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

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Answer 1 · 2017-05-15T03:57:01

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

$L=int_a^bsqrt(1+(dy/dx)^2)dx$

Here, to find $dy/dx$ , we have to apply the second Fundamental Theorem of Calculus.

$y=int_0^(x^2)(sqrtt+1)^-2dt$

$dy/dx=(sqrt(x^2)+1)^-2d/dx(x^2)=(x+1)^-2(2x)$

$(dy/dx)^2=(4x^2)/(x+1)^4$

So the arc length is given by:

$L=int_0^1sqrt(1+(4x^2)/(x+1)^4)dx$

Putting this into a calculator or Wolfram Alpha:

$L=1.07943$

How do you find the lengths of the curve y=int (sqrtt+1)^-2y=int (sqrtt+1)^-2 from [0,x^2][0,x^2] for the interval 0<=x<=10<=x<=1?