The problem involves evaluating definite integrals of a piecewise-defined function $g(x)$, represented by a graph. Here's the breakdown of the situation and the steps to calculate the remaining part (c):

Observations

1. The graph is composed of:
   • Two straight lines between $x = 0$ to $x = 10$ and $x = 10$ to $x = 20$.
   • A semicircle from $x = 20$ to $x = 35$.

Given:

• $\int_0^{10} g(x) \, dx = 100$
• $\int_{10}^{30} g(x) \, dx = -50\pi$

For part (c), $\int_0^{35} g(x) \, dx$, the total integral can be broken down as: $\int_0^{35} g(x) \, dx = \int_0^{10} g(x) \, dx + \int_{10}^{30} g(x) \, dx + \int_{30}^{35} g(x) \, dx$

Steps:

1. Use the first segment:
   $\int_0^{10} g(x) \, dx = 100$ (already given).

2. Semicircle calculation (middle segment):
   $\int_{10}^{30} g(x) \, dx = -50\pi$ (already given).

3. Calculate the last segment ($x = 30$ to $x = 35$): The function appears to return to 0 along a straight line. The area under the curve for this part is triangular:

   • Height = 10 (from $y = 0$ to $y = 10$)
   • Base = 5 (from $x = 30$ to $x = 35$) Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 10 = 25$.

4. Add all components:
   $\int_0^{35} g(x) \, dx = 100 + (-50\pi) + 25$

Final Answer:

$\int_0^{35} g(x) \, dx = 125 - 50\pi$

Would you like detailed explanations of specific parts or have additional questions?

Related Questions:

1. How do you compute definite integrals for piecewise functions?
2. Why does the semicircle contribute a negative value to the integral?
3. What is the geometric interpretation of definite integrals in real-world contexts?
4. How can areas of basic shapes (triangles, circles) help in definite integration?
5. What happens if the function crosses the x-axis multiple times?

Tip:

Always separate a complex graph into manageable geometric or functional components before integrating.