To evaluate the integral $\int_0^4 g(t) \, dt$, we need to compute the area under the curve $y = g(t)$ from $t = 0$ to $t = 4$. This involves summing up the areas of the triangular and trapezoidal regions, considering the signs of the areas.

Steps to Solve:

1. Divide the graph into geometric shapes:

   • From $t = 0$ to $t = 1$: A triangle below the $t$-axis.
   • From $t = 1$ to $t = 3$: A larger triangle above the $t$-axis.
   • From $t = 3$ to $t = 4$: A smaller triangle below the $t$-axis.

2. Calculate the areas:

   • Triangle from $t = 0$ to $t = 1$:

     • Base = $1 - 0 = 1$
     • Height = $-1$ (below the axis, so negative)
     • Area = $\frac{1}{2} \times 1 \times (-1) = -\frac{1}{2}$

   • Triangle from $t = 1$ to $t = 3$:

     • Base = $3 - 1 = 2$
     • Height = $2$ (above the axis, so positive)
     • Area = $\frac{1}{2} \times 2 \times 2 = 2$

   • Triangle from $t = 3$ to $t = 4$:

     • Base = $4 - 3 = 1$
     • Height = $-2$ (below the axis, so negative)
     • Area = $\frac{1}{2} \times 1 \times (-2) = -1$

3. Add the areas together: $\int_0^4 g(t) \, dt = -\frac{1}{2} + 2 - 1 = \frac{1}{2}$

Final Answer:

$\int_0^4 g(t) \, dt = \frac{1}{2}$

Let me know if you have further questions or need clarification!

Related Questions:

1. What is the significance of negative and positive areas in integration?
2. How can you calculate the integral for non-linear functions?
3. What happens if we integrate beyond $t = 4$ in this case?
4. How can this integral be verified using numerical approximation methods?
5. What is the relationship between definite integrals and total displacement in physics?

Tip: Always identify the geometric shapes under the curve when evaluating integrals graphically.