91Ó°ÊÓ

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

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

let us use the formula

${∫}_a^b f\left(x\right)dx≈\frac{\mathrm{Δ}x}{3}\left[f\left(x_0\right)+4f\left(x_1\right)+2f\left(x_2\right)+…+2f\left(x_{n−2}\right)+4f\left(x_{n−1}\right)+f\left(x_n\right)\right]$

Where $∆x=\frac{b-a}{n}$

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

Therefore,

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

Where,

role="math" localid="1650703952394" $\begin{array}{r}f\left(x_0=0\right)=\frac{1}{1+0^3}\\ =1\end{array}$

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,

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

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

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

Implies that

$I≈0.43327$

$\frac{1}{1−x}=\underset{k=0}{\overset{\mathrm{∞}}{∑}} x^k$

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,

$\frac{1}{1+x^3}=\underset{k=0}{\overset{\mathrm{∞}}{∑}} {\left(−x^3\right)}^k$

Implies that

$\frac{1}{1+x^3}=\underset{k=0}{\overset{\mathrm{∞}}{∑}} {(−1)}^k{x}^{3k}$

Therefore,

$\begin{array}{r}I={∫}_0^{05} \left[\underset{k=0}{\overset{\mathrm{∞}}{∑}} (−1{)}^k{x}^{3k}\right]dx\\ =\underset{k=0}{\overset{\mathrm{∞}}{∑}} (−1{)}^k{∫}_0^{0.5} x^{3k}dx\\ =\underset{k=0}{\overset{\mathrm{∞}}{∑}} (−1{)}^k{\left[\frac{x^{3k+1}}{3k+1}\right]}_0^{0.5}\\ =\underset{k=0}{\overset{\mathrm{∞}}{∑}} \frac{(−1{)}^k}{3k+1}(0.5{)}^{3k+1}\end{array}$

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,

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

Since $0.00009765<<0.001$

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

Therefore,

$I≈0.5−0.015625+0.0011607$

That is,