To solve this problem, we need to determine the antiderivative $F(x)$ of $f(x) = 2 - \cos(x)$ that satisfies the condition $F\left(\frac{\pi}{2}\right) = 2$, and then find $F(\pi)$.

Step-by-Step Solution

1. Find the Antiderivative $F(x)$: Since $f(x) = 2 - \cos(x)$, we can integrate each term individually: $\int (2 - \cos(x)) \, dx = \int 2 \, dx - \int \cos(x) \, dx$

   • The antiderivative of $2$ is $2x$.
   • The antiderivative of $-\cos(x)$ is $-\sin(x)$.

   Therefore, we get: $F(x) = 2x - \sin(x) + C$ where $C$ is the constant of integration.

2. Use the Condition $F\left(\frac{\pi}{2}\right) = 2$ to Solve for $C$: Substitute $x = \frac{\pi}{2}$ and $F\left(\frac{\pi}{2}\right) = 2$: $2 = F\left(\frac{\pi}{2}\right) = 2 \cdot \frac{\pi}{2} - \sin\left(\frac{\pi}{2}\right) + C$ Simplify: $2 = \pi - 1 + C$ Solving for $C$: $C = 3 - \pi$

3. Write the Complete Expression for $F(x)$: Now we have: $F(x) = 2x - \sin(x) + (3 - \pi)$

4. Find $F(\pi)$: Substitute $x = \pi$: $F(\pi) = 2 \cdot \pi - \sin(\pi) + (3 - \pi)$ Since $\sin(\pi) = 0$, this simplifies to: $F(\pi) = 2\pi + 3 - \pi = \pi + 3$

Conclusion

The value of $F(\pi)$ is $\pi + 3$.

Answer

The correct answer is: $\boxed{3 + \pi}$

Would you like further details on any part of this solution?

Here are some related questions for further practice:

1. How do you find the antiderivative of trigonometric functions like $\cos(x)$ and $\sin(x)$?
2. How can we verify that $F(x)$ is indeed the antiderivative of $f(x)$ by differentiating it?
3. What happens if the initial condition $F\left(\frac{\pi}{2}\right) = 2$ is different? How would that affect $C$?
4. How would the solution change if $f(x)$ had a different form, such as $f(x) = 2 + \cos(x)$?
5. How can definite integrals be used to find $F(\pi) - F\left(\frac{\pi}{2}\right)$ directly?

Tip: Always check your antiderivative by differentiating it to ensure it matches the original function.