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

Mar 18, 2018

The area is $= 1 {u}^{2}$

#### Explanation:

We need

$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(x \ne - 1\right)$

The area is

$\mathrm{dA} = y \mathrm{dx}$

$A = {\int}_{\frac{1}{2}}^{1} \frac{\mathrm{dx}}{\sqrt{2 x - 1}}$

$= {\int}_{\frac{1}{2}}^{1} {\left(2 x - 1\right)}^{- \frac{1}{2}} \left(\mathrm{dx}\right)$

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

$= \left(1\right) - \left(0\right)$

$= 1 {u}^{2}$

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

Mar 18, 2018

The Area is $1 \setminus {\text{unit}}^{2}$

#### Explanation:

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

$A = {\int}_{\alpha}^{\beta} \setminus f \left(x\right) \setminus \mathrm{dx}$

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

$I = {\int}_{\frac{1}{2}}^{1} \frac{1}{\sqrt{2 x - 1}} \setminus \mathrm{dx}$

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

$u = 2 x - 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2$

And we must also change the limits of integration:

When $x = \left\{\begin{matrix}\frac{1}{2} \\ 1\end{matrix}\right. \implies u = \left\{\begin{matrix}0 \\ 1\end{matrix}\right.$

And so we can write:

$I = \frac{1}{2} \setminus {\int}_{\frac{1}{2}}^{1} \frac{1}{\sqrt{2 x - 1}} \setminus \left(2\right) \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{2} \setminus {\int}_{0}^{1} \frac{1}{\sqrt{u}} \setminus \mathrm{du}$

$\setminus \setminus = \frac{1}{2} \setminus {\left[2 \sqrt{u}\right]}_{0}^{1}$

$\setminus \setminus = {\left[\sqrt{u}\right]}_{0}^{1}$

$\setminus \setminus = \sqrt{1} - \sqrt{0}$

$\setminus \setminus = 1$