Find the mass of the thin bar with the given density function. p(x) = 1 + sinx, for π/3 <= x <= 3π/4 (Type exact answers.) The mass of the thin bar is units. (Type an exact answer.)

To find the mass of the thin bar with the given density function \(\rho(x) = 1 + \sin x\) over the interval \(\left[\frac{\pi}{3}, \frac{3\pi}{4}\right]\), we need to evaluate the following integral: \[ m = \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} \rho(x) \, dx = \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} (1 + \sin x) \, dx. \] This integral can be broken down into two parts: \[ m = \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} 1 \, dx + \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} \sin x \, dx. \] Calculating the first integral: \[ \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} 1 \, dx = \left[ x \right]_{\frac{\pi}{3}}^{\frac{3\pi}{4}} = \frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi}{12} - \frac{4\pi}{12} = \frac{5\pi}{12}. \] Now, we calculate the second integral: \[ \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} \sin x \, dx = \left[ -\cos x \right]_{\frac{\pi}{3}}^{\frac{3\pi}{4}} = -\cos\left(\frac{3\pi}{4}\right) + \cos\left(\frac{\pi}{3}\right). \] Evaluating \(-\cos\left(\frac{3\pi}{4}\right)\): \(-\cos\left(\frac{3\pi}{4}\right) = -\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\). And evaluating \(\cos\left(\frac{\pi}{3}\right)\): \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). So the second integral becomes: \[ \int_{\frac{\pi}{3}}^{\frac{3\pi}{4}} \sin x \, dx = \frac{1}{\sqrt{2}} + \frac{1}{2}. \] Now we find a common denominator to combine: \[ \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{2}{2\sqrt{2}} + \frac{\sqrt{2}}{2\sqrt{2}} = \frac{2 + \sqrt{2}}{2\sqrt{2}}. \] Consolidating everything, we have: \[ m = \frac{5\pi}{12} + \left(\frac{2 + \sqrt{2}}{2\sqrt{2}}\right). \] The mass of the thin bar is: \( m = \frac{5\pi}{12} + \frac{2 + \sqrt{2}}{2\sqrt{2}} \).