1. **Problem statement:** We are given a function $g(x) = \int_0^x f(t) \, dt$ where $f$ is a piecewise linear function described by its graph. 2. **Goal:** Evaluate $g(0)$, $g(1)$, $g(2)$, $g(3)$, and $g(6)$. 3. **Formula and concept:** The function $g(x)$ is the definite integral of $f(t)$ from 0 to $x$. This means $g(x)$ represents the area under the curve of $f(t)$ from $t=0$ to $t=x$. Positive areas add to $g(x)$, negative areas subtract. 4. **Step-by-step evaluation:** - $g(0) = \int_0^0 f(t) \, dt = 0$ because the integral over zero width is zero. - $g(1) = \int_0^1 f(t) \, dt$. From the graph, $f(t) = 1$ for $0 \leq t \leq 1$, so area is a rectangle of height 1 and width 1: $$g(1) = 1 \times 1 = 1$$ - $g(2) = g(1) + \int_1^2 f(t) \, dt$. From $t=1$ to $t=2$, $f(t)$ rises linearly from 1 to a peak (assumed 2 for calculation since exact peak not given, but graph suggests about 2). The area under this segment is a trapezoid: $$\text{Area} = \frac{(f(1) + f(2))}{2} \times (2-1) = \frac{1 + 2}{2} \times 1 = 1.5$$ So, $$g(2) = g(1) + 1.5 = 1 + 1.5 = 2.5$$ - $g(3) = g(2) + \int_2^3 f(t) \, dt$. From $t=2$ to $t=3$, $f(t)$ sharply declines through zero. Assume linear drop from 2 at $t=2$ to 0 at $t=3$. Area is a triangle below the $t$-axis: $$\text{Area} = \frac{1}{2} \times 1 \times 2 = 1$$ Since $f(t)$ goes from positive to zero, the area from 2 to 3 is positive above $t$-axis, but graph says sharp decline through zero, so area from 2 to 3 is positive area above $t$-axis minus area below. Assuming linear drop to zero, area is positive triangle: $$g(3) = g(2) + 1 = 2.5 + 1 = 3.5$$ - $g(6) = g(3) + \int_3^6 f(t) \, dt$. From $t=3$ to $t=5$, $f(t)$ declines below zero to a minimum, stays constant near minimum until $t=5$, then rises again by $t=6$. Approximate the area from 3 to 5 as a rectangle below $t$-axis with height about -1 and width 2: $$\text{Area} = -1 \times 2 = -2$$ From 5 to 6, $f(t)$ rises from minimum (-1) to about 0 or slightly above, approximate area as triangle: $$\text{Area} = \frac{1}{2} \times 1 \times 1 = 0.5$$ So total from 3 to 6: $$-2 + 0.5 = -1.5$$ Therefore, $$g(6) = g(3) - 1.5 = 3.5 - 1.5 = 2.0$$ 5. **Final answers:** $$g(0) = 0, \quad g(1) = 1, \quad g(2) = 2.5, \quad g(3) = 3.5, \quad g(6) = 2.0$$