Chapter 8, Problem 46RQ

To determine

To Find: The value of ${\displaystyle {\int }_0^1\sqrt{1+x^4}dx}$ correct tot two decimal places.

Answer to Problem 46RQ

The value of the integral is $1.09$ .

Explanation of Solution

Calculation:

Consider the binomial series as,

  ${\left(1+x\right)}^4=1+kx+\frac{k\left(k-1\right)}{2!}{x}^2+\frac{k\left(k-1\right)\left(k-2\right)}{3!}{x}^3+\mathrm{..}$

Substitute $k=\frac{1}{2}$ and replace $x^4$ for $x$ as,

  $\begin{array}{c}{\left(1+x^4\right)}^{\frac{1}{2}}=1+\frac{1}{2}{x}^4+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}{\left(x^4\right)}^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}{\left(x^4\right)}^3+\mathrm{....}\\ =1+\frac{x^4}{2}-\frac{1}{2^{2}2!}{x}^8+\frac{1\cdot 3}{2^{4}4!}{x}^{16}+\frac{1\cdot 3\cdot 5}{2^{4}4!}{x}^{16}+\mathrm{...}\end{array}$

Then, the value of the integral ${\displaystyle {\int }_0^1\sqrt{1+x^4}dx}$ is evaluated as,

  $\begin{array}{c}{\displaystyle {\int }_0^1\sqrt{1+x^4}dx}=\left[1+\frac{x^4}{2}-\frac{1}{2^{2}2!}{x}^8+\frac{1\cdot 3}{2^{4}4!}{x}^{16}+\frac{1\cdot 3\cdot 5}{2^{4}4!}{x}^{16}+\mathrm{...}\right]dx\\ =1+\left(\frac{1}{10}-\frac{1}{72}+\frac{3}{624}-\frac{15}{6528}+\mathrm{..}\right)\end{array}$

Then, by the use of the approximate the sum by the use of alternate series estimation theorem,

  $\frac{15}{6528}=2.3\times {10}^{-3}$

Then,

  $\begin{array}{c}{\displaystyle {\int }_0^1\sqrt{1+x^4}dx}=1+\left(\frac{1}{10}-\frac{1}{72}+\frac{3}{624}\right)\\ =1.09\end{array}$