Two curves with equations $y = x^3 - 4x^2 + 3x + 1$ and $y = x^3 - 3x + 1$ intersect as shown in the diagram - Scottish Highers Maths - Question 10 - 2017

To calculate the shaded area between the two curves, we follow these steps:

1. Identify Intersection Points: First, we determine the points where the two curves intersect by setting their equations equal to each other: $x^3 - 4x^2 + 3x + 1 = x^3 - 3x + 1$ Simplifying this gives: $-4x^2 + 3x + 3x = 0$ which simplifies to $-4x^2 + 6x = 0$. Factoring out the common term: $2x(3 - 2x) = 0$. Thus, the intersection points are: $x = 0$ and $x = \frac{3}{2}$.

2. Set Up the Integral: The area between the curves can be found by integrating the difference between the upper curve and the lower curve: $\text{Area} = \int_{0}^{\frac{3}{2}} [(x^3 - 4x^2 + 3x + 1) - (x^3 - 3x + 1)] \,dx$. This simplifies to: $\int_{0}^{\frac{3}{2}} (-4x^2 + 6x) \,dx$.

3. Integrate the Function: Calculating the integral: $= \int (-4x^2 + 6x) \,dx \Rightarrow -\frac{4}{3}x^3 + 3x^2$.

4. Substitute Limits: We evaluate the integral from 0 to (\frac{3}{2}):

   $\left[-\frac{4}{3}\left(\frac{3}{2}\right)^3 + 3\left(\frac{3}{2}\right)^2\right] - [0]$ $= -\frac{4}{3}\left(\frac{27}{8}\right) + 3\left(\frac{9}{4}\right)$ $= -\frac{9}{2} + \frac{27}{4} = \frac{9}{4}$.

5. Evaluate Total Area: The area is: $\text{Area} = \frac{9}{4}$.