To evaluate the definite integral $\int_0^1 -3f(x) \, dx$, we use geometric reasoning based on the graph of $f(x)$.

Steps:

1. Identify the segment of interest:

   • From $x = 0$ to $x = 1$, $f(x)$ is a linear segment starting at $(0, -1)$ and ending at $(1, 0)$.

2. Compute the area under the curve:

   • The region forms a right triangle with:

     • Base = 1 (from $x = 0$ to $x = 1$),
     • Height = 1 (from $y = -1$ to $y = 0$).

   • The area of the triangle is: $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}.$

3. Adjust for the negative sign in $-3f(x)$:

   • Since $f(x)$ is below the $x$-axis on this interval, the signed area is $-\frac{1}{2}$.
   • Multiplying by $-3$, the integral becomes: $\int_0^1 -3f(x) \, dx = -3 \cdot \left(-\frac{1}{2}\right) = \frac{3}{2}.$

Final Answer:

$\boxed{\frac{3}{2}}$

Would you like me to evaluate additional integrals or clarify the process further?

Here are 5 related questions you might explore:

1. What is the value of $\int_0^1 f(x) \, dx$ without the scaling factor $-3$?
2. How does the integral $\int_0^3 -3f(x) \, dx$ compare to this result?
3. How do the signs of the definite integrals change if $f(x)$ is above or below the $x$-axis?
4. Can we compute definite integrals geometrically for any piecewise linear function?
5. How would the integral change if the scaling factor were $-2$ instead of $-3$?

Tip: Always pay attention to the regions above and below the $x$-axis when calculating definite integrals geometrically.