To solve this problem, we interpret $F'(x) = f(x)$, and $F(0) = 0$, to find $F(b)$ at the given points $b = 2, 4, 7,$ and $10$. The function $F(x)$ is the integral of $f(x)$, so:

$F(b) = \int_0^b f(x) \, dx$

The graph of $f(x)$ is a piecewise linear function, and the area under the curve is calculated geometrically (e.g., using areas of triangles and rectangles). Let’s calculate $F(b)$:

Step 1: Break the integral into parts

1. For $b = 2$, we compute the area under $f(x)$ from $x = 0$ to $x = 2$.
2. Similarly for $b = 4, 7, 10$, calculate areas up to these points.

Step 2: Compute the areas geometrically

1. From $x = 0$ to $x = 2$: The graph shows a triangle with base 2 and height 2. $\text{Area} = \frac{1}{2} \times 2 \times 2 = 2$ Therefore, $F(2) = 2$.

2. From $x = 0$ to $x = 4$: Add the rectangle from $x = 2$ to $x = 4$ (base 2, height 2). $\text{Area of rectangle} = 2 \times 2 = 4$ Total $F(4) = 2 + 4 = 6$.

3. From $x = 0$ to $x = 7$: Include the triangle from $x = 4$ to $x = 7$ (base 3, height 3). $\text{Area of triangle} = \frac{1}{2} \times 3 \times 3 = 4.5$ Total $F(7) = 6 + 4.5 = 10.5$.

4. From $x = 0$ to $x = 10$: Subtract the area below the x-axis (from $x = 7$ to $x = 10$, base 3, height -3). $\text{Negative area of triangle} = -\frac{1}{2} \times 3 \times 3 = -4.5$ Total $F(10) = 10.5 - 4.5 = 6$.

Final Table: