Evaluate the definite integral. Use a graphing utility to verify your result. (Round your answer to three decimal places.) \[ \int_{0}^{\pi / 2} x \sin (2 x) d x \]

To evaluate the definite integral \[ \int_{0}^{\pi / 2} x \sin(2x) \, dx, \] you can use integration by parts. Let \( u = x \) and \( dv = \sin(2x) \, dx \). Then, differentiate and integrate accordingly: 1. \( du = dx \) 2. \( v = -\frac{1}{2} \cos(2x) \) Now, applying integration by parts: \[ \int u \, dv = uv - \int v \, du, \] we have: \[ \int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) \bigg|_{0}^{\pi/2} + \frac{1}{2} \int \cos(2x) \, dx. \] Next, compute the remaining integral: \[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x). \] Putting it all together, we get: \[ \int x \sin(2x) \, dx = -\frac{1}{2} x \cos(2x) \bigg|_{0}^{\pi/2} + \frac{1}{2} \cdot \frac{1}{2} \sin(2x) \bigg|_{0}^{\pi/2}. \] Evaluating each part: 1. For \( -\frac{1}{2} x \cos(2x) \bigg|_{0}^{\pi/2} \): At \( x = \frac{\pi}{2} \): \( -\frac{1}{2} \cdot \frac{\pi}{2} \cdot \cos(\pi) = -\frac{1}{2} \cdot \frac{\pi}{2} \cdot (-1) = \frac{\pi}{4} \). At \( x = 0 \): \( 0 \). Thus, this part evaluates to \( \frac{\pi}{4} - 0 = \frac{\pi}{4} \). 2. For \( \frac{1}{4} \sin(2x) \bigg|_{0}^{\pi/2} \): At \( x = \frac{\pi}{2} \): \( \sin(\pi) = 0 \). At \( x = 0 \): \( \sin(0) = 0 \). Thus, this integral evaluates to \( 0 \). So, combining the results gives: \[ \int_{0}^{\pi/2} x \sin(2x) \, dx = \frac{\pi}{4} + 0 = \frac{\pi}{4}. \] Evaluating \( \frac{\pi}{4} \) gives approximately \( 0.785 \). Thus, rounding it to three decimal places: \[ \int_{0}^{\pi/2} x \sin(2x) \, dx \approx 0.785. \] To verify using a graphing utility, you can plot \( y = x \sin(2x) \) and evaluate the definite integral numerically from \( 0 \) to \( \frac{\pi}{2} \). The results should match closely with \( 0.785 \).