Q. 43 - Calculus

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.

$f\left(x\right)=x^2-\frac{1}{8}\ln x$, $\left[a,b\right]=\left[1,2\right]$ 

Step-by-Step Solution

Answer

The arc length is $3.087$.  

Step 1. Given information.

Consider the function is $f\left(x\right)=x^2-\frac{1}{8}\ln x$, $\left[a,b\right]=\left[1,2\right]$.

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.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_1^2\sqrt{1+{\left(\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{x}}^2-\frac{1}{8}\mathrm{lnx}\right)\right)}^2\mathrm{dx}}\\ & =& {\int }_1^2\sqrt{1+{\left(2\mathrm{x}-\frac{1}{8\mathrm{x}}\left(1\right)\right)}^2\mathrm{dx}}\\ & =& {\int }_1^2\sqrt{1+\frac{{\left(16{\mathrm{x}}^2-1\right)}^2}{64{\mathrm{x}}^2}\mathrm{dx}}\\ & =& {\int }_1^2\sqrt{\frac{64{\mathrm{x}}^2+{\left(16{\mathrm{x}}^2-1\right)}^2}{64{\mathrm{x}}^2}\mathrm{dx}}\\ & =& {\int }_1^2\frac{1+16{\mathrm{x}}^2}{8\mathrm{x}}\mathrm{dx}\\ & =& {\left[\frac{1}{8}\mathrm{lnx}+\frac{2{\mathrm{x}}^2}{2}\right]}_1^2\\ & =& \left[\frac{1}{8}\ln \left(2\right)+\frac{2{\left(2\right)}^2}{2}\right]-\left[\frac{1}{8}\ln \left(1\right)+\frac{2{\left(1\right)}^2}{2}\right]\\ & =& \left[\frac{1}{8}\ln \left(2\right)+4\right]-\left[0+1\right]\\ & =& \frac{1}{8}\ln \left(2\right)+3\\ & \approx & 3.087\end{array}$