The problem asks to evaluate the definite integral $K = \int_{0}^{\pi} x \sin x \, dx$.

1. Problem Description

The problem asks to evaluate the definite integral $K = \int_{0}^{\pi} x \sin x \, dx$.

2. Solution Steps

To evaluate the integral $\int x \sin x \, dx$, we use integration by parts. The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

Let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = \int \sin x \, dx = -\cos x$. Applying the integration by parts formula, we get:

$\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x + C$

Now we can evaluate the definite integral:

$\int_{0}^{\pi} x \sin x \, dx = [-x \cos x + \sin x]_{0}^{\pi}$

$= (-\pi \cos \pi + \sin \pi) - (-0 \cos 0 + \sin 0)$

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

$= \pi$

3. Final Answer

The final answer is $\pi$.