Math Calculators, Lessons and Formulas

It is time to solve your math problem

mathportal.org
Applications Of The Definite Integrals: (lesson 3 of 3)

## Arc Length

$\text{Arc Length} = \int\limits_a^b {\sqrt {1 + {{\left[ {f'(x)} \right]}^2}} dx}$

Example 1:

Find the arc length of the graph of $y = \frac{x^3}{6} + \frac{1}{2x}$ on $\left[ \frac{1}{2} , 2 \right]$.

Solution:

\begin{aligned} \color{blue}{f'(x)} &\color{blue}{= \frac{1}{2}\left( {{x^2} - \frac{1}{{{x^2}}}} \right) =} \\ \color{blue}{s} &\color{blue}{= \int\limits_a^b {\sqrt {1 + {{\left[ {f'(x)} \right]}^ \to }} dx}} = \int\limits_{\frac{1}{2}}^2 {\sqrt {1 + {{\left[ {\frac{1}{2}\left( {{x^2} - \frac{1}{{{x^2}}}} \right)} \right]}^2}} dx} = \\ &= \int\limits_{\frac{1}{2}}^2 {\sqrt {\left[ {\frac{1}{4}\left( {{x^4} + 2x + \frac{1}{{{x^4}}}} \right)} \right]} dx} = \int\limits_{\frac{1}{2}}^2 {\sqrt {{{\left[ {\frac{1}{2}\left( {{x^2} + \frac{1}{{{x^2}}}} \right)} \right]}^2}} dx} = \\ &= \int\limits_{\frac{1}{2}}^2 {\frac{1}{2}\left( {{x^2} + \frac{1}{{{x^2}}}} \right)dx} = \left[ {\frac{1}{2}\left( {\frac{{{x^3}}}{3} - \frac{1}{x}} \right)} \right]_{\frac{1}{2}}^2 = \\ &= \frac{33}{16} \end{aligned}