4. Evaluate the iterated integral \[ \int_{0}^{\pi / 2} \int_{0}^{x} x \sin y d y d x \]

To evaluate the given iterated integral \[ I = \int_{0}^{\pi / 2} \int_{0}^{x} x \sin y \, dy \, dx, \] we will first evaluate the inner integral: \[ \int_{0}^{x} x \sin y \, dy. \] Since \( x \) is constant with respect to \( y \), we can factor it out: \[ \int_{0}^{x} x \sin y \, dy = x \int_{0}^{x} \sin y \, dy. \] Next, we calculate the integral \( \int_{0}^{x} \sin y \, dy \): \[ \int \sin y \, dy = -\cos y. \] Evaluating from \( 0 \) to \( x \): \[ \int_{0}^{x} \sin y \, dy = -\cos x - (-\cos 0) = -\cos x + 1 = 1 - \cos x. \] Thus, we have: \[ \int_{0}^{x} x \sin y \, dy = x (1 - \cos x). \] Now substituting this back into our original iterated integral, we have: \[ I = \int_{0}^{\pi / 2} x (1 - \cos x) \, dx. \] Next, we can split this integral into two parts: \[ I = \int_{0}^{\pi / 2} x \, dx - \int_{0}^{\pi / 2} x \cos x \, dx. \] Calculating the first integral: \[ \int_{0}^{\pi / 2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\pi / 2} = \frac{(\pi / 2)^2}{2} = \frac{\pi^2}{8}. \] Now we will evaluate the second integral \( \int_{0}^{\pi / 2} x \cos x \, dx \) using integration by parts. Let: - \( u = x \) (hence \( du = dx \)), - \( dv = \cos x \, dx \) (thus \( v = \sin x \)). Using integration by parts \( \int u \, dv = uv - \int v \, du \): \[ \int_{0}^{\pi / 2} x \cos x \, dx = \left[ x \sin x \right]_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \sin x \, dx. \] Evaluating \( \left[ x \sin x \right]_{0}^{\pi / 2} \): \[ \left[ x \sin x \right]_{0}^{\pi / 2} = \left( \frac{\pi}{2} \cdot 1 - 0 \cdot 0 \right) = \frac{\pi}{2}. \] Now calculating \( \int_{0}^{\pi / 2} \sin x \, dx \): \[ \int \sin x \, dx = -\cos x, \] evaluating from \( 0 \) to \( \pi / 2 \): \[ \int_{0}^{\pi / 2} \sin x \, dx = -\cos\left(\frac{\pi}{2}\right) - \left(-\cos(0)\right) = 0 + 1 = 1. \] Now combining these results: \[ \int_{0}^{\pi / 2} x \cos x \, dx = \frac{\pi}{2} - 1. \] Substituting back into the original iterated integral: \[ I = \frac{\pi^2}{8} - \left( \frac{\pi}{2} - 1 \right). \] Rearranging gives: \[ I = \frac{\pi^2}{8} - \frac{\pi}{2} + 1. \] Combining the terms, we convert \( \frac{\pi}{2} \) to a fraction over \( 8 \): \[ I = \frac{\pi^2}{8} - \frac{4\pi}{8} + 1 = \frac{\pi^2 - 4\pi + 8}{8}. \] Thus, the final answer for the evaluated iterated integral is \[ \boxed{\frac{\pi^2 - 4\pi + 8}{8}}. \]