To solve this problem, we need to evaluate the integral $g(-2) = \int_{-6}^{-2} f(t) \, dt$ by analyzing the graph of $f(t)$.

Step-by-Step Solution

1. Identify the Function: $g(x)$ is defined as an integral of $f(t)$ from $t = -6$ to $t = x$, specifically here for $g(-2)$.

2. Evaluate $g(-2)$ as the Area: The integral $\int_{-6}^{-2} f(t) \, dt$ represents the area under $f(t)$ from $t = -6$ to $t = -2$. This area may be positive or negative based on the position of $f(t)$ relative to the $t$-axis.

3. Analyze the Graph:

   • From $t = -6$ to $t = -4$, $f(t)$ is above the $t$-axis, creating a triangular region.
   • From $t = -4$ to $t = -2$, $f(t)$ is below the $t$-axis, creating another triangular region.

4. Calculate the Areas:

   • First Triangle (from $t = -6$ to $t = -4$):

     • Base = 2 units (from $-6$ to $-4$)
     • Height = 2
     • Area = $\frac{1}{2} \times 2 \times 2 = 2$

   • Second Triangle (from $t = -4$ to $t = -2$):

     • Base = 2 units (from $-4$ to $-2$)
     • Height = 2 (below the $t$-axis, so it’s negative)
     • Area = $\frac{1}{2} \times 2 \times 2 = 2$, but since it’s below the axis, this area is $-2$.

5. Add the Areas:

   • Total area from $t = -6$ to $t = -2$: $2 + (-2) = 0$.

Conclusion

The value of $g(-2) = 0$.

Answer

The correct answer is not listed in the options, but the calculation shows: $g(-2) = 0$

Would you like more details or have any questions?

Related Questions

1. How do you evaluate definite integrals with regions above and below the $x$-axis?
2. How does the integral's value change if the graph has multiple peaks and valleys?
3. What is the significance of the integral’s limits in determining the area?
4. Can integrals yield negative values, and if so, what does it signify?
5. How do we handle integrals when a function is partially positive and partially negative?

Tip

When finding areas under curves, always consider the sign based on the curve's position relative to the $x$-axis.