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#