`Lt_(n rarr oo) Sigma_(r=1)^(n-1) (pi)/(n) sin ((r pi)/(n))`=

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the limit of the summation given in the question, we will convert the summation into an integral. Let's break down the solution step by step.

### Step 1: Rewrite the Summation
We start with the expression:
\[ \lim_{n \to \infty} \sum_{r=1}^{n-1} \frac{\pi}{n} \sin\left(\frac{r \pi}{n}\right) \]
This summation can be interpreted as a Riemann sum, which approximates the integral of a function over an interval.

### Step 2: Identify the Limits of Integration
As \( n \to \infty \):
- The lower limit of \( r/n \) as \( r \) approaches 1 becomes \( \frac{1}{n} \to 0 \).
- The upper limit of \( r/n \) as \( r \) approaches \( n-1 \) becomes \( \frac{n-1}{n} \to 1 \).

Thus, we can rewrite the summation as:
\[ \lim_{n \to \infty} \sum_{r=1}^{n-1} \frac{\pi}{n} \sin\left(\frac{r \pi}{n}\right) \approx \int_0^1 \pi \sin(\pi x) \, dx \]
where we let \( x = \frac{r}{n} \).

### Step 3: Set Up the Integral
Now we can express the limit of the summation as an integral:
\[ \int_0^1 \pi \sin(\pi x) \, dx \]

### Step 4: Evaluate the Integral
To evaluate the integral, we can use the following formula:
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C \]
In our case, \( k = \pi \):
\[ \int \sin(\pi x) \, dx = -\frac{1}{\pi} \cos(\pi x) \]
Thus, we have:
\[ \int_0^1 \pi \sin(\pi x) \, dx = \pi \left[-\frac{1}{\pi} \cos(\pi x)\right]_0^1 \]

### Step 5: Apply the Limits
Now we evaluate the limits:
\[ = \pi \left[-\frac{1}{\pi} \cos(\pi) + \frac{1}{\pi} \cos(0)\right] \]
Calculating the cosine values:
- \( \cos(\pi) = -1 \)
- \( \cos(0) = 1 \)

Substituting these values in:
\[ = \pi \left[-\frac{1}{\pi}(-1) + \frac{1}{\pi}(1)\right] = \pi \left[\frac{1}{\pi} + \frac{1}{\pi}\right] = \pi \cdot \frac{2}{\pi} = 2 \]

### Final Result
Thus, the limit of the summation is:
\[ \boxed{2} \]