The diagram shows part of the curve with equation $y = x^3 - 2x^2 - 4x + 1$ and the line with equation $y = x - 5$ - Scottish Highers Maths - Question 8 - 2023

The limits of integration are set from $x = -2$ to $x = 1$.

Now we simplify the integrand: $(x - 5) - (x^3 - 2x^2 - 4x + 1) = -x^3 + 2x^2 + 5 - 4x$

Thus, the integral becomes: $\int_{-2}^{1} (-x^3 + 2x^2 - 4x + 4) \, dx$

Next, we calculate the definite integral: $\int (-x^3 + 2x^2 - 4x + 4) \, dx = -\frac{1}{4}x^4 + \frac{2}{3}x^3 - 2x^2 + 4x$

Substituting the limits: $= \left[-\frac{1}{4}(1)^4 + \frac{2}{3}(1)^3 - 2(1)^2 + 4(1)\right] - \left[-\frac{1}{4}(-2)^4 + \frac{2}{3}(-2)^3 - 2(-2)^2 + 4(-2)\right]$

Calculating these expressions:

For $x = 1$: $-\frac{1}{4}(1) + \frac{2}{3}(1) - 2(1) + 4(1) = -\frac{1}{4} + \frac{2}{3} - 2 + 4 = \frac{9}{12} + \frac{6}{12} - \frac{24}{12} = \frac{-9}{12}$

For $x = -2$: $-\frac{1}{4}(16) + \frac{2}{3}(-8) - 2(4) - 8 = -4 - \frac{16}{3} - 8 = -4 + 8 - \frac{16}{3} = -\frac{12}{3} - \frac{16}{3} = \frac{-28}{3}$

Finally, the total area is: $\left( -\frac{9}{12} - (-\frac{28}{3}) \right) = (-\frac{9}{12} + \frac{112}{12}) = \frac{103}{12}$