# int int_R e^(-(xy)/2)dA where R:the region bounded by y=1/4x,y=2x,y=1/x,y=4/x Evaluate integrals?

## $\int {\int}_{R} {e}^{- \frac{x y}{2}} \mathrm{dA}$ where $R$:the region bounded by $y = \frac{1}{4} x , y = 2 x , y = \frac{1}{x} , y = \frac{4}{x}$

Jun 29, 2018

$= \frac{\ln 8 \left({e}^{\frac{3}{2}} - 1\right)}{4 {e}^{2}}$

#### Explanation:

graph{(y-2x)(y-1/4 x)(y-1/x)(y-4/x)=0 [-0.5, 5, -0.5, 3]}

You will see from the drawing that you cannot just write down the limits and integrate here, so I have used the first sub that came to mind:

• $\left\{\begin{matrix}u = x y \\ v = \frac{y}{x}\end{matrix}\right. q \quad \left\{\begin{matrix}x = \sqrt{\frac{u}{v}} \\ y = \sqrt{u v}\end{matrix}\right.$

$\implies \left\{\begin{matrix}y = \frac{1}{4} x \\ y = 2 x \\ y = \frac{1}{x} \\ y = \frac{4}{x}\end{matrix}\right. \to \left\{\begin{matrix}v = \frac{1}{4} \\ v = 2 \\ u = 1 \\ u = 4\end{matrix}\right.$

Jacobian:

$\frac{\partial \left(x , y\right)}{\partial \left(u , v\right)} = \det \left[\begin{matrix}{x}_{u} & {x}_{v} \\ {y}_{u} & {y}_{v}\end{matrix}\right]$

$= \det \left[\begin{matrix}\frac{1}{2 \sqrt{u v}} & - \frac{1}{2} \sqrt{\frac{u}{v} ^ 3} \\ \frac{1}{2} \sqrt{\frac{v}{u}} & \frac{1}{2} \sqrt{\frac{u}{v}}\end{matrix}\right] = \frac{1}{2 v}$

$\therefore \int {\int}_{R} {e}^{- \frac{x y}{2}} \setminus \mathrm{dx} \mathrm{dy}$

$= {\int}_{\frac{1}{4}}^{2} \setminus {\int}_{1}^{4} q \quad \frac{1}{2 v} {e}^{- \frac{u}{2}} \setminus \mathrm{du} \setminus \mathrm{dv}$

$= \frac{1}{2} {\int}_{\frac{1}{4}}^{2} \frac{1}{v} \setminus \mathrm{dv} \cdot {\int}_{1}^{4} {e}^{- \frac{u}{2}} \mathrm{du}$

$= \frac{1}{2} {\left[\ln v\right]}_{\frac{1}{4}}^{2} \cdot {\left[- \frac{1}{2} {e}^{- \frac{u}{2}}\right]}_{1}^{4}$

$= - \frac{1}{4} \left[\ln 2 - \ln \left(\frac{1}{4}\right)\right] \cdot \left[{e}^{- 2} - {e}^{- \frac{1}{2}}\right]$

$= \frac{\ln 8 \left({e}^{\frac{3}{2}} - 1\right)}{4 {e}^{2}}$

$\left[\approx 0.245\right]$

I made this up on the hoof so make sure you're happy with it.