Question

Find the length of an arc of the curve $y=\frac{1}{6}{x}^3+\frac{1}{2}{x}^{-1}$ from $x=1$ to $x=2$

• A. $\frac{17}{12}$
• B. $\frac{12}{17}$
• C. $\frac{3}{7}$
• D. $\frac{16}{17}$

Answer 1

The correct option is A $\frac{17}{12}$

Given $f\left(x\right)=y=\frac{1}{6}{x}^3+\frac{1}{2}{x}^{-1}$ on $[1,2]$
The length of the arc is given by $L={\int }_a^b\sqrt{f^{\prime }\left(x\right){)}^2+1}dx$
We have $f\left(x\right)=\frac{1}{6}{x}^3+\frac{1}{2}{x}^{-1}$
$\therefore f^{\prime }\left(x\right)=\frac{3x^2}{6}-\frac{1}{2x^2}=\frac{x^2}{2}-\frac{1}{2x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^4-1}{2x^2}$
$\therefore L={\int }_1^2\sqrt{{\left(\frac{x^4-1}{2x^2}\right)}^2+1}dx$
$={\int }_1^2\sqrt{\frac{(x^4-1{)}^2}{4x^4}+1}dx$
$={\int }_1^2\sqrt{\frac{(x^4-1{)}^2+4x^4}{4x^4}}dx$
$={\int }_1^2\sqrt{\frac{x^8+1+2x^4}{4x^4}}dx$
$={\int }_1^2\sqrt{\frac{(x^4+1{)}^2}{(2x^2{)}^2}}dx$
$={\int }_1^2\frac{x^4+1}{2x^2}dx$
$=\frac{1}{2}{\int }_1^2\frac{x^4+1}{x^2}dx$
$=\frac{1}{2}{\int }_1^2(x^2+\frac{1}{x^2})dx$
$=\frac{1}{2}{[\frac{x^3}{3}-\frac{1}{x}]}_1^2$
$=\frac{1}{2}[(\frac{8}{3}-\frac{1}{2})-(\frac{1}{3}-1)]$
$=\frac{1}{2}\left[\frac{17}{6}\right]$
$=\frac{17}{12}$
$\therefore L=\frac{17}{12}$