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 ... #