Q. 46 - 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.

$f\left(x\right)=\frac{x^4+3}{6x}$, $\left[a,b\right]=\left[1,3\right]$

Step-by-Step Solution

Answer

The arc length is $\begin{array}{rcl}& & 7.883\end{array}$.

Step 1. Given information.

Consider the function is $f\left(x\right)=\frac{x^4+3}{6x}$, $\left[a,b\right]=\left[1,3\right]$.

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.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{length}& =& {\int }_1^3\sqrt{1+{\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{{\mathrm{x}}^4+3}{6\mathrm{x}}\right)\right)}^2}\mathrm{dx}\\ & =& {\int }_1^3\sqrt{1+{\left(\frac{\left(6\mathrm{x}\right)\left(4{\mathrm{x}}^3+0\right)+\left({\mathrm{x}}^4+3\right)\left(6\right)}{{\left(6\mathrm{x}\right)}^2}\right)}^2}\mathrm{dx}\\ & =& {\int }_1^3\sqrt{1+{\left(\frac{24{\mathrm{x}}^4+6{\mathrm{x}}^4+18}{36{\mathrm{x}}^2}\right)}^2}\mathrm{dx}\\ & =& {\int }_1^3\sqrt{1+{\left(\frac{5{\mathrm{x}}^4+3}{6{\mathrm{x}}^2}\right)}^2}\mathrm{dx}\\ & =& {\int }_1^3\sqrt{\frac{36{\mathrm{x}}^4+{\left(5{\mathrm{x}}^4+3\right)}^2}{36{\mathrm{x}}^4}}\mathrm{dx}\\ & \approx & 7.883\end{array}$