The graphs of $y = x^2 + 2x + 3$ and $y = 2x^2 + x + 1$ are shown below - Scottish Highers Maths - Question 8 - 2019

Now, we evaluate the integral:

$A = \int_{-1}^{2} (x^2 - x - 2) \, dx$

First, calculate the antiderivative:

$\int (x^2 - x - 2) \, dx = \frac{1}{3}x^3 - \frac{1}{2}x^2 - 2x$

Next, we substitute the limits $-1$ and $2$:

$A = \left[ \frac{1}{3}(2)^3 - \frac{1}{2}(2)^2 - 2(2) \right] - \left[ \frac{1}{3}(-1)^3 - \frac{1}{2}(-1)^2 - 2(-1) \right]$

Evaluating this gives:

$A = \left[ \frac{8}{3} - 2 - 4 \right] - \left[ -\frac{1}{3} - \frac{1}{2} + 2 \right]$

Calculating the first part:

$= \frac{8}{3} - 6 = \frac{8}{3} - \frac{18}{3} = -\frac{10}{3}$

And now the second part:

$-\left(-\frac{1}{3} - \frac{1}{2} + 2\right) = -\left(-\frac{1}{3} - \frac{3}{6} + \frac{12}{6}\right) = -\left(\frac{8}{6} - \frac{1}{3}\right) = \frac{10}{3}$

Thus, the total area is:

$A = \frac{10}{3} + \frac{10}{3} = \frac{20}{3}$

So, the shaded area between the curves is:

$A = 9 \; \text{units}^2$