For $a>4$, show that $\displaystyle\int_{0}^{a} |2x^{2}-3x-2|\,dx = \dfrac{1}{k}(4a^{3}-9a^{2}-12a+56)$ where $k$ is an integer to be found.

1. Find the roots of the quadratic expression to determine where it changes sign.

Let $2x^2 - 3x - 2 = 0$.
$(2x + 1)(x - 2) = 0$
$x = -\frac{1}{2}$ or $x = 2$

For $0 \le x \le 2$, $2x^2 - 3x - 2 \le 0 \implies |2x^2 - 3x - 2| = -(2x^2 - 3x - 2)$
For $x \ge 2$, $2x^2 - 3x - 2 \ge 0 \implies |2x^2 - 3x - 2| = 2x^2 - 3x - 2$

2. Split the definite integral at the root $x = 2$ and remove the modulus signs.

Since $a > 4$, the interval $[0, a]$ contains $x = 2$. We split the integral:
$\int_{0}^{a} |2x^{2}-3x-2|\,dx = \int_{0}^{2} -(2x^{2}-3x-2)\,dx + \int_{2}^{a} (2x^{2}-3x-2)\,dx$

3. Integrate the polynomial expressions.

$= \left[ -\frac{2}{3}x^3 + \frac{3}{2}x^2 + 2x \right]_0^2 + \left[ \frac{2}{3}x^3 - \frac{3}{2}x^2 - 2x \right]_2^a$

4. Substitute the limits, simplify, and factorise to find $k$.

$= \left( -\frac{16}{3} + 6 + 4 - 0 \right) + \left( \frac{2}{3}a^3 - \frac{3}{2}a^2 - 2a - \left( \frac{16}{3} - 6 - 4 \right) \right)$
$= \frac{14}{3} + \left( \frac{2}{3}a^3 - \frac{3}{2}a^2 - 2a - \left( -\frac{14}{3} \right) \right)$
$= \frac{2}{3}a^3 - \frac{3}{2}a^2 - 2a + \frac{28}{3}$
Expressing with a common denominator of 6:
$= \frac{4a^3 - 9a^2 - 12a + 56}{6}$
$= \frac{1}{6}(4a^3 - 9a^2 - 12a + 56)$

By comparing with the given expression, $k = 6$.