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:

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

## Random Quote

Creative mathematicians now, as in the past, are inspired by the art of mathematics rather than by any prospect of ultimate usefulness.

Eric Temple Bell

## Random Quote

Creative mathematicians now, as in the past, are inspired by the art of mathematics rather than by any prospect of ultimate usefulness.

Eric Temple Bell