The curves $y = (x - 1)^2$ and $y = 5 - x^2$ intersect at two points, as shown in the diagram - HSC - SSCE Mathematics Advanced - Question 14 - 2024 - Paper 1

The area A between the curves can be calculated using the integral:

$A = \int_{-1}^{2} [(5 - x^2) - (x - 1)^2] \: dx$

First, simplify the integrand:

$5 - x^2 - (x^2 - 2x + 1) = 5 - 2x^2 + 2x - 1 = 4 - 2x^2 + 2x$

Now the integral becomes:

$A = \int_{-1}^{2} (4 - 2x^2 + 2x) \: dx$

Calculating the integral:

$= \left[ 4x - \frac{2}{3}x^3 + x^2 \right]_{-1}^{2}$

Evaluating at the limits:

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

$= (8 - \frac{16}{3} + 4) - (-4 + \frac{2}{3} + 1)$

Simplify:

$= (12 - \frac{16}{3}) - (-3 + \frac{2}{3})$

Combine:

$= \left( 12 + 3 - \frac{16}{3} - \frac{2}{3} \right) = \frac{39 - 18}{3} = \frac{21}{3} = 7$

Thus, the area enclosed by the two curves is:

$A = 9$