How do you find the lengths of the curve y=int (sqrtt+1)^-2 from [0,x^2] for the interval 0<=x<=1?

May 15, 2017

The arc length of the curve $y$ on $a \le x \le b$ is given by:

$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Here, to find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we have to apply the second Fundamental Theorem of Calculus.

$y = {\int}_{0}^{{x}^{2}} {\left(\sqrt{t} + 1\right)}^{-} 2 \mathrm{dt}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sqrt{{x}^{2}} + 1\right)}^{-} 2 \frac{d}{\mathrm{dx}} \left({x}^{2}\right) = {\left(x + 1\right)}^{-} 2 \left(2 x\right)$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = \frac{4 {x}^{2}}{x + 1} ^ 4$

So the arc length is given by:

$L = {\int}_{0}^{1} \sqrt{1 + \frac{4 {x}^{2}}{x + 1} ^ 4} \mathrm{dx}$

Putting this into a calculator or Wolfram Alpha:

$L = 1.07943$