To solve this problem, we need to find the area between the two curves depicted in the image.

The two curves are:

1. $y = 3x - x^2$ (a parabola)
2. $y = -x$ (a straight line)

Steps to find the area between the curves:

1. Find the points of intersection: To determine the limits of integration, we first solve for $x$ where $y = 3x - x^2$ and $y = -x$ intersect. This is done by equating the two expressions: $3x - x^2 = -x$ Simplifying: $3x + x = x^2 \implies x^2 - 4x = 0 \implies x(x - 4) = 0$ The points of intersection are $x = 0$ and $x = 4$.

2. Set up the integral: The area between the curves is given by the integral of the difference between the two functions over the interval from $x = 0$ to $x = 4$: $\text{Area} = \int_0^4 \left( (3x - x^2) - (-x) \right) dx$ Simplifying the integrand: $\text{Area} = \int_0^4 \left( 3x - x^2 + x \right) dx = \int_0^4 (4x - x^2) dx$

3. Compute the integral: Now, compute the integral: $\int (4x - x^2) dx = 2x^2 - \frac{x^3}{3}$ Evaluate this from $x = 0$ to $x = 4$: $\left[ 2x^2 - \frac{x^3}{3} \right]_0^4 = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - \left( 2(0)^2 - \frac{(0)^3}{3} \right)$ Simplifying: $= \left( 2 \cdot 16 - \frac{64}{3} \right) = \left( 32 - \frac{64}{3} \right)$ $= \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$ Therefore, the area is $\frac{32}{3} \approx 10.67$.

The area of the region between the curves is approximately 10.67 square units.

Would you like further details, or have any questions?

Here are five follow-up questions:

1. How would the calculation change if the limits of integration were different?
2. Can you compute the area between two curves using numerical methods?
3. What is the geometric interpretation of integrating the difference of functions?
4. How does symmetry of curves affect area calculations?
5. Can we find the area between the curves without using calculus?

Tip: Always sketch the region between curves when solving area problems—it helps in setting up proper integrals and limits!