# How do you find the length of the curve y = ( 2 x + 1 ) <mrow class="MJX-T

How do you find the length of the curve $y=\left(2x+1{\right)}^{\frac{3}{2}},0\le x\le 2$?
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Ordettyreomqu
The length of a curve between a and b values for x is given by
$L={\int }_{a}^{b}\sqrt{1+{y}^{\prime 2}}dx$
${y}^{\prime 2}=18x+9$
$L={\int }_{0}^{2}\sqrt{18x+10}dx=\frac{1}{27}\left(18x+10{\right)}^{\frac{3}{2}}{\right]}_{0}^{2}=10.38$
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pablos28spainzd
The Arc Length of a curve y=f(x) from x=a to x=b is given by:
$L={\int }_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}{\right)}^{2}}dx$
So, for the given curve $y=\left(2x+1{\right)}^{\frac{3}{2}}$ for $x\in \left[0,2\right]$ we form the derivative using the power rule for differentiation in conjunction with the chain rule:
$\frac{dy}{dx}=\left(\frac{3}{2}\right)\left(2x+1{\right)}^{\frac{3}{2}-1}\left(2\right)$
$=3\left(2x+1{\right)}^{\frac{1}{2}}$
So then, the arc length is:
$L={\int }_{0}^{2}\sqrt{1+\left(3\left(2x+1{\right)}^{\frac{1}{2}}{\right)}^{2}}dx$
$={\int }_{0}^{2}\sqrt{1+9\left(2x+1\right)}dx$
$={\int }_{0}^{2}\sqrt{1+18x+9}dx$
$={\int }_{0}^{2}\sqrt{18x+10}dx$
And using the power rule for integration, we can integrate to get:
$L=\left[\frac{\left(18x+10{\right)}^{\frac{3}{2}}}{\frac{3}{2}}\ast \frac{1}{18}{\right]}_{0}^{2}$
$=\left[\frac{1}{27}\left(18x+10{\right)}^{\frac{3}{2}}{\right]}_{0}^{2}$
$=\frac{1}{27}\left[\left(36+10{\right)}^{\frac{3}{2}}-{10}^{\frac{3}{2}}\right]$
$=\frac{1}{27}\left[46\sqrt{46}-10\sqrt{10}\right]$
$\approx 10.384\left(3dp\right)$