Let's solve the problem by evaluating the definite integrals based on the graph.

Step 1: Evaluate $\int_{0}^{2} f(x) \, dx$

From the graph:

• $f(x)$ is constant and equal to $-2$ for $x \in [0, 2]$.
• The integral in this interval is the area of the rectangle between $y = -2$ and the x-axis from $x = 0$ to $x = 2$.

$\int_{0}^{2} f(x) \, dx = \text{Area of rectangle} = \text{(base)} \times \text{(height)} = 2 \times (-2) = -4$

Step 2: Evaluate $\int_{2}^{7} f(x) \, dx$

Break the interval $[2, 7]$ into smaller sections based on changes in the graph.

a) Interval $[2, 3]$

• $f(x)$ increases linearly from $-2$ to $0$.
• This is a right triangle with:
  • Base = $1$ (from $x = 2$ to $x = 3$),
  • Height = $2$ (from $-2$ to $0$).

Area: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1$

b) Interval $[3, 6]$

• $f(x)$ increases linearly from $0$ to $3$.
• This is a right triangle with:
  • Base = $3$ (from $x = 3$ to $x = 6$),
  • Height = $3$.

Area: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5$

c) Interval $[6, 7]$

• $f(x)$ is constant at $3$.
• This is a rectangle with:
  • Base = $1$ (from $x = 6$ to $x = 7$),
  • Height = $3$.

Area: $\text{Area} = \text{base} \times \text{height} = 1 \times 3 = 3$

Total for $[2, 7]$:

$\int_{2}^{7} f(x) \, dx = 1 + 4.5 + 3 = 8.5$

Final Answers:

1. $\int_{0}^{2} f(x) \, dx = -4$
2. $\int_{2}^{7} f(x) \, dx = 8.5$

Let me know if you would like a detailed explanation of any part!

Follow-up Questions:

1. How do you calculate the area under a curve using definite integrals?
2. Why is the area below the x-axis considered negative?
3. Can you explain why a triangle's area is calculated as $\frac{1}{2} \times \text{base} \times \text{height}$?
4. What happens when the graph has both positive and negative regions? How does it affect the integral?
5. How do we approximate areas when the function is not linear or constant?

Tip: Always check whether the area is above or below the x-axis to apply the correct sign to your integral.