Q. 17 - Calculus

Question

(a)  Explain why the definite integral$I={\int }_0^{0.5} \frac{dx}{1+x^3}$ exists.

(b)  Explain how to use Simpson’s method to approximate I to within 0.001 of its actual value. 

(c) Use substitution in the Maclaurin series for $\frac{1}{1-x}$ to find a Maclaurin series for $\frac{1}{1+x^3}$.

 (d) Explain how to use Theorems 7.38 and 8.12 to approximate I to within 0.001 of its actual value. 

Step-by-Step Solution

Answer

Ans:  

(a)  The integral I exist, since the function $\frac{1}{1+x^3}$ is continuous in the interval [0,0.5]

(b)   $I\approx 0.43327$

(c)   $\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{3k+1}(0.5{)}^{3k+1}$

(d)   $I\approx 0.4855$

Step 1. Given information.

given,   $I={\int }_0^{0.5} \frac{dx}{1+x^3}$

Step 2. Consider the definite integral I = ∫ 0 0.5   d x 1 + x 3

 (a)  The integral I exist, since the function $\frac{1}{1+x^3}$ is continuous in the interval [0,0.5]

Step 3. (b) To use the Simpson’s Approximation to approximate the integral I within 0.001 of its actual value,

let us use the formula 

    ${\int }_a^b f\left(x\right)dx\approx \frac{\mathrm{\Delta }x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+\dots +2f\left(x_{n-2}\right)+4f\left(x_{n-1}\right)+f\left(x_n\right)\right]$

Where $∆x=\frac{b-a}{n}$

In the integral $I={\int }_0^{0.5} \frac{dx}{1+x^3},a=0,b=0.5\text{ and }n=3$ (Since we need to approximate the integral I to within 0.001 of its actual value)

Therefore,

     $I\approx \frac{\mathrm{\Delta }x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+f\left(x_3\right)\right]$

 Where,

      $\begin{array}{r}f\left(x_0=0\right)=\frac{1}{1+0^3}\\ =1 \end{array}$

And

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

Again,

     $\begin{array}{r}f\left(x_2-\frac{1}{3}\right)=\frac{1}{1+{\left(\frac{1}{3}\right)}^3}\\ =\frac{27}{28} \end{array}$   

Similarly   

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

Step 4. Hence,

    $\begin{array}{r}I\approx \frac{\frac{1}{6}}{3}\left[1+4\left(\frac{216}{217}\right)+2\left(\frac{27}{28}\right)+\frac{8}{9}\right] \\ \approx \frac{1}{18}[1+3.98156+1.92857+0.8888]\end{array}$

Implies that

     $I\approx 0.43327$

Step 3. (c) Since the Maclaurin series for 1 1 - x is

   $\frac{1}{1-x}=\sum _{k=0}^{\mathrm{\infty }} x^k$

So, the Maclaurin series for $\frac{1}{1-x^3}$ can be found by substituting x by $-x^3$ in the Maclaurin series of $\frac{1}{1-x}$

Therefore,

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

Implies that 

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

 

Step 6. (d) Since I = ∫ 0 0.5   d x 1 + x 3

Therefore,

   $\begin{array}{r}I={\int }_0^{05} \left[\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k{x}^{3k}\right]dx\\ =\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k{\int }_0^{0.5} x^{3k}dx\\ =\sum _{k=0}^{\mathrm{\infty }} (-1{)}^k{\left[\frac{x^{3k+1}}{3k+1}\right]}_0^{0.5}\\ =\sum _{k=0}^{\mathrm{\infty }} \frac{(-1{)}^k}{3k+1}(0.5{)}^{3k+1}\end{array}$

Step 7. Now,

to evaluate the definite integral of the series I from 0 to 0.5, let us find the sum of the terms that are greater than 0.001. The resultant term is an approximation.

Thus,

   $\begin{array}{r}I=1\left(0.5\right)-\frac{1}{4}(0.5{)}^4+\frac{1}{7}(0.5{)}^7-\frac{1}{10}(0.5{)}^{10}+\frac{1}{13}(0.5{)}^{13}-\dots \\ =0.5-0.015625+0.0011607-0.00009765+\dots \end{array}$

Since $0.00009765<<0.001$

So, we consider the first three terms to approximate I to within 0.001 of its actual value.

Therefore,

     $I\approx 0.5-0.015625+0.0011607$

That is,

     $I\approx 0.4855$