Here we have to use the following formula
[tex] l=\int\limits^a_b {\sqrt{1+(x')^2} \, dy [/tex]
x' is the derivative with respect to y
Given value of x is
[tex] x = y^{1/2} [/tex]
Differentiating with respet to y
[tex] x ' = \frac{1}{2y^{1/2}} [/tex]
Substituting the value of x' in the formula, we will get
[tex] L=\int\limits^4_0 { \sqrt{1+\frac{1}{4y}}} \, dy [/tex]
[tex] L = 4.65 [/tex]
