This problem asks for the behavior of $g(x) = \int_0^x f(t) \, dt$, given the graph of $f(x)$, for the interval $0 < x < 3$.

Key Steps:

1. Relation Between $f(x)$ and $g(x)$:

   • The value of $g(x)$ is the definite integral of $f(t)$ from $0$ to $x$.
   • $g'(x) = f(x)$: The derivative of $g(x)$ is equal to $f(x)$.
   • $g''(x) = f'(x)$: The second derivative of $g(x)$ is equal to the slope of $f(x)$.

2. Analyze the Graph:

   • For $0 < x < 3$, $f(x)$ is positive (above the x-axis). Hence, $g(x)$ is increasing.
   • The graph of $f(x)$ is decreasing (the slope of $f(x)$ is negative). This means $g'(x) = f(x)$ is decreasing, and $g(x)$ is concave down.

Conclusion:

For $0 < x < 3$, $g(x)$ is increasing and concave down.

Correct Answer:

D) Increasing and concave down.

Would you like me to explain any part of this in more detail or explore related concepts?

5 Related Questions:

1. How does the Fundamental Theorem of Calculus connect $f(x)$ and $g(x)$?
2. What does the concavity of a function tell us about its behavior?
3. How can we determine intervals of increase or decrease for a function using derivatives?
4. Why does the sign of $f(x)$ determine whether $g(x)$ is increasing or decreasing?
5. How does the second derivative test relate to concavity?

Tip:

Always consider the behavior of $f(x)$ when analyzing $g(x) = \int f(t) \, dt$. Positive $f(x)$ means $g(x)$ increases, and the slope of $f(x)$ determines concavity.