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$f(x)$ is proportional to x on $0 \leq x \leq 1$, and $f(x)=0$ otherwise. What is $P\{X \in (\frac{1}{3},\frac{7}{9})\}$

Now I know that the density function is $$\int_{0}^{1} f(x)\;\mathrm dx$$ and $f(x) =0$ otherwise. Where I'm stuck is what should $f(x)$ be.

If someone could help finding $f(x)$, so I can continue.

$$\int_{0}^{1} c\,x \;\mathrm dx= c\left[\frac{x^2}{2}\right]^1_0 \\~\\\therefore\; c=2$$

Then $$\int_{{1}/{3}}^{{7}/{9}}2\,x \;\mathrm dx= \frac{40}{81}$$

Is this correct?

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    Proportional to $x$ means $f(x) = cx$ for some $c$. Now use the fact that $\int_0^1 f(x) dx= 1$ to find $c$. Then integrate $f$ over $(\frac 13,\frac 79)$. – 2017-02-23
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    Oh okay, that was the point I needed that $f(x) =cx$, I will work on the problem and update it. Thank you – 2017-02-23
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    The update was correct (formatting a *bit* unclear but identifiable; fixed) – 2017-02-24
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    Thanks, so everything is good right? – 2017-02-24

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