The random variable X has pdf $f(x) = \frac{6}{5}(x^2 + x)$ where $0 \leq x \leq 1$
Find C.D.F of X
Attempt:
$$F(X \leq x) = \int_{-\infty}^{x} f(t)dt = \int_{-\infty}^{0} f(t)dt + \int_{0}^{x}f(t)dt$$
$$=0 + \frac{6}{5}\bigg(\frac{x^3}{3}+\frac{x^2}{2}\bigg) \text{ Because -infinity and 0 arent in the range, so it's 0}$$
How do I get $F(x)$?
What you have computed is $F(x)=P(X\leq x)$ when $0\leq x\leq 1 $ i.e. $$ F(x)=P(X\leq x)=\int_{0}^x f(t)\, dt=\frac{6}{5}\bigg(\frac{x^3}{3}+\frac{x^2}{2}\bigg). $$ To finish off the description of $F$ one can also say $F(x)=P(X\leq x)=0$ for $x<0$ and $F(x)=P(X\leq x)=1$ for $x>1$. Compactly $$ F(x)=\begin{cases}0 & x<0\\ \frac{6}{5}\bigg(\frac{x^3}{3}+\frac{x^2}{2}\bigg) &0\leq x\leq 1\\ 1& x>1 \end{cases} $$