Question

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

Answer 1

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]}

Answer 2

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`