Question

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

$f\left(x\right)=\frac{1}{3}{x}^{3/2}-x^{1/2}$,$\left[a,b\right]=\left[0,1\right]$

Short answer

The arc length is$\frac{4}{3}$.

Step-by-step solution

Step 1. Given information. 

Consider the function is $f\left(x\right)=\frac{1}{3}{x}^{3/2}-x^{1/2}$,$\left[a,b\right]=\left[0,1\right]$.

Step 2. Use arc length formula.

The formula for a function to find the arc length from $x=a$to $x=b$is given by ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find the arc length. 

$\begin{array}{rcl}\mathrm{Arc}\mathrm{length}& =& {\int }_0^1\sqrt{1+{\left(\frac{d}{dx}\left(\frac{1}{3}{x}^{3/2}-x^{1/2}\right)\right)}^2}dx\\ & =& {\int }_0^1\sqrt{1+{\left(\frac{1}{3}\left(\frac{3}{2}\right)x^{1/2}-\left(\frac{1}{2}\right)x^{-1/2}\right)}^2}dx\\ & =& {\int }_0^1\sqrt{1+{\left(\frac{1}{2}{x}^{1/2}-\left(\frac{1}{2}\right)x^{-1/2}\right)}^2}dx\\ & =& {\int }_0^1\sqrt{1+\frac{1}{4}{\left(\frac{x-1}{\sqrt{x}}\right)}^2}dx\\ & =& {\int }_0^1\sqrt{1+\frac{1}{4}\left(\frac{{\left(x-1\right)}^2}{x}\right)}dx\\ & =& {\int }_0^1\sqrt{\frac{4x+{\left(x-1\right)}^2}{4x}}dx\\ & =& \frac{1}{2}{\int }_0^1\frac{x+1}{\sqrt{x}}dx\\ & =& \frac{1}{2}\left({\int }_0^1\sqrt{x}dx+{\int }_0^1\frac{1}{\sqrt{x}}dx\right)\\ & =& \frac{1}{2}\left({\left[\frac{x^{3/2}}{3/2}\right]}_0^1+{\left[\frac{x^{1/2}}{1/2}\right]}_0^1\right)\\ & =& \frac{1}{2}\left(\left[\frac{1^{3/2}}{3/2}-\frac{0^{3/2}}{3/2}\right]+\left[\frac{1^{1/2}}{1/2}-\frac{0^{1/2}}{1/2}\right]\right)\\ & =& \frac{1}{2}\left(\frac{2}{3}+2\right)\\ & =& \frac{4}{3}\end{array}$