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.

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

Short answer

The arc length is$\frac{3}{2}$.

Step-by-step solution

Step 1. Given information. 

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

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 role="math" localid="1649223576634" ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(x\right)\right)}^2}dx$.

Step 3. Find the arc length.  

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