Write integral from 0 to () pi sin x dx in the form

Write $\int_0^\pi \sin x \, dx$ in the form of an infinite series (Riemann sum limit) and find the limit of that series.

We will express the integral as a Riemann sum and then find its limit.

Step 1: Partition the interval

Divide the interval $[0,\pi]$ into $n$ equal subintervals. The width of each subinterval is:

\[\Delta x = \frac{\pi - 0}{n} = \frac{\pi}{n}\]

The partition points are:

\[x_i = 0 + i \cdot \Delta x = \frac{i\pi}{n}, \quad i = 0, 1, 2, \ldots, n\]

Step 2: Choose the sample points

We use the right endpoint of each subinterval:

\[\xi_i = x_i = \frac{i\pi}{n}\]

Step 3: Form the Riemann sum

The Riemann sum is:

\[S_n = \sum_{i=1}^{n} f(\xi_i) \Delta x = \sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right) \cdot \frac{\pi}{n} = \frac{\pi}{n} \sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right)\]

Step 4: Express as an infinite series

As $n \to \infty$, the Riemann sum becomes:

\[\int_0^\pi \sin x \, dx = \lim_{n \to \infty} \frac{\pi}{n} \sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right)\]

This can be written as:

\[\int_0^\pi \sin x \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right) \cdot \frac{\pi}{n}\]

Step 5: Simplify the sum using a trigonometric identity

To evaluate $\sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right)$, we use the formula:

\[\sum_{i=1}^{n} \sin(i\theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right) \sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\]

With $\theta = \frac{\pi}{n}$:

\[\sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right) = \frac{\sin\left(\frac{(n+1)\pi}{2n}\right) \sin\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2n}\right)} = \frac{\sin\left(\frac{(n+1)\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}\]

Step 6: Calculate the Riemann sum

\[S_n = \frac{\pi}{n} \cdot \frac{\sin\left(\frac{(n+1)\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}\]

Step 7: Take the limit as $n \to \infty$

As $n \to \infty$:

\[\frac{(n+1)\pi}{2n} = \frac{\pi}{2}\left(1 + \frac{1}{n}\right) \to \frac{\pi}{2}\]

So $\sin\left(\frac{(n+1)\pi}{2n}\right) \to \sin\left(\frac{\pi}{2}\right) = 1$

Also, $\lim_{n \to \infty} \frac{\sin\left(\frac{\pi}{2n}\right)}{\frac{\pi}{2n}} = 1$, which means:

\[\lim_{n \to \infty} \frac{\frac{\pi}{2n}}{\sin\left(\frac{\pi}{2n}\right)} = 1\]

Therefore:

\[\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{\pi}{n} \cdot \frac{1}{\sin\left(\frac{\pi}{2n}\right)} = \lim_{n \to \infty} \frac{\pi}{n} \cdot \frac{\frac{\pi}{2n}}{\sin\left(\frac{\pi}{2n}\right)} \cdot \frac{2n}{\pi} = \frac{\pi}{2} \cdot 2 = 2\]

Verification using antiderivative:

\[\int_0^\pi \sin x \, dx = \left[-\cos x\right]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) + 1 = 2\]

Final answer:

\[\boxed{\int_0^\pi \sin x \, dx = \lim_{n \to \infty} \frac{\pi}{n} \sum_{i=1}^{n} \sin\left(\frac{i\pi}{n}\right) = 2}\]