AMU 2026 — Mathematics PYQ

The mean (or expected value) $E(X)$ of a continuous random variable with a PDF $f(x)$ is calculated using the following integral formula:

$$E(X)={\int }_{-\infty }^{\infty }x\cdot f(x)dx$$

Given the range 0 < x < 1 and $f(x)=12x(1-x{)}^2$:

1. Set up the integral:

   $$E(X)={\int }_0^{1}x\cdot [12x(1-x{)}^2]dx$$

   $$E(X)=12{\int }_0^1{x}^2(1-x{)}^{2}dx$$

2. Expand the expression inside the integral:

   $$(1-x{)}^2=1-2x+x^2$$

   $$x^2(1-2x+x^2)=x^2-2x^3+x^4$$

3. Integrate:

   $$E(X)=12{\int }_0^1(x^2-2x^3+x^4)dx$$

   $$E(X)=12{\left[\frac{x^3}{3}-\frac{2x^4}{4}+\frac{x^5}{5}\right]}_0^1$$

   $$E(X)=12\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)$$

4. Simplify the calculation:

   To solve the fraction, find a common denominator ($30$):

   $$\frac{1}{3}-\frac{1}{2}+\frac{1}{5}=\frac{10-15+6}{30}=\frac{1}{30}$$

   $$E(X)=12\times \left(\frac{1}{30}\right)=\frac{12}{30}=\frac{2}{5}$$