The diagram shows two parabolas $y = 4x - x^2$ and $y = ax^2$, where $a > 0$ - HSC - SSCE Mathematics Advanced - Question 30 - 2020 - Paper 1

To determine the value of $a$, we need to calculate the shaded area between the two curves:

$\int_0^{\frac{4}{a+1}} \left( (4x - x^2) - (ax^2) \right) dx = \frac{16}{3}$

Calculating the integral:

$\int_0^{\frac{4}{a+1}} (4x - (1 + a)x^2) dx$

Calculating the limits:

$\left[2x^2 - \frac{(1 + a)x^3}{3}\right]_0^{\frac{4}{a+1}}$

Substituting the upper limit:

$2\left(\frac{4}{a+1}\right)^2 - \frac{(1 + a)\left(\frac{4}{a+1}\right)^3}{3} = \frac{16}{3}$

Expanding and simplifying:

$\frac{32}{(a + 1)^2} - \frac{64(1 + a)}{3(a + 1)^3} = \frac{16}{3}$

To solve for $a$, multiply through by $3(a + 1)^3$ and simplify to find:

$a + 1)^3 - 32(a + 1) + 64(1 + a) = 16\Rightarrow a + 1 = 2$

Thus, $a = 1$.