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Math Help - How do I find this constant?

  1. #1
    Member billym's Avatar
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    How do I find this constant?

    f(x,y)=\left\{ <br />
\begin{array}{lr} <br />
cx^2+y& 0\le x\le 1,0\le y\le 1,y\le x\\ <br />
0&elsewhere,<br />
\end{array} <br />
\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:

    1=\int_{0}^{1}\int_{0}^{x}x^2y \, dydx=\frac{c}{5}\Longrightarrow c=5
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  2. #2
    MHF Contributor matheagle's Avatar
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    you copied the density incorrectly, there is a + between the variables.

    1=\int_{0}^{1}\int_{0}^{x}\biggl(cx^2+y \biggr)\, dydx
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  3. #3
    Member billym's Avatar
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    Quote Originally Posted by billym View Post
    f(x,y)=\left\{ <br />
\begin{array}{lr} <br />
cx^2y& 0\le x\le 1,0\le y\le 1,y\le x\\ <br />
0&elsewhere,<br />
\end{array} <br />
\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:

    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...
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  4. #4
    Member Focus's Avatar
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    Quote Originally Posted by billym View Post
    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.
    <br />
\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}<br />
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  5. #5
    Member billym's Avatar
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    was up too late. All makes perfect sense now.
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