Give a definite integral which repersents the length of the curve H(x)=x^3-1 for 0<x<2.?

Question

Show your work if possible. Also if possible could you explain how you would use a calculator to do this so that is would be accurate to three decimal places.

Answer 1

Let's call the distance between the points (x,y) and (x+dx,y+dy) ds, which is given by

ds = (dx^2 + dy^2)^(1/2).

The length of a curve between two points would then be the integral of ds.

In this case, y(x) = x^3 - 1, so dy = 3x^2 dx, and ds = (dx^2 + 9 x^4 dx^2)^(1/2) = (1 + 9 x^4)^(1/2) dx.

The length of the curve between x = 0 and x = 2 is then

Int[ds,x,0,2] = Int[(1 + 9 x^4)^(1/2),x,0,2].

Insert this exact expression into your calculator. Mine gives 8.63033.

Answer 2

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Answer 3

ds^2 =dx^2+ dy^2
so ds = sqrt(1+y´^2)dx and
s= int sqrt( 1+9x^4) dx between 0 and 2 as y´=3x^2