# How do you find the length of the curve #y=(2x+1)^(3/2), 0<=x<=2#?

##### 2 Answers

Length is 10.38 units. See details below

#### Explanation:

The length of a curve between a and b values for x is given by

# 1/27{46sqrt(46)-10sqrt(10)} ~~ 10.384 \ (3dp)#

#### Explanation:

The Arc Length of a curve

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

Sop, for the given curve

# dy/dx = (3/2)(2x+1)^(3/2-1)(2) #

# \ \ \ \ \ \ = 3(2x+1)^(1/2) #

So then, the arc length is:

# L = int_0^2 \ sqrt(1+(3(2x+1)^(1/2))^2 ) \ dx #

# \ \ = int_0^2 \ sqrt(1+9(2x+1) ) \ dx #

# \ \ = int_0^2 \ sqrt(1+18x+9 ) \ dx #

# \ \ = int_0^2 \ sqrt(18x+10 ) \ dx #

And using the power rule for integration, we can integrate to get:

# L = [((18x+10 )^(3/2) )/(3/2) * 1/18]_0^2 #

# \ \ = [1/27(18x+10 )^(3/2)]_0^2 #

# \ \ = 1/27{(36+10)^(3/2)-10^(3/2)} #

# \ \ = 1/27{46sqrt(46)-10sqrt(10)} #

# \ \ ~~ 10.384 \ (3dp)#