To solve the problem, let’s analyze the function $F(x) = \int_0^x f(t) \, dt$ for the given values of $x$ based on the graph provided for $f(t)$.

Approach:

Since $F(x)$ represents the area under the curve of $f(t)$ from $t = 0$ to $t = x$, we can find each value by calculating the area under the graph up to the given $x$-values.

Step-by-Step Solution:

1. Determine $F(1)$:

   • From $t = 0$ to $t = 1$, the function $f(t)$ forms a triangle with base $1$ and height $1$.
   • Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = 0.5$.
   • Therefore, $F(1) = 0.5$.

2. Determine $F(4)$:

   • From $t = 0$ to $t = 4$, there are two regions:
     • From $t = 0$ to $t = 2$, a triangle with base $2$ and height $2$: $\text{Area} = \frac{1}{2} \times 2 \times 2 = 2$.
     • From $t = 2$ to $t = 4$, another triangle but now decreasing, with base $2$ and height $2$: $\text{Area} = \frac{1}{2} \times 2 \times 2 = 2$.
   • The total area from $t = 0$ to $t = 4$ is $2 + 2 = 4$.
   • Thus, $F(4) = 4$.

3. Determine $F(6)$:

   • From $t = 4$ to $t = 6$, there is a triangle below the x-axis (negative area) with base $2$ and height $2$: $\text{Area} = -\frac{1}{2} \times 2 \times 2 = -2$.
   • Adding this to the previous area $F(4)$, we get $F(6) = 4 + (-2) = 2$.

4. Determine $F(7)$:

   • From $t = 6$ to $t = 7$, the function $f(t) = 0$, so there is no additional area.
   • Therefore, $F(7) = F(6) = 2$.

Final Answers:

• a. $F(1) = 0.5$
• b. $F(4) = 4$
• c. $F(6) = 2$
• d. $F(7) = 2$

Would you like a detailed breakdown of each step or have any questions?

Related Questions:

1. How would $F(x)$ change if the function $f(t)$ was scaled by a factor of 2?
2. What if $f(t)$ had a constant positive offset added? How would $F(x)$ be affected?
3. How do we interpret the behavior of $F(x)$ if $f(t)$ crosses the x-axis multiple times?
4. Can you generalize the effect of negative areas on the integral $F(x)$?
5. How would the values change if the area from $t = 6$ to $t = 7$ was non-zero?

Tip:

When evaluating integrals of piecewise functions, breaking down the areas into simple shapes like triangles or rectangles can simplify calculations significantly.