To solve the given problem, we will find the area between the two functions $f(x) = -x + 3$ and $g(x) = 2$ over the interval $0 \leq x \leq 3$.

Step 1: Identify the intersection point

The functions intersect when $f(x) = g(x)$: $-x + 3 = 2$ $x = 1$

Step 2: Divide the area into two regions

From $x = 0$ to $x = 1$, $f(x)$ is above $g(x)$. From $x = 1$ to $x = 3$, $g(x)$ is above $f(x)$.

Step 3: Set up the integrals

• For $0 \leq x \leq 1$, the area is: $\int_0^1 \big[ (-x + 3) - 2 \big] \, dx$
• For $1 \leq x \leq 3$, the area is: $\int_1^3 \big[ 2 - (-x + 3) \big] \, dx$

Step 4: Evaluate the integrals

1. Compute the first integral: [ \int_0^1 \big[ -x + 3 - 2 \big] , dx = \int_0^1 \big[ -x + 1 \big] , dx ] $= \left[ -\frac{x^2}{2} + x \right]_0^1 = \left( -\frac{1^2}{2} + 1 \right) - \left( -\frac{0^2}{2} + 0 \right)$ $= -\frac{1}{2} + 1 = \frac{1}{2}$

2. Compute the second integral: [ \int_1^3 \big[ 2 - (-x + 3) \big] , dx = \int_1^3 \big[ 2 + x - 3 \big] , dx = \int_1^3 \big[ x - 1 \big] , dx ] $= \left[ \frac{x^2}{2} - x \right]_1^3 = \left( \frac{3^2}{2} - 3 \right) - \left( \frac{1^2}{2} - 1 \right)$ $= \left( \frac{9}{2} - 3 \right) - \left( \frac{1}{2} - 1 \right) = \frac{3}{2} + \frac{1}{2} = 2$

Step 5: Add the areas

The total area is: $\frac{1}{2} + 2 = \frac{5}{2} = 2.5$

Would you like more details on any of these steps? Here are five related questions:

1. How do you determine the points of intersection between two functions?
2. Can you explain in detail why the area is split into two integrals?
3. How do you verify if one function is above or below another graphically?
4. Could you explain the integration process step-by-step?
5. How would this change if the boundaries of integration were different?

Tip: When setting up definite integrals, ensure that you consistently subtract the lower function from the upper function to avoid negative areas!