$f(x) = \frac{(1+\alpha x)}{2} $ for $-1 \leq x \leq 1$ and $f(x) = 0$ otherwise, where $-1\leq \alpha \leq 1$. Show that $f$ is a density and find the cdf.
I am mainly having trouble with finding the cdf. By definition the cdf of a density function is defined as $$F(x) = \int_{-\infty}^{x} f(u) du $$ but i am having difficulty solving this particular one, specifically splitting up the limits of integration correctly. I tried this:
$$\int_{-\infty}^{-1} f(u) du + \int_{-1}^{x} f(u) du $$
To which i got:
$$\frac{x-1}{2} + \frac{\alpha(x^2 + 1)}{4} $$
I mean it differentiates out correctly to get the density function, but it doesn't mean that my integration was correct.
Best Answer
I don't understand why you must split up the integral, why not say $$ \int_{-\infty}^x f(u) du = \int_{-1}^x f(u) du = \left.\frac{u + \alpha u^2/2}{2}\right|_{-1}^x ... $$