The problem is asking for the value of the definite integral:

$\int_0^\pi \sin(x) \, dx$

Let's solve it step by step.

Step 1: Find the Antiderivative

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

Step 2: Apply the Fundamental Theorem of Calculus

We need to evaluate the antiderivative at the bounds $x = 0$ and $x = \pi$:

$\int_0^\pi \sin(x) \, dx = \left[-\cos(x)\right]_0^\pi$

Step 3: Calculate the Values at the Bounds

Evaluate $-\cos(x)$ at $x = \pi$ and $x = 0$:

$-\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 - (-1) = 1 + 1 = 2$

Final Answer

The value of the integral is:

$\boxed{2}$

Would you like to know more details or have any other questions?

Here are some related questions you might find interesting:

1. What is the significance of the Fundamental Theorem of Calculus?
2. How would you compute the integral $\int_0^\pi \cos(x) \, dx$?
3. Can you explain the relationship between integrals and areas under curves?
4. How does the sine function behave over the interval $[0, \pi]$?
5. What are some real-world applications of definite integrals?

Tip: When dealing with trigonometric integrals, knowing the basic antiderivatives is crucial for quick and accurate computation.