1. **Problem statement:** Evaluate the definite integrals of the piecewise linear function $f(x)$ given by: - $f(x) = 0$ for $0 \leq x \leq 2$ - $f(x)$ increases linearly from 0 to 3 on $[2,3]$ - $f(x)$ increases linearly from 3 to 7 on $[3,10]$ 2. **Formula and approach:** The definite integral $\int_a^b f(x) dx$ represents the area under the curve $f(x)$ from $x=a$ to $x=b$. Since $f(x)$ is piecewise linear, we can find areas of geometric shapes (rectangles, triangles, trapezoids) formed by the graph and the x-axis. 3. **Calculate each integral:** **a) $\int_0^3 f(x) dx$** - From $0$ to $2$, $f(x)=0$, so area is 0. - From $2$ to $3$, $f(x)$ goes from 0 to 3 linearly, forming a triangle with base 1 and height 3. - Area of triangle = $\frac{1}{2} \times 1 \times 3 = \frac{3}{2} = 1.5$ - So, $\int_0^3 f(x) dx = 0 + 1.5 = 1.5$ **b) $\int_3^4 f(x) dx$** - From $3$ to $4$, $f(x)$ increases linearly from 3 to approximately 3.57 (since from 3 to 10 it goes from 3 to 7, slope $= \frac{7-3}{10-3} = \frac{4}{7} \approx 0.571$) - So, $f(4) = 3 + 0.571 \times (4-3) = 3 + 0.571 = 3.571$ - Area under $f(x)$ from 3 to 4 is a trapezoid with bases 3 and 3.571 and height 1. - Area = $\frac{1}{2} (3 + 3.571) \times 1 = \frac{6.571}{2} = 3.2855$ **c) $\int_0^4 f(x) dx$** - Sum of parts a) and b): $1.5 + 3.2855 = 4.7855$ **d) $\int_4^{10} f(x) dx$** - From $4$ to $10$, $f(x)$ increases linearly from 3.571 to 7. - Area is trapezoid with bases 3.571 and 7, height 6. - Area = $\frac{1}{2} (3.571 + 7) \times 6 = \frac{10.571}{2} \times 6 = 5.2855 \times 6 = 31.713$ **e) $\int_0^{10} -4 f(x) dx$** - Use linearity: $-4 \int_0^{10} f(x) dx$ - $\int_0^{10} f(x) dx = \int_0^4 f(x) dx + \int_4^{10} f(x) dx = 4.7855 + 31.713 = 36.4985$ - Multiply by $-4$: $-4 \times 36.4985 = -145.994$ 4. **Final answers:** $$\int_0^3 f(x) dx = 1.5$$ $$\int_3^4 f(x) dx \approx 3.29$$ $$\int_0^4 f(x) dx \approx 4.79$$ $$\int_4^{10} f(x) dx \approx 31.71$$ $$\int_0^{10} -4 f(x) dx \approx -145.99$$