The graph shown has equation $y = x^3 - 5x^2 + 2x + 8$ - Scottish Highers Maths - Question 4 - 2022

To find the shaded area above the x-axis, we need to determine the definite integral of the function from the x-values where the curve intersects the x-axis.

1. Find the points of intersection with the x-axis: Set the equation equal to zero:

   $x^3 - 5x^2 + 2x + 8 = 0$

   This equation can be solved using numerical or graphical methods to find the roots, which occur approximately at $x hickapprox -1$ and $x hickapprox 4$.

2. Calculate the integral: The area above the x-axis can be calculated with the integral:

   $A = \int_{-1}^{4} (x^3 - 5x^2 + 2x + 8) \, dx$

   Start by computing the indefinite integral:

   $\int (x^3 - 5x^2 + 2x + 8) \, dx = \frac{x^4}{4} - \frac{5x^3}{3} + x^2 + 8x + C$

   Now, substitute the limits:

   $\left[ \frac{(4)^4}{4} - \frac{5(4)^3}{3} + (4)^2 + 8(4) \right] - \left[ \frac{(-1)^4}{4} - \frac{5(-1)^3}{3} + (-1)^2 + 8(-1) \right]$

   Evaluating these gives:

   • For $x = 4$:

   $\frac{256}{4} - \frac{5(64)}{3} + 16 + 32 = 64 - \frac{320}{3} + 48 = 64 + 48 - \frac{320}{3} = \frac{384 - 320}{3} = \frac{64}{3}$

   • For $x = -1$:

   $\frac{1}{4} - \frac{5(-1)}{3} + 1 - 8 = \frac{1}{4} + \frac{5}{3} - 7 = \frac{1}{4} + \frac{20}{12} - \frac{84}{12} = -\frac{63}{12} = -\frac{21}{4}$

   Finally, the area above the x-axis is:

   $A = \left( \frac{64}{3} - \left(-\frac{21}{4}\right)\right)\text{squared units} = \frac{256}{12} + \frac{63}{12} = \frac{319}{12} \text{squared units}$