Questions: Find the mass of the thin bar with the given density function. ρ(x)=1+sin x ; for 0 ≤ x ≤ π/3 Set up the integral that gives the mass of the thin bar.

Solution Steps

To find the mass of the thin bar with the given density function, we need to integrate the density function over the given interval. The mass \( M \) of the bar can be found by evaluating the integral of the density function \(\rho(x)\) from \(0\) to \(\frac{\pi}{3}\).

The density function of the thin bar is given by \(\rho(x) = 1 + \sin x\). We need to find the mass of the bar over the interval \(0 \leq x \leq \frac{\pi}{3}\).

To find the mass \(M\) of the bar, we set up the integral of the density function over the given interval: \[ M = \int_{0}^{\frac{\pi}{3}} (1 + \sin x) \, dx \]

Evaluate the integral: \[ M = \left[ x - \cos x \right]_{0}^{\frac{\pi}{3}} \]

Substitute the limits of integration: \[ M = \left( \frac{\pi}{3} - \cos\left(\frac{\pi}{3}\right) \right) - \left( 0 - \cos(0) \right) \] \[ M = \left( \frac{\pi}{3} - \frac{1}{2} \right) - (-1) \] \[ M = \frac{\pi}{3} - \frac{1}{2} + 1 \] \[ M = \frac{\pi}{3} + \frac{1}{2} \]

Final Answer