How do you find the area of the region under the curve #y=1/sqrt(2x-1)# from x=1/2, to x=1?

2 Answers
Mar 18, 2018

Answer:

The area is #=1u^2#

Explanation:

We need

#intx^ndx=x^(n+1)/(n+1)+C(x!=-1)#

The area is

#dA=ydx#

#A=int_(1/2)^1 (dx)/(sqrt(2x-1))#

#=int_(1/2)^1 (2x-1)^(-1/2)(dx)#

#=[1/2((2x-1)^(1/2)/(1/2)]_(1/2)^1#

#=(1)-(0)#

#=1u^2#

graph{1/sqrt(2x-1) [-5.546, 5.55, -2.773, 2.774]}

Mar 18, 2018

Answer:

The Area is #1\ "unit"^2 #

Explanation:

The (net) area bounded by a curve #y=f(x)# and two #x#-coordinates, #x=alpha# and #x=beta# is given by:

# A =int_alpha^beta \ f(x) \ dx #

So in this case we seek the value of the definite integral:

# I = int_(1/2)^(1) 1/sqrt(2x-1) \ dx #

We can perform a change of variable (equivalent of a translation) by a substitution. Let:

# u = 2x-1 => (du)/dx = 2 #

And we must also change the limits of integration:

When #x={ (1/2),(1) :} => u={ (0),(1) :} #

And so we can write:

# I = 1/2 \ int_(1/2)^(1) 1/sqrt(2x-1) \ (2) \ dx #

# \ \ = 1/2 \ int_(0)^(1) 1/sqrt(u) \ du #

# \ \ = 1/2 \ [2sqrt(u)]_(0)^(1) #

# \ \ = [sqrt(u)]_(0)^(1) #

# \ \ = sqrt(1) - sqrt(0) #

# \ \ = 1 #