# Thread: How do I find this constant?

1. ## How do I find this constant?

$\displaystyle f(x,y)=\left\{ \begin{array}{lr} cx^2+y& 0\le x\le 1,0\le y\le 1,y\le x\\ 0&elsewhere, \end{array} \right.$

I can't figure out how to set this up. I have to show that c = 10, but every time I do it I get c = 5. What am I doing wrong:

$\displaystyle 1=\int_{0}^{1}\int_{0}^{x}x^2y \, dydx=\frac{c}{5}\Longrightarrow c=5$

2. you copied the density incorrectly, there is a + between the variables.

$\displaystyle 1=\int_{0}^{1}\int_{0}^{x}\biggl(cx^2+y \biggr)\, dydx$

3. Originally Posted by billym
$\displaystyle f(x,y)=\left\{ \begin{array}{lr} cx^2y& 0\le x\le 1,0\le y\le 1,y\le x\\ 0&elsewhere, \end{array} \right.$

I can't figure out how to set this up. I have to show that c = 10, but every time I do it I get c = 5. What am I doing wrong:

$\displaystyle 1=\int_{0}^{1}\int_{0}^{x}x^2y \, dydx=\frac{c}{5}\Longrightarrow c=5$
Sorry, I actually wrote it down wrong the first time. The correct function is cx^2y. I still can't get ten out of it...

4. Originally Posted by billym
Sorry, I actually wrote it down wrong the first time. The correct function is cx^2y. I still can't get ten out of it...
Thats because you computed the integral wrong.
$\displaystyle \int_0^1 \int_0^x cx^2y \, dy dx=\int_0^1 \frac{1}{2}cx^2x^2 \, dx=\int_0^1 \frac{1}{2}cx^4\, dx=\frac{c}{10}$

5. was up too late. All makes perfect sense now.