Question

Find the length of the curve x=(y^3/3)+1/(4y) from y=1 to y=3

Answer 1

$x=\dfrac{y^3}3+\dfrac1{4y}$
$\dfrac{\mathrm dx}{\mathrm dy}=y^2-\dfrac1{4y^2}$

The curve's length along the interval [1,3] is

$\displaystyle\int_1^3\sqrt{1+\left(\dfrac{\mathrm dx}{\mathrm dy}\right)^2}\,\mathrm dy=\int_1^3\sqrt{1+\left(y^2-\dfrac1{4y^2}\right)^2}\,\mathrm dy=\int_1^3\sqrt{y^4+\dfrac12+\dfrac1{16y^4}}\,\mathrm dy$

Since

$y^4+\dfrac12+\dfrac1{16y^4}=\left(y^2+\dfrac1{4y^2}\right)^2$

you have

$\displaystyle\int_1^3\left(y^2+\dfrac1{4y^2}\right)\,\mathrm dy=\dfrac{y^3}3-\dfrac1{4y}\bigg|_{y=1}^{y=3}=\dfrac{107}{12}-\dfrac1{12}=\dfrac{53}6$