To solve these integrals using the graph provided, we can calculate the area under the curve $y = f(x)$ between the specified limits. Let's go through each question step-by-step.

(a) $\int_0^2 f(x) \, dx$

From $x = 0$ to $x = 2$:

• The function is constant at $f(x) = -1$.
• The area of this rectangle is $\text{width} \times \text{height} = 2 \times (-1) = -2$.

So, the answer is -2.

(b) $\int_0^4 f(x) \, dx$

From $x = 0$ to $x = 2$:

• We already calculated this area as $-2$.

From $x = 2$ to $x = 4$:

• The function is constant at $f(x) = 1$.
• The area of this rectangle is $\text{width} \times \text{height} = 2 \times 1 = 2$.

Adding both areas: $-2 + 2 = 0$

So, the answer is 0.

(c) $\int_0^8 f(x) \, dx$

Breaking down the integral from $x = 0$ to $x = 8$:

• From $x = 0$ to $x = 2$: area = $-2$ (as calculated).
• From $x = 2$ to $x = 4$: area = $2$ (as calculated).
• From $x = 4$ to $x = 6$: the function is at $f(x) = 2$.
  • Area = $\text{width} \times \text{height} = 2 \times 2 = 4$.
• From $x = 6$ to $x = 8$: the function is back to $f(x) = -1$.
  • Area = $\text{width} \times \text{height} = 2 \times (-1) = -2$.

Summing all areas: $-2 + 2 + 4 - 2 = 2$

So, the answer is 2.

Summary of Answers

• (a) $-2$
• (b) $0$
• (c) $2$

Would you like further details or have any questions about the solution?

Additional Related Questions

1. How do we handle areas below the x-axis in definite integrals?
2. What changes in the result if we switch the limits of integration, say from $\int_2^0 f(x) \, dx$ instead of $\int_0^2 f(x) \, dx$?
3. Can we compute these integrals without directly counting the areas if we have the function's expression?
4. What is the impact of negative areas in integration, and how do they affect cumulative totals?
5. How would the integral values change if the function extended beyond $x = 8$ with similar values?

Tip: When calculating areas under curves, remember to consider both the magnitude and direction (positive above the x-axis, negative below) to get accurate integral values.