1. The problem asks to evaluate $g(x) = \int_0^x f(t) \, dt$ at $x=0,1,2,3,6$ where $f(t)$ is a piecewise linear function given by the graph. 2. Recall that $g(x)$ is the area under the curve $f(t)$ from $0$ to $x$. Positive areas add to $g(x)$ and negative areas subtract. 3. Evaluate $g(0)$: $$g(0) = \int_0^0 f(t) \, dt = 0$$ 4. Evaluate $g(1)$: From $t=0$ to $1$, $f(t)=1$ (constant), so area is a rectangle of height 1 and width 1: $$g(1) = 1 \times 1 = 1$$ 5. Evaluate $g(2)$: From $t=1$ to $2$, $f(t)$ increases linearly from 1 to 2. Area is a trapezoid with bases 1 and 2, height 1: $$\text{Area} = \frac{1+2}{2} \times 1 = \frac{3}{2} = 1.5$$ Add to previous area: $$g(2) = g(1) + 1.5 = 1 + 1.5 = 2.5$$ 6. Evaluate $g(3)$: From $t=2$ to $3$, $f(t)$ decreases from 2 to approximately 0 (since at 3.5 it is -1, at 3 it is about 0.5, but problem states steep decrease through 0 to -1 at 3.5, so approximate linearly from 2 at 2 to 0 at 3): Area is a trapezoid with bases 2 and 0, height 1: $$\text{Area} = \frac{2+0}{2} \times 1 = 1$$ Add to previous area: $$g(3) = g(2) + 1 = 2.5 + 1 = 3.5$$ 7. Evaluate $g(6)$: From $t=3$ to $3.5$, $f(t)$ decreases from 0 to -1: Area is a triangle below the axis: $$\text{Area} = \frac{1}{2} \times 0.5 \times 1 = 0.25$$ (negative area) From $3.5$ to $5$, $f(t) = -1$ constant: $$\text{Area} = -1 \times 1.5 = -1.5$$ From $5$ to $6$, $f(t)$ increases linearly from -1 to 0: Area is a triangle below axis: $$\text{Area} = \frac{1}{2} \times 1 \times 1 = 0.5$$ (negative area) Sum negative areas: $$-0.25 - 1.5 - 0.5 = -2.25$$ Add to $g(3)$: $$g(6) = 3.5 - 2.25 = 1.25$$ Final answers: $$g(0) = 0, \quad g(1) = 1, \quad g(2) = 2.5, \quad g(3) = 3.5, \quad g(6) = 1.25$$