Answer:
(a)
c=1/2
cdf is [tex]F(x) = x^2[/tex]
Step-by-step explanation:
Remember that if [tex]f[/tex] is the pdf of a random variable X .
[tex]\int\limits_{-\infty}^{\infty} f(x) \,dx = 1[/tex]
Then,
(a)
[tex]\int\limits_{-\infty}^{\infty} 4xc \,dx = \int\limits_{0}^{1} 4xc \, d x = 2c = 1[/tex]
Therefore
c=1/2
and
[tex]f(x)=2x[/tex]
then to compute the cdf, we have
[tex]F(x) = P(X \leq x) = \int\limits_{-\infty}^{x} 2y \,dy = \int\limits_{0}^{x} 2y \, dy = x^2[/tex]