Q.18 - Calculus

Question

(a) Explain why the definite integral $I={\int }_0^2\frac{dx}{1+x^3}$ exists.

(b) Explain how to use Simpson's method to approximate the value of I to within a few percent $0.001$of its actual value.

(c) Explain why you cannot use the method of Exercise 17 (b) and (c) to approximate $I$.

Step-by-Step Solution

Answer

(a) The integral $I$ exists, since the function $\frac{1}{1+x^3}$ is continuous in the interval$[0,2]$.

(b) Using  Simpson's method ,

$$\begin{gathered} f\left(x_1=\frac{1}{6}\right)=\frac{1}{1+{\left(\frac{1}{6}\right)}^3} \\ =\frac{216}{217} \end{gathered}$$

(c) The method of Exercise 17 (b) and (c) can not be used to approximate $I$ because the term will continue to grow and the value will be increased more than the given value.

Part (a) Step 1: Given information

The given integral is $\mathrm{I}={\int }_0^2\frac{\mathrm{dx}}{1+{\mathrm{x}}^3}$

Part (a) Step 2 : Calculations

Because the function $\frac{1}{1+{\mathrm{x}}^3}$ is continuous in the interval, the integral  $\left[0,2\right]$ where I exists.

Part (b) Step 1: Given information

The given integral and values are $\mathrm{I}={\int }_0^2\frac{\mathrm{dx}}{1+{\mathrm{x}}^3},\mathrm{a}=0,\mathrm{b}=2\text{ and }\mathrm{n}=3$

Part (b) Step 2: Calculations

Let us apply the formula to approximate the integral I to within $0.001$ of its true value using Simpson's approximation.

${\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\approx \frac{\mathrm{\Delta x}}{3}\left[\mathrm{f}\left({\mathrm{x}}_0\right)+4\mathrm{f}\left({\mathrm{x}}_1\right)+2\mathrm{f}\left({\mathrm{x}}_2\right)+\dots +2\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}-2}\right)+4\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}-1}\right)+\mathrm{f}\left({\mathrm{x}}_{\mathrm{n}}\right)\right]$

For $\mathrm{\Delta x}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}$

In the integral  $\mathrm{I}={\int }_0^2 \frac{\mathrm{dx}}{1+{\mathrm{x}}^3},\mathrm{a}=0,\mathrm{b}=2\text{ and }\mathrm{n}=3$(because the integral I must be approximated to within 0.001 of its true value),

$\begin{array}{r}\mathrm{\Delta }x=\frac{2-0}{3}\\ =\frac{2}{3}\end{array}$

Implies that,

$\mathrm{I}\approx \frac{\mathrm{\Delta x}}{3}\left[\mathrm{f}\left({\mathrm{x}}_0\right)+4\mathrm{f}\left({\mathrm{x}}_1\right)+2\mathrm{f}\left({\mathrm{x}}_2\right)+\mathrm{f}\left({\mathrm{x}}_3\right)\right]$

and 

$\begin{array}{r}\mathrm{f}\left({\mathrm{x}}_1=\frac{1}{6}\right)=\frac{1}{1+{\left(\frac{1}{6}\right)}^3}\\ =\frac{216}{217}\end{array}$

Parallelly,

$\begin{array}{r}\mathrm{f}\left({\mathrm{x}}_2=\frac{1}{3}\right)=\frac{1}{1+{\left(\frac{1}{3}\right)}^3}\\ =\frac{27}{28}\end{array}$

Furthermore,

$\begin{array}{r}f\left(x_2=\frac{1}{2}\right)=\frac{1}{1+{\left(\frac{1}{2}\right)}^3}\\ =\frac{8}{9}\end{array}$

Finally,

$\begin{array}{r}\mathrm{I}\approx \frac{2}{3}\left[1+4\left(\frac{216}{217}\right)+2\left(\frac{27}{28}\right)+\frac{8}{9}\right]\\ \approx \frac{2}{9}[1+3.98156+1.92857+0.8888]\end{array}$

That is, $\text{I≈1.733}$

Part (c) Step 1: Given information

The given Maclaurin series $\frac{1}{1-x}$ is $\frac{1}{1-\mathrm{x}}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{x}}^{\mathrm{k}}$

Part (c) Step 2 : Calculations

For the first time $\frac{1}{1-x}$since the Maclaurin series, is $\frac{1}{1-\mathrm{x}}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\mathrm{x}}^{\mathrm{k}}$

As a result, the Maclaurin series for $\frac{1}{1+{\mathrm{x}}^3}$ may be determined by replacing x with $-{\mathrm{x}}^3$ in the Maclaurin series for $\frac{1}{1-\mathrm{x}}$

That is,

$\frac{1}{1+{\mathrm{x}}^3}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} {\left(-{\mathrm{x}}^3\right)}^{\mathrm{k}}$

Implies that,

$\frac{1}{1+{\mathrm{x}}^3}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }} (-1{)}^2{\mathrm{x}}^{3\mathrm{k}}$

Therefore,$\mathrm{I}={\int }_0^2 \frac{\mathrm{dx}}{1+{\mathrm{x}}^3}$

$\begin{array}{r}\mathrm{I}={\int }_0^2 \left[\sum _{\mathrm{k}=0}^{\mathrm{\infty }} (-1{)}^{\mathrm{k}}{\mathrm{x}}^{3\mathrm{k}}\right]\mathrm{dx}\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} (-1{)}^{\mathrm{k}}{\int }_0^2 {\mathrm{x}}^{3\mathrm{k}}\mathrm{dx}\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} (-1{)}^{\mathrm{k}}{\left[\frac{{\mathrm{x}}^{3\mathrm{k}+1}}{3\mathrm{k}+1}\right]}_0^2\\ =\sum _{\mathrm{k}=0}^{\mathrm{\infty }} \frac{(-1{)}^{\mathrm{k}}}{3\mathrm{k}+1}(2{)}^{3\mathrm{k}+1}\end{array}$

Find the sum of the terms that are bigger than$0.001$ to get the definite integral of the series / from 0 to 0.5. The approximation is the resultant term.

Thus,

$\mathrm{I}=2-\frac{1}{4}(2{)}^4+\frac{1}{7}(2{)}^7-\frac{1}{10}(2{)}^{10}+\frac{1}{13}(2{)}^{13}-\dots$

We can see that as we go farther into the approximation, the term will keep increasing, and no term in the approximation of I will be lower than $0.001$