The diagram below shows the graphs of \(y = \left| {\frac{3}{2}x - 3} \right|,{\text{ }}y = 3\) and a quadratic function, that all intersect in the same two points.

Given that the minimum value of the quadratic function is −3, find an expression for the area of the shaded region in the form \(\int_0^t {(a{x^2} + bx + c){\text{d}}x} \), where the constants a, b, c and t are to be determined. (Note: The integral does not need to be evaluated.)

\(\left| {\frac{3}{2}x - 3} \right| = 0\) when x = 2     (A1)

the equation of the parabola is \(y = p{(x - 2)^2} - 3\)     (M1)

through \((0,{\text{ }}3) \Rightarrow 3 = 4p - 3 \Rightarrow p = \frac{3}{2}\)     (M1)

the equation of the parabola is \(y = \frac{3}{2}{(x - 2)^2} - 3{\text{ }}\left( { = \frac{3}{2}{x^2} - 6x + 3} \right)\)     A1

area \( = 2\int_0^2 {\left( {3 - \frac{3}{2}x} \right) - \left( {\frac{3}{2}{x^2} - 6x + 3} \right){\text{d}}x} \)     M1M1A1

Note: Award M1 for recognizing symmetry to obtain \(2\int_0^2 , \)

M1 for the difference,

A1 for getting all parts correct.

\( = \int_0^2 {( - 3{x^2} + 9x){\text{d}}x} \)     A1

[8 marks]

This was a difficult question and, although many students obtained partial marks, there were few completely correct solutions.