Calculus with Applications (10th Edition)

$\frac{1}{63}x^{3}-\frac{27}{98}$
We are given $f(x)=\frac{3}{98}x^{2}$ for $3\leq x \leq 5$ The cumulative distribution function is given by $F(x)=P(X\leq x)=\int^{x}_{3}(\frac{3}{98}t^{2})dt$ $=\frac{1}{98}t^{3}|^{x}_{3}$ $=\frac{1}{63}x^{3}-\frac{27}{98}$