We are asked to find the sum of the areas of the three regions enclosed by the parabola $y = x^2 - 2x$, the x-axis, and the vertical lines $x = -1$ and $x = 3$.

1. Problem Description

We are asked to find the sum of the areas of the three regions enclosed by the parabola $y = x^2 - 2x$, the x-axis, and the vertical lines $x = -1$ and $x = 3$.

2. Solution Steps

First, we need to find the x-intercepts of the parabola $y = x^2 - 2x$. To do this, we set $y = 0$:

$x^2 - 2x = 0$

$x(x - 2) = 0$

So the x-intercepts are $x = 0$ and $x = 2$.

Since the region is bounded by $x = -1$ and $x = 3$, we need to split the integral into three parts:

1. From $x = -1$ to $x = 0$

2. From $x = 0$ to $x = 2$

3. From $x = 2$ to $x = 3$

In the interval $[-1, 0]$, $x^2 - 2x > 0$, so the area is

$A_1 = \int_{-1}^0 (x^2 - 2x) dx = [\frac{x^3}{3} - x^2]_{-1}^0 = (0 - 0) - (\frac{(-1)^3}{3} - (-1)^2) = 0 - (-\frac{1}{3} - 1) = 0 - (-\frac{4}{3}) = \frac{4}{3}$.

In the interval $[0, 2]$, $x^2 - 2x < 0$, so the area is

$A_2 = \int_{0}^2 |x^2 - 2x| dx = -\int_{0}^2 (x^2 - 2x) dx = -[\frac{x^3}{3} - x^2]_{0}^2 = -(\frac{2^3}{3} - 2^2 - (0 - 0)) = -(\frac{8}{3} - 4) = -(\frac{8}{3} - \frac{12}{3}) = -(-\frac{4}{3}) = \frac{4}{3}$.

In the interval $[2, 3]$, $x^2 - 2x > 0$, so the area is

$A_3 = \int_{2}^3 (x^2 - 2x) dx = [\frac{x^3}{3} - x^2]_{2}^3 = (\frac{3^3}{3} - 3^2) - (\frac{2^3}{3} - 2^2) = (\frac{27}{3} - 9) - (\frac{8}{3} - 4) = (9 - 9) - (\frac{8}{3} - \frac{12}{3}) = 0 - (-\frac{4}{3}) = \frac{4}{3}$.

The total area is $A = A_1 + A_2 + A_3 = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = \frac{12}{3} = 4$.

3. Final Answer

4