# 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 ... #