# How do you find the arc length of the curve f(x)=x^3/6+1/(2x) over the interval [1,3]?

Feb 15, 2017

The arc length is $\frac{14}{3}$ units.

#### Explanation:

The arc length of a curve on the interval $\left[a , b\right]$ is given by evaluating ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.

The derivative of $f ' \left(x\right)$, given by the power rule, is

$f ' \left(x\right) = \frac{1}{2} {x}^{2} - \frac{1}{2 {x}^{2}} = \frac{{x}^{4} - 1}{2 {x}^{2}}$

Substitute this into the above formula.

${\int}_{1}^{3} \sqrt{1 + {\left(\frac{{x}^{4} - 1}{2 {x}^{2}}\right)}^{2}} \mathrm{dx}$

Expand.

${\int}_{1}^{3} \sqrt{1 + \frac{{x}^{8} - 2 {x}^{4} + 1}{4 {x}^{4}}} \mathrm{dx}$

Put on a common denominator.

int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx

Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form ${\left(a x\right)}^{2}$.

${\int}_{1}^{3} \sqrt{{\left({x}^{4} + 1\right)}^{2} / {\left(2 {x}^{2}\right)}^{2}} \mathrm{dx}$

Eliminate the square root using ${\left({a}^{2}\right)}^{\frac{1}{2}} = a$

${\int}_{1}^{3} \frac{{x}^{4} + 1}{2 {x}^{2}} \mathrm{dx}$

Factor out a $\frac{1}{2}$ and put it in front of the integral.

$\frac{1}{2} {\int}_{1}^{3} \frac{{x}^{4} + 1}{x} ^ 2 \mathrm{dx}$

Separate into different fractions.

$\frac{1}{2} {\int}_{1}^{3} {x}^{4} / {x}^{2} + \frac{1}{x} ^ 2 \mathrm{dx}$

Simplify using ${a}^{n} / {a}^{m} = {a}^{n - m}$ and $\frac{1}{a} ^ n = {a}^{-} n$.

$\frac{1}{2} {\int}_{1}^{3} {x}^{2} + {x}^{-} 2 \mathrm{dx}$

Integrate using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$, with $n \in \mathbb{R} , n \ne - 1$.

$\frac{1}{2} {\left[\frac{1}{3} {x}^{3} - \frac{1}{x}\right]}_{1}^{3}$

Evaluate using the second fundamental theorem of calculus, which states that for ${\int}_{a}^{b} F \left(x\right) = f \left(b\right) - f \left(a\right)$, if $f \left(x\right)$ is continuous on $\left[a , b\right]$ and where $f ' \left(x\right) = F \left(x\right)$.

$\frac{1}{2} \left(\frac{1}{3} {\left(3\right)}^{3} - \frac{1}{3} - \left(\frac{1}{3} {\left(1\right)}^{3} - \frac{1}{1}\right)\right)$

Combine fractions and simplify.

$\frac{1}{2} \left(9 - \frac{1}{3} - \frac{1}{3} + 1\right)$

$\frac{1}{2} \left(10 - \frac{2}{3}\right)$

$5 - \frac{1}{3}$

$\frac{14}{3}$

Hopefully this helps!