To solve the given integral:

$\int_0^{1/2} \sin(2\pi x) \, dx$

we can proceed as follows:

Step 1: Set up the Integral

The integral is:

$\int_0^{1/2} \sin(2\pi x) \, dx$

Step 2: Integrate $\sin(2\pi x)$

The antiderivative of $\sin(2\pi x)$ with respect to $x$ is:

$-\frac{1}{2\pi} \cos(2\pi x)$

So, we have:

$\int \sin(2\pi x) \, dx = -\frac{1}{2\pi} \cos(2\pi x)$

Step 3: Apply the Limits

Now we evaluate this from $x = 0$ to $x = \frac{1}{2}$:

$\int_0^{1/2} \sin(2\pi x) \, dx = \left[ -\frac{1}{2\pi} \cos(2\pi x) \right]_0^{1/2}$

Substitute $x = \frac{1}{2}$ and $x = 0$:

1. For $x = \frac{1}{2}$: $-\frac{1}{2\pi} \cos\left(2\pi \cdot \frac{1}{2}\right) = -\frac{1}{2\pi} \cos(\pi) = -\frac{1}{2\pi} \cdot (-1) = \frac{1}{2\pi}$

2. For $x = 0$: $-\frac{1}{2\pi} \cos(2\pi \cdot 0) = -\frac{1}{2\pi} \cos(0) = -\frac{1}{2\pi} \cdot 1 = -\frac{1}{2\pi}$

Step 4: Calculate the Result

Now, we subtract the values:

$\frac{1}{2\pi} - \left(-\frac{1}{2\pi}\right) = \frac{1}{2\pi} + \frac{1}{2\pi} = \frac{1}{\pi}$

Final Answer:

The correct answer is:

$\boxed{\frac{1}{\pi}}$

So, the correct option is $\text{d. } \frac{1}{\pi}$.

Would you like more details on this solution, or do you have any questions?

Related Questions:

1. How do we find the antiderivative of trigonometric functions like $\sin$ and $\cos$?
2. Why does $\cos(\pi)$ equal $-1$?
3. How does the Fundamental Theorem of Calculus apply to definite integrals?
4. What are the basic properties of definite integrals over symmetric intervals?
5. How would the result change if the limits of integration were different?

Tip: Remember that trigonometric integrals often involve using the antiderivatives of $\sin$ and $\cos$, which are $-\cos(x)$ and $\sin(x)$, respectively.