Chapter 8.6, Problem 27E

To determine

To find: To determine the approximation value of the definite integral to six decimal places.

Answer to Problem 27E

  ${\displaystyle {\int }_0^{0.2}\frac{1}{1+x^4}dx}=0.399999\left[\text{correct upto }6\text{ decimal places}\right]$

Explanation of Solution

Given information: ${\displaystyle {\int }_0^{0.2}\frac{1}{1+x^5}dx}$

Calculation:

  $\begin{array}{l}\frac{1}{1-x}={\displaystyle \sum _{n=0}^{\infty }{x}^n}\left|x\right|<1\\ \Rightarrow \frac{1}{1+x^5}=\frac{1}{1-\left(-x^5\right)}\\ \Rightarrow \frac{1}{1+x^5}={\displaystyle \sum _{n=0}^{\infty }{\left(-x^5\right)}^n}\left|x\right|<1\\ \Rightarrow \frac{1}{1+x^5}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}{x}^{5n}\end{array}$

Now,

  $\begin{array}{l}{\displaystyle {\int }_0^{0.2}\frac{1}{1+x^5}}dx\\ ={\displaystyle {\int }_0^{0.2}{\displaystyle \sum _{n=0}^{\infty }\left(-1\right)x^{5n}dx}}\\ ={\left[{\displaystyle \sum _{n=0}^{\infty }\left(-1\right)\frac{x^{5n+1}}{5n+1}}\right]}_0^{0.2}+c\left[c\text{ is Integrating Constant}\right]\end{array}$

Put, $x=0$

Then,

  $\begin{array}{l}{\displaystyle {\int }_0^{0.2}\frac{1}{1+0}dx}=c\\ \Rightarrow c={\displaystyle {\int }_0^{0.2}dx}\\ \Rightarrow c=0.2\end{array}$

  $\begin{array}{l}\therefore {\displaystyle {\int }_0^{0.2}\frac{1}{1+x^4}dx}\\ ={\left[{\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{x^{5n+1}}{5n+1}}\right]}_0^{0.2}+0.2\\ ={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\left(0.2\right)}^{5n+1}}{5n+1}}+0.2\\ =0.2+\left[\left(0.2\right)-\frac{{\left(0.2\right)}^6}{6}+\frac{{\left(0.2\right)}^{11}}{11}-\mathrm{.......}\right]\\ =0.4-0.00000107+0\left[\text{correct upto }6\text{ decimal places}\right]\\ =0.39999\end{array}$

Hence, ${\displaystyle {\int }_0^{0.2}\frac{1}{1+x^4}dx}=0.399999\left[\text{correct upto }6\text{ decimal places}\right]$