1. **Problem Statement:** Evaluate the definite integrals of the piecewise linear function $f(x)$ given the graph description: - $f(x)=0$ from $x=0$ to $x=2$ - At $x=3$, $f(3)=1$ - $f(x)$ increases linearly from $(3,1)$ to $(10,3)$ We need to find: (a) $\int_0^3 f(x) \, dx$ (b) $\int_3^4 f(x) \, dx$ (c) $\int_0^4 f(x) \, dx$ (d) $\int_4^{10} f(x) \, dx$ (e) $\int_0^{10} -4 f(x) \, dx$ --- 2. **Step 1: Understand the function and geometry** - From $0$ to $2$, $f(x)=0$, so area under curve is zero. - At $x=3$, $f(3)=1$. - From $3$ to $10$, $f(x)$ is linear from $1$ to $3$. 3. **Step 2: Calculate $\int_0^3 f(x) \, dx$** - From $0$ to $2$, $f(x)=0$, so area is $0$. - From $2$ to $3$, the graph jumps from $0$ to $1$ at $x=3$. - Since no value is given between $2$ and $3$, assume $f(x)=0$ for $2 \leq x < 3$ (based on description). - So area from $0$ to $3$ is $0$. 4. **Step 3: Calculate $\int_3^4 f(x) \, dx$** - From $3$ to $10$, $f(x)$ is linear from $1$ to $3$. - Find equation of line between $(3,1)$ and $(10,3)$: $$m=\frac{3-1}{10-3}=\frac{2}{7}$$ $$f(x)=1 + \frac{2}{7}(x-3)$$ - Evaluate integral from $3$ to $4$: $$\int_3^4 \left(1 + \frac{2}{7}(x-3)\right) dx = \int_3^4 1 \, dx + \int_3^4 \frac{2}{7}(x-3) \, dx$$ Calculate each: $$\int_3^4 1 \, dx = 4-3=1$$ $$\int_3^4 \frac{2}{7}(x-3) \, dx = \frac{2}{7} \int_3^4 (x-3) \, dx = \frac{2}{7} \left[ \frac{(x-3)^2}{2} \right]_3^4 = \frac{2}{7} \times \frac{(4-3)^2 - 0}{2} = \frac{2}{7} \times \frac{1}{2} = \frac{1}{7}$$ Sum: $$1 + \frac{1}{7} = \frac{8}{7} \approx 1.1429$$ 5. **Step 4: Calculate $\int_0^4 f(x) \, dx$** - This is sum of $\int_0^3 f(x) \, dx$ and $\int_3^4 f(x) \, dx$: $$0 + \frac{8}{7} = \frac{8}{7}$$ 6. **Step 5: Calculate $\int_4^{10} f(x) \, dx$** - Use the same linear function: $$\int_4^{10} \left(1 + \frac{2}{7}(x-3)\right) dx = \int_4^{10} 1 \, dx + \int_4^{10} \frac{2}{7}(x-3) \, dx$$ Calculate each: $$\int_4^{10} 1 \, dx = 10-4=6$$ $$\int_4^{10} \frac{2}{7}(x-3) \, dx = \frac{2}{7} \left[ \frac{(x-3)^2}{2} \right]_4^{10} = \frac{2}{7} \times \frac{(10-3)^2 - (4-3)^2}{2} = \frac{2}{7} \times \frac{49 - 1}{2} = \frac{2}{7} \times 24 = \frac{48}{7}$$ Sum: $$6 + \frac{48}{7} = \frac{42}{7} + \frac{48}{7} = \frac{90}{7} \approx 12.8571$$ 7. **Step 6: Calculate $\int_0^{10} -4 f(x) \, dx$** - Use linearity of integral: $$\int_0^{10} -4 f(x) \, dx = -4 \int_0^{10} f(x) \, dx$$ - Calculate $\int_0^{10} f(x) \, dx = \int_0^3 f(x) \, dx + \int_3^{10} f(x) \, dx = 0 + \int_3^{10} f(x) \, dx$ - Calculate $\int_3^{10} f(x) \, dx$: $$\int_3^{10} \left(1 + \frac{2}{7}(x-3)\right) dx = \int_3^{10} 1 \, dx + \int_3^{10} \frac{2}{7}(x-3) \, dx = 7 + \frac{2}{7} \times \frac{(10-3)^2 - 0}{2} = 7 + \frac{2}{7} \times \frac{49}{2} = 7 + 7 = 14$$ - So: $$\int_0^{10} -4 f(x) \, dx = -4 \times 14 = -56$$ --- **Final answers:** (a) $0$ (b) $\frac{8}{7}$ (c) $\frac{8}{7}$ (d) $\frac{90}{7}$ (e) $-56$