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

Note that the probability density function $f$ has $[1,3]$ as its domain. Since all probabilities must to integrate to $1$, $f$ satisfies

$\int _{1}^{3}f(x)dx=1$

We first evaluate the integral on the left hand side;

$\int _{1}^{3}f(x)dx=\int _{1}^{2}f(x)dx+\int _{2}^{3}f(x)dx=\int _{1}^{2}{\frac {12}{31}}dx+\int _{2}^{3}ax^{2}dx={\frac {12}{31}}+{\frac {1}{3}}a(3^{3}-2^{3})={\frac {12}{31}}+{\frac {19}{3}}a.$

Plugging this result into the equation, we get ${\frac {12}{31}}+{\frac {19}{3}}a=1$ and therefore ${\frac {19}{3}}a={\frac {19}{31}}\iff a={\frac {3}{31}}$

Answer: $\color {blue}a={\frac {3}{31}}$