Question

The value of \( \int_0^\pi x \sin x \, dx \) is .......

Hint:

Use integration by parts for products of functions like \( x \sin x \). Remember to carefully evaluate boundary terms.

Answer 1

Correct Answer: 1

Step 1: Setting up the integral.
We are tasked with finding the value of the integral: \[ \int_0^\pi x \sin x \, dx \] Step 2: Integration by parts.
We use the method of integration by parts, which is given by the formula: \[ \int u \, dv = uv - \int v \, du \] Let \( u = x \) and \( dv = \sin x \, dx \). Then, \( du = dx \) and \( v = -\cos x \). Applying the integration by parts formula: \[ \int_0^\pi x \sin x \, dx = \left[ -x \cos x \right]_0^\pi + \int_0^\pi \cos x \, dx \] Evaluating the boundary terms: \[ \left[ -x \cos x \right]_0^\pi = -( \pi \cos \pi - 0 \cos 0 ) = \pi \] Now integrating \( \int_0^\pi \cos x \, dx \): \[ \int_0^\pi \cos x \, dx = \left[ \sin x \right]_0^\pi = \sin \pi - \sin 0 = 0 \] Step 3: Conclusion.
Thus, the value of the integral is: \[ \pi + 0 = 2 \]