# Area under two curves by integration

• Nov 6th 2007, 03:25 AM
rbenito
Area under two curves by integration
I want to find the area between two curves (see graph attached). But I want to find the area in function of "r", can anybody help?. The first curve is a line (x+y=1), and the second has the equation xy=r
• Nov 6th 2007, 03:36 AM
DivideBy0
$r$ is just a constant, isn't it?
Then simply solve both equations for $y$:
$y = 1-x$

$y=\frac{r}{x}$
The graphs intersect at $\lbrace{(x,y): \left(\frac{1+\sqrt{1-4r}}{2},\frac{1-\sqrt{1-4r}}{2}\right),\left(\frac{1-\sqrt{1-4r}}{2},\frac{1+\sqrt{1-4r}}{2}\right)\rbrace}$

So you need to evaluate:

$\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} (1-x)\,dx-\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} \frac{r}{x}\,dx$

Hahahah... I might be wrong

On second thought it might be better to solve for r instead...

x = 0.65, y = .2 seems like a good spot.

$(0.65)(0.2)=r \Rightarrow r = 0.13$
• Nov 7th 2007, 06:47 AM
rbenito
r is a constant
but varies from 0 to 0.25, that means that r can be any number between [0,0.25]
• Nov 7th 2007, 07:45 AM
topsquark
Quote:

Originally Posted by DivideBy0
$r$ is just a constant, isn't it?
Then simply solve both equations for $y$:
$y = 1-x$

$y=\frac{r}{x}$
The graphs intersect at $\lbrace{(x,y): \left(\frac{1+\sqrt{1-4r}}{2},\frac{1-\sqrt{1-4r}}{2}\right),\left(\frac{1-\sqrt{1-4r}}{2},\frac{1+\sqrt{1-4r}}{2}\right)\rbrace}$

So you need to evaluate:

$\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} (1-x)\,dx-\int_{\frac{1-\sqrt{1-4r}}{2}}^{\frac{1+\sqrt{1-4r}}{2}} \frac{r}{x}\,dx$

I see no problem here. The final answer will depend on r.

-Dan