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Find the length of the curve $x = \frac{y^3}{3} + \frac{1}{4y}$ from y = 1 to y = 2
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Solution:

We have $x=\frac{y^{3}}{3}+\frac{1}{4 y}$

Diff w.r.t. y, we get

$\frac{d x}{d y}=y^{2}-\frac{1}{4 y^{2}}$

$1+\left(\frac{d x}{d y}\right)^{2}=1+\left(y^{2}-\frac{1}{4 y^{2}}\right)^{2}=y^{4}+\frac{1}{2}+\frac{1}{16 y^{4}}=\left(y^{2}+\frac{1}{4 y^{2}}\right)^{2}$

We know that,

$s=\int_{1}^{2} \sqrt{1+\left(\frac{d x}{d y}\right)^{2}} d y$

$s=\int_{1}^{2} \sqrt{\left(y^{2}+\frac{1}{4 y^{2}}\right)^{2} d y}$

$\begin{aligned} s &=\left[\frac{y^{3}}{3}-\frac{1}{4 y}\right]_{1}^{2} \\ s &=\frac{8}{3}-\frac{1}{8}-\left(\frac{1}{3}-\frac{1}{4}\right) \\ s &=\frac{59}{24} \end{aligned}$

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