Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $f(x) = \int_a^x g(t) dt$, then $f'(x) = g(x)$.
• First Derivative Test: If $f'(x)$ changes sign from positive to negative at $x=c$, then $f(x)$ has a local maximum at $x=c$. If $f'(x)$ changes sign from negative to positive at $x=c$, then $f(x)$ has a local minimum at $x=c$.

Step-by-Step Solution

Step 1: Find the First Derivative, $f'(x)$

We are given the function $f(x) = \int_0^x \sqrt{t} \sin t \, dt$. To find the critical points, we need to find the first derivative $f'(x)$. Using the Fundamental Theorem of Calculus (Part 1), we can directly differentiate the integral: $f'(x) = \sqrt{x} \sin x$ This is because the derivative of the integral from a constant to $x$ of a function is simply the function evaluated at $x$.

Step 2: Find the Critical Points

Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. We have $f'(x) = \sqrt{x} \sin x$. Since $x \in (0, \frac{5\pi}{2})$, $\sqrt{x}$ is always defined and positive. Thus, we only need to find where $\sin x = 0$. $\sin x = 0$ In the interval $(0, \frac{5\pi}{2})$, the solutions are: $x = \pi, 2\pi$ Note that $x=0$ is not included in the interval.

Step 3: Analyze the Sign of $f'(x)$ using the First Derivative Test

We need to determine the sign of $f'(x) = \sqrt{x} \sin x$ in the intervals determined by the critical points: $(0, \pi)$, $(\pi, 2\pi)$, and $(2\pi, \frac{5\pi}{2})$. Since $\sqrt{x}$ is always positive in the interval $(0, \frac{5\pi}{2})$, the sign of $f'(x)$ is determined by the sign of $\sin x$.

• Interval $(0, \pi)$: $\sin x > 0$, so $f'(x) > 0$.
• Interval $(\pi, 2\pi)$: $\sin x < 0$, so $f'(x) < 0$.
• Interval $(2\pi, \frac{5\pi}{2})$: $\sin x > 0$, so $f'(x) > 0$.

At $x = \pi$, $f'(x)$ changes from positive to negative, so $f(x)$ has a local maximum at $x = \pi$. At $x = 2\pi$, $f'(x)$ changes from negative to positive, so $f(x)$ has a local minimum at $x = 2\pi$.

Therefore, $f$ has a local maximum at $\pi$ and a local minimum at $2\pi$.

Common Mistakes & Tips

• Remember to consider the given domain when finding critical points. Solutions outside the domain are not relevant.
• Be careful with the sign of $\sin x$ in different quadrants to correctly apply the first derivative test.
• The Fundamental Theorem of Calculus is crucial for differentiating integrals with variable limits.

Summary

We found the derivative of the function $f(x)$ using the Fundamental Theorem of Calculus. Then, we found the critical points by setting the derivative equal to zero. Finally, we used the first derivative test to determine that $f(x)$ has a local maximum at $x=\pi$ and a local minimum at $x=2\pi$.

Final Answer

The final answer is \boxed{local maximum at \pi and local minimum at 2\pi}, which corresponds to option (C).