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In an exercise I am asked:

Consider the function $f(x)=kx$ if $1

What value must the real $k$ take so that $f(x)$ is a function of uniform density?

For $f(x)$ to be uniform, for any value of $x$ of the range, $f(x)$ must always be equal. But from there on, I do not know how to solve it.

I hope you can help me. Thank you very much.

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    There are two conditions any density function has to respect: 1.) $f(x)\geq 0 $ for all $x\in \mathbb{R}$ and 2.) $\int_{-\infty}^\infty f(x) \, dx =1$. One of these conditions will implie the value $k$. edit: f(x) will never be the density of a uniform random variable (its value should be constant on $1 – 2017-02-23

1 Answers 1

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$$ \int^5_1 kx~ dx =1$$

$$ \frac{kx^2}{2}|^5_1=1$$

$$ \frac{25k}{2}-\frac{k}{2}=1$$

$$12k=1$$

$$ \therefore k=\frac{1}{12}$$