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.

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

Short answer

The arc length is$\pi$.

Step-by-step solution

Step 1. Given information. 

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

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. 

$\begin{array}{rcl}\mathrm{Arc}\mathrm{length}& =& {\int }_{-1}^1\sqrt{1+\frac{\mathrm{d}}{\mathrm{dx}}{\left(\sqrt{1-{\mathrm{x}}^2}\right)}^2}\mathrm{dx}\\ & =& {\int }_{-1}^1\sqrt{1+{\left(\frac{(-2\mathrm{x})}{2\sqrt{1-{\mathrm{x}}^2}}\right)}^2}\mathrm{dx}\\ & =& {\int }_{-1}^1\sqrt{\frac{1-{\mathrm{x}}^2+{\mathrm{x}}^2}{1-{\mathrm{x}}^2}}\mathrm{dx}\\ & =& {\int }_{-1}^1\sqrt{\frac{1}{1-{\mathrm{x}}^2}}\mathrm{dx}\\ & =& {\left[\mathrm{arcsinx}\right]}_{-1}^1\\ & =& \left[\arcsin \left(1\right)\right]-\left[\arcsin (-1)\right]\\ & =& \left[\frac{\mathrm{\pi }}{2}\right]-\left[-\frac{\mathrm{\pi }}{2}\right]\\ & =& \mathrm{\pi }\\ & & \end{array}$