For each ositive integer n, define a function `f _(n) on [0,1]` as follows:
`f _(n((x)={{:(0, if , x =0),(sin ""(pi)/(2n), if , 0 lt x le 1/n),( sin ""(2pi)/(2n) , if , 1/n lt x le 2/n), (sin ""(3pi)/(2pi), if, 2/n lt x le 3/n), (sin "'(npi)/(2pi) , if, (n-1)/(n) lt x le 1):}` Then the value of `lim _(x to oo)int _(0)^(1) f_(n) (x) dx ` is:

Options

A. `pi`
B. `pi/2`
C. `1/pi`
D. `2/pi`

Answer

To solve the problem, we need to evaluate the limit of the integral of the function \( f_n(x) \) as \( n \) approaches infinity. The function \( f_n(x) \) is defined piecewise on the interval \([0, 1]\) as follows:

\[ f_n(x) = \begin{cases} 0 & \text{if } x = 0 \\ \sin\left(\frac{\pi}{2n}\right) & \text{if } 0 < x \leq \frac{1}{n} \\ \sin\left(\frac{2\pi}{2n}\right) & \text{if } \frac{1}{n} < x \leq \frac{2}{n} \\ \sin\left(\frac{3\pi}{2n}\right) & \text{if } \frac{2}{n} < x \leq \frac{3}{n} \\ \vdots & \vdots \\ \sin\left(\frac{n\pi}{2n}\right) & \text{if } \frac{n-1}{n} < x \leq 1 \end{cases} \]

### Step 1: Set up the integral
We need to evaluate the limit:
\[ \lim_{n \to \infty} \int_0^1 f_n(x) \, dx \]
We can break this integral into segments based on the definition of \( f_n(x) \):
\[ \int_0^1 f_n(x) \, dx = \sum_{k=1}^{n} \int_{\frac{k-1}{n}}^{\frac{k}{n}} f_n(x) \, dx \]

### Step 2: Evaluate each segment
For each segment \( \left[\frac{k-1}{n}, \frac{k}{n}\right] \), the function \( f_n(x) \) is constant and equals \( \sin\left(\frac{k\pi}{2n}\right) \):
\[ \int_{\frac{k-1}{n}}^{\frac{k}{n}} f_n(x) \, dx = \sin\left(\frac{k\pi}{2n}\right) \cdot \left(\frac{k}{n} - \frac{k-1}{n}\right) = \sin\left(\frac{k\pi}{2n}\right) \cdot \frac{1}{n} \]

### Step 3: Combine the segments
Thus, we can rewrite the integral as:
\[ \int_0^1 f_n(x) \, dx = \sum_{k=1}^{n} \sin\left(\frac{k\pi}{2n}\right) \cdot \frac{1}{n} \]

### Step 4: Recognize the Riemann sum
As \( n \to \infty \), this sum approaches the integral:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \sin\left(\frac{k\pi}{2n}\right) \cdot \frac{1}{n} = \int_0^{\frac{\pi}{2}} \sin(x) \, dx \]

### Step 5: Evaluate the integral
The integral \( \int_0^{\frac{\pi}{2}} \sin(x) \, dx \) can be computed as:
\[ -\cos(x) \bigg|_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

### Step 6: Final limit
Thus, we have:
\[ \lim_{n \to \infty} \int_0^1 f_n(x) \, dx = 1 \]

### Conclusion
The final value of the limit is:
\[ \boxed{1} \]