The problem asks us to evaluate definite integrals using the graph and geometric formulas. Here's the step-by-step breakdown of the two parts:

(a) $\int_0^1 -2f(x) \, dx$

1. The graph from $x = 0$ to $x = 1$ shows $f(x)$ is constant and equal to $y = -1$.
2. The integral becomes: $\int_0^1 -2f(x) \, dx = \int_0^1 -2(-1) \, dx = \int_0^1 2 \, dx.$
3. The value of $\int_0^1 2 \, dx$ represents the area of a rectangle with height 2 and width 1: $\text{Area} = 2 \cdot 1 = 2.$
4. Thus, the result is: $\int_0^1 -2f(x) \, dx = 2.$

(b) $\int_3^4 3f(x) \, dx$

1. The graph from $x = 3$ to $x = 4$ shows $f(x)$ is constant and equal to $y = 2$.
2. The integral becomes: $\int_3^4 3f(x) \, dx = \int_3^4 3(2) \, dx = \int_3^4 6 \, dx.$
3. The value of $\int_3^4 6 \, dx$ represents the area of a rectangle with height 6 and width 1: $\text{Area} = 6 \cdot 1 = 6.$
4. Thus, the result is: $\int_3^4 3f(x) \, dx = 6.$

Final Results:

• (a): $\int_0^1 -2f(x) \, dx = 2.$
• (b): $\int_3^4 3f(x) \, dx = 6.$

Would you like additional clarification on geometric integration or any other part of this problem?

Related Questions:

1. How do you calculate the integral of linear segments on graphs using geometric formulas?
2. Why does multiplying the function inside an integral scale the area?
3. What are the common steps to compute integrals over a piecewise linear function?
4. How do definite integrals differ when the function has negative values?
5. What geometric shapes are typically used in evaluating integrals graphically?

Tip:

For definite integrals of piecewise functions, always identify the corresponding segments and check their values before integrating!