From the uploaded image, you are tasked with evaluating definite integrals based on a graph consisting of line segments. Each definite integral can be evaluated using geometric formulas by finding the area under the graph of $f(x)$ multiplied by the given scalar for each case. Let's solve each part step by step:

Analysis of Graph:

The graph is segmented into linear pieces with defined vertices. Key points are marked as:

• $(0, 0)$
• $(3, 2)$
• $(4, 2)$
• $(8, -2)$
• $(11, 1)$

Geometric Formulas:

For these problems, calculate the areas of triangles and rectangles formed by the graph and consider the multiplier (e.g., $2f(x)$, $8f(x)$).

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

• $f(x) = 0$ from $x = 0$ to $x = 1$. Since $f(x) = 0$, the integral evaluates to 0.

(b) $\int_3^4 8f(x) \, dx$

• From $x = 3$ to $x = 4$, $f(x) = 2$.
• The integral is the area of a rectangle with height $2$ and width $1$, multiplied by 8: $\int_3^4 8f(x) \, dx = 8 \cdot (1 \cdot 2) = 16$

(c) $\int_0^7 6f(x) \, dx$

• Break the interval into segments:
  • $x = 0$ to $3$: Triangle with base $3$ and height $2$: $\text{Area} = \frac{1}{2}(3)(2) = 3$
  • $x = 3$ to $4$: Rectangle with width $1$ and height $2$: $\text{Area} = 2$
  • $x = 4$ to $7$: Triangle with base $3$ and height $-2$: $\text{Area} = \frac{1}{2}(3)(-2) = -3$
• Total weighted area: $6 \cdot (3 + 2 - 3) = 6 \cdot 2 = 12$

(d) $\int_5^{11} -3f(x) \, dx$

• Break into segments:
  • $x = 5$ to $8$: Triangle with base $3$ and height $-2$: $\text{Area} = \frac{1}{2}(3)(-2) = -3$
  • $x = 8$ to $11$: Triangle with base $3$ and height $1$: $\text{Area} = \frac{1}{2}(3)(1) = 1.5$
• Total weighted area: $-3 \cdot (-3 + 1.5) = -3 \cdot -1.5 = 4.5$

(e) $\int_0^{11} 2f(x) \, dx$

• Add all areas:
  • $x = 0$ to $3$: $3$
  • $x = 3$ to $4$: $2$
  • $x = 4$ to $8$: $-3$
  • $x = 8$ to $11$: $1.5$
• Total area: $2 \cdot (3 + 2 - 3 + 1.5) = 2 \cdot 3.5 = 7$

(f) $\int_4^{10} 2f(x) \, dx$

• Add areas:
  • $x = 4$ to $8$: $-3$
  • $x = 8$ to $10$: Part of the triangle $x = 8$ to $x = 11$:
    • Height $1$, base $3$, consider $\frac{2}{3}$ of the area: $\text{Area} = \frac{1}{2}(1)(2) = 1$
• Total area: $2 \cdot (-3 + 1) = 2 \cdot -2 = -4$

Final Answers:

(a) $0$
(b) $16$
(c) $12$
(d) $4.5$
(e) $7$
(f) $-4$

Would you like step-by-step elaboration on any specific part or further explanation?

Expanding Thoughts:

1. How does the scaling constant affect the result of an integral?
2. Why is the area below the x-axis considered negative?
3. What is the importance of geometric interpretation in definite integrals?
4. How can such problems apply to physics or economics?
5. What happens when the limits of integration are reversed?

Tip: Always sketch graphs when solving definite integrals, especially piecewise functions!