How do you set up an integral for the length of the curve $y=sqrtx, 1<=x<=2$ ?

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Answer 1 · 2017-07-27T20:47:09

arc length is:

$L = int_1^2 \ sqrt(1+1/(4x)) \ dx$

The arc length of a curve $y=f(x)$ over an interval $[a,b]$ is given by:

$L = int_a^b \ sqrt(1+(dy/dx)^2) \ dx$

So for the given function:

$y = sqrt(x)$

Then differentiating wrt $x$ we get

$dy/dx = 1/(2sqrt(x))$

So then the arc length is:

$L = int_1^2 \ sqrt(1+(1/(2sqrt(x)))^2) \ dx$
$\ \ = int_1^2 \ sqrt(1+1/(4x)) \ dx$

NB:
If we evaluate this integral we get:

$L = 1.08306427952 ...$

How do you set up an integral for the length of the curve y=sqrtx, 1<=x<=2y=sqrtx, 1<=x<=2?