Wondering how to find the value of the constant k for this. I know you have to integrate, so I integrated both functions, however I'm not quite sure on the steps afterwards. Any help would be greatly appreciated!
1 Answer
$\begingroup$We know that the integral over the density function must be equal to $1$.
That is:
$$\int_0^1 kx^2~ dx+\int_1^2 k(2-x)~ dx=1$$
Computing the two integrals gives:$$\frac{k}{2}+\frac{k}{3}=1 \iff \frac{5k}{6}=1$$ and so we have $$k=\frac{6}{5}$$
$\endgroup$ 2
