Q. 35 - 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.

$f\left(x\right)={(2x+3)}^{3/2}$, $\left[a,b\right]=\left[-1,1\right]$ 

Step-by-Step Solution

Answer

The arc length is $\frac{1}{27}{\left(46\right)}^{3/2}-\frac{1}{27}{\left(10\right)}^{3/2}$.

Step 1. Given information.

Consider the given function $f\left(x\right)={(2x+3)}^{3/2}$, $\left[a,b\right]=\left[-1,1\right]$. 

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 ${\int }_a^b\sqrt{1+(f^{\text{'}}{\left(x\right))}^2}dx$.

Step 3. Find the arc length.

$\begin{array}{rcl}\mathrm{Arc} \mathrm{lenght}& =& {\int }_{-1}^1\sqrt{1+{\left(\frac{d}{dx}{(2x+3)}^{3/2}\right)}^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+{\left(\frac{3}{2}{(2x+3)}^{1/2}\left(2\right)\right)}^2}dx\\ & =& {\int }_{-1}^1\sqrt{1+9(2x+3)}dx\\ & =& {\int }_{-1}^1\sqrt{28+18x}dx\end{array}$

Step 4. Simply obtained integral by simple substitution.

Substitute $28+18x=u$, $dx=\frac{1}{18}du$ into $\begin{array}{rcl}& & {\int }_{-1}^1\sqrt{28+18x}dx\end{array}$.

$\begin{array}{rcl}{\int }_{10}^{46}\frac{\sqrt{u}}{18}du& =& \frac{1}{18}{\int }_{10}^{46}\sqrt{u}du\\ & =& \frac{1}{18}{\left[\frac{u^{3/2}}{3/2}\right]}_{10}^{46}\\ & =& \frac{1}{27}\left[{46}^{3/2}-{10}^{3/2}\right]\\ & =& \frac{1}{27}{\left(46\right)}^{3/2}-\frac{1}{27}{\left(10\right)}^{3/2}\end{array}$