To solve the problem, we need to calculate the definite integrals by interpreting the graph of $g(x)$. The integral of $g(x)$ represents the net area under the curve, taking areas above the $x$-axis as positive and areas below as negative.

(a) $\int_0^5 g(x) \, dx$

1. Break the interval $[0, 5]$ into segments where $g(x)$ is linear or constant.

   • $[0, 1]$: Rectangle with height $-1$ and width $1$. Area = $-1 \times 1 = -1$.
   • $[1, 2]$: Triangle below the $x$-axis with height $-2$ and base $1$. Area = $\frac{1}{2} \times 1 \times -2 = -1$.
   • $[2, 4]$: Triangle below the $x$-axis with height $-3$ and base $2$. Area = $\frac{1}{2} \times 2 \times -3 = -3$.
   • $[4, 5]$: Triangle above the $x$-axis with height $1$ and base $1$. Area = $\frac{1}{2} \times 1 \times 1 = 0.5$.

   Total Area = $-1 - 1 - 3 + 0.5 = -4.5$.

   Answer: $-4.5$

(b) $\int_5^6 g(x) \, dx$

1. $[5, 6]$: Rectangle above the $x$-axis with height $1$ and width $1$. Area = $1 \times 1 = 1$.

   Answer: $1$ (this is already correct in the image).

(c) $\int_0^6 g(x) \, dx$

1. Combine the results of $\int_0^5 g(x) \, dx$ and $\int_5^6 g(x) \, dx$.

   • From (a): $\int_0^5 g(x) \, dx = -4.5$.
   • From (b): $\int_5^6 g(x) \, dx = 1$.

   Total = $-4.5 + 1 = -3.5$.

   Answer: $-3.5$

Final Answers:

• (a) $-4.5$
• (b) $1$
• (c) $-3.5$

Would you like detailed explanations for how the areas were calculated or any clarification on integral interpretation?

Related Questions:

1. What does the definite integral represent geometrically in a graph?
2. How do you calculate the area of a triangle for definite integrals?
3. Why are areas below the $x$-axis considered negative?
4. How would the integral change if $g(x)$ were shifted upwards by 2 units?
5. Can you compute integrals of more complex piecewise functions graphically?

Tip:

Always break the integration range into manageable pieces where the function behavior is consistent (e.g., linear or constant).