bound formula:

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

Since

  $f\left(x\right)=x^3-2$

Then

First derivative:

  $f\text{'}\left(x\right)=3x^2$

Second derivative:

  $f\text{'}\text{'}\left(x\right)=6x$

Third derivative:

  $f^{\left(3\right)}\left(x\right)=6$

Fourth derivative:

  $f^{\left(4\right)}\left(x\right)=0$

Hence,

  $M_{f\left(4\right)}=0$

Therefore,

  $\left|E_S\right|\le \frac{b-a}{180}{h}^4\left(0\right)=0$

(d)

To determine

Whether it is possible to make general statement about using Simpson’s Rule to approximate integrals of cubic polynomials.

Answer to Problem 19E

Simpson rule will always give exact values for all cubic polynomials.

Explanation of Solution

Given information:

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

For any cubic polynomial,

If $f^{\left(4\right)}=0$ ,

Then

  $M_{f\left(4\right)}=0$ for all cubic polynomial.

Then

The error − bound formula would give

  $\left|E_S\right|\le 0$

Such that

Error is always zero.

Therefore,

Simpson rule will always give exact values for all cubic polynomials.