How do you set up an integral for the length of the curve #y=sqrtx, 1<=x<=2#?
1 Answer
Jul 27, 2017
arc length is:
# L = int_1^2 \ sqrt(1+1/(4x)) \ dx#
Explanation:
The arc length of a curve
# L = int_a^b \ sqrt(1+(dy/dx)^2) \ dx#
So for the given function:
# y = sqrt(x) #
Then differentiating wrt
# 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 ... #