To find the correct values for $g(8)$, $g(12)$, and $g(24)$, let us evaluate the integrals step by step, using the given function $f(t)$ from the second image. Since $g(x) = \int_0^x f(t) \, dt$, the value of $g(x)$ corresponds to the area under the curve of $f(t)$ from $t = 0$ to $t = x$.

Step-by-step evaluation

1. $g(8)$

• From $t = 0$ to $t = 4$: The area is a rectangle with a base of 4 and a height of 4. $\text{Area} = 4 \cdot 4 = 16.$
• From $t = 4$ to $t = 8$: The area is a triangle with a base of 4 and a height of 8. $\text{Area} = \frac{1}{2} \cdot 4 \cdot 8 = 16.$
• Adding these together: $g(8) = 16 + 16 = 32.$

2. $g(12)$

• From $t = 0$ to $t = 8$: Already computed as $g(8) = 32$.
• From $t = 8$ to $t = 12$: The area is a triangle with a base of 4 and a height of $-4$ (area below the x-axis is negative). $\text{Area} = \frac{1}{2} \cdot 4 \cdot (-4) = -8.$
• Adding these: $g(12) = 32 + (-8) = 24.$

3. $g(24)$

• From $t = 0$ to $t = 8$: $g(8) = 32$.
• From $t = 8$ to $t = 20$:
  • From $t = 8$ to $t = 12$: Area = $-8$ (computed above).
  • From $t = 12$ to $t = 20$: This is a trapezoid with bases of $-4$ and $-8$, and a height of 8. $\text{Area} = \frac{1}{2} \cdot 8 \cdot (-4 + -8) = \frac{1}{2} \cdot 8 \cdot (-12) = -48.$ $\text{Total area from } t = 8 \text{ to } t = 20 = -8 + (-48) = -56.$
• From $t = 20$ to $t = 24$: The area is a triangle with a base of 4 and a height of 4. $\text{Area} = \frac{1}{2} \cdot 4 \cdot 4 = 8.$
• Adding these: $g(24) = 32 + (-56) + 8 = -16.$

Final Correct Values:

1. $g(8) = 32$ (Correct)
2. $g(12) = 24$ (Corrected)
3. $g(24) = -16$ (Corrected)

Would you like further details on any step? Here are five related questions to expand on:

1. How do we handle integrals of piecewise functions geometrically?
2. Why is the area below the x-axis subtracted in integrals?
3. Can you explain how the trapezoid area formula works in this context?
4. What is the significance of $g(x)$ as the integral of $f(t)$?
5. How could these integrals be solved analytically without geometry?

Tip: Always ensure that areas below the x-axis are treated as negative to maintain accuracy in definite integrals!