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 $t=0$ to $t=x$. 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 $t=1$, $f(t) = 1$ (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 $t=2$, $f(t)$ increases linearly from 1 to 3. Area from 1 to 2 is a trapezoid with bases 1 and 3 and height 1: $$\text{Area} = \frac{1+3}{2} \times 1 = 2$$ Add to previous area: $$g(2) = g(1) + 2 = 1 + 2 = 3$$ 6. Evaluate $g(3)$: From $t=2$ to $t=3$, $f(t)$ decreases linearly from 3 to 0. Area is a triangle with base 1 and height 3: $$\text{Area} = \frac{1 \times 3}{2} = 1.5$$ Add to previous area: $$g(3) = g(2) + 1.5 = 3 + 1.5 = 4.5$$ 7. Evaluate $g(6)$: From $t=3$ to $t=5$, $f(t)$ decreases linearly from 0 to -2. Area is a trapezoid below the axis with bases 0 and -2 and width 2: $$\text{Area} = \frac{0 + (-2)}{2} \times 2 = -2$$ From $t=5$ to $t=6$, $f(t) = -2$ constant: $$\text{Area} = -2 \times 1 = -2$$ Total area from 3 to 6: $$-2 + (-2) = -4$$ Add to previous area: $$g(6) = g(3) + (-4) = 4.5 - 4 = 0.5$$ Final answers: $$g(0) = 0, \quad g(1) = 1, \quad g(2) = 3, \quad g(3) = 4.5, \quad g(6) = 0.5$$