If \(\int \limits_0^{2 \pi} |x \: \sin x| dx=k\pi\) then the value of \(k\) is…

Compute the integral from 0 to 2π of |x sin x| dx and express it as kπ; find k.

• Note that sin x ≥ 0 on [0, π] and sin x ≤ 0 on [π, 2π]. Therefore |x sin x| = x sin x on [0, π] and |x sin x| = -x sin x on [π, 2π].

• Split the integral accordingly: integral = ∫_0^π x sin x dx + ∫_π^{2π} (-x sin x) dx = ∫_0^π x sin x dx - ∫_π^{2π} x sin x dx.

• Find an antiderivative: ∫ x sin x dx = -x cos x + sin x + C.

• Evaluate on [0, π]: ∫_0^π x sin x dx = [-x cos x + sin x]_0^π = (-π cos π + sin π) - (0) = π.

• Evaluate on [π, 2π]: ∫_π^{2π} x sin x dx = [-x cos x + sin x]_π^{2π} = (-2π cos 2π + sin 2π) - (-π cos π + sin π) = (-2π) - (π) = -3π.

• Therefore the original integral = π - (-3π) = 4π, so k = 4.

Answer: k = 4