Answer
$=\frac{1}{63}x^{3}-\frac{1}{63}$
Work Step by Step
We are given $f(x)=\frac{x^{2}}{21}$
for $1\leq x \leq 4$
The cumulative distribution function is given by
$F(x)=P(X\leq x)=\int^{x}_{1}(\frac{1}{21}t^{2})dt$
$=\frac{1}{63}t^{3}|^{x}_{1}$
$=\frac{1}{63}x^{3}-\frac{1}{63}$
