# Science:Math Exam Resources/Courses/MATH105/April 2016/Question 04 (a)/Solution 1

Note that the probability density function ${\displaystyle f}$ has ${\displaystyle [1,3]}$ as its domain. Since all probabilities must to integrate to ${\displaystyle 1}$, ${\displaystyle f}$ satisfies
${\displaystyle \int _{1}^{3}f(x)dx=1}$
${\displaystyle \int _{1}^{3}f(x)dx=\int _{1}^{2}f(x)dx+\int _{2}^{3}f(x)dx=\int _{1}^{2}{\frac {12}{31}}dx+\int _{2}^{3}ax^{2}dx={\frac {12}{31}}+{\frac {1}{3}}a(3^{3}-2^{3})={\frac {12}{31}}+{\frac {19}{3}}a.}$
Plugging this result into the equation, we get ${\displaystyle {\frac {12}{31}}+{\frac {19}{3}}a=1}$ and therefore ${\displaystyle {\frac {19}{3}}a={\frac {19}{31}}\iff a={\frac {3}{31}}}$
Answer: ${\displaystyle \color {blue}a={\frac {3}{31}}}$