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.
Transcript text: Find the mass of the thin bar with the given density function.
\[
\rho(x)=1+\sin x ; \text { for } 0 \leq x \leq \frac{\pi}{3}
\]
Set up the integral that gives the mass of the thin bar.
Solution
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 \(\frac{\pi}{3}\).
Step 1: Define the Density Function and Limits
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 \(0 \leq x \leq \frac{\pi}{3}\).
Step 2: Set Up the Integral
To find the mass \(M\) of the bar, we set up the integral of the density function over the given interval:
\[
M = \int_{0}^{\frac{\pi}{3}} (1 + \sin x) \, dx
\]
Step 3: Evaluate the Integral
Evaluate the integral:
\[
M = \left[ x - \cos x \right]_{0}^{\frac{\pi}{3}}
\]
Step 4: Calculate the Definite Integral
Substitute the limits of integration:
\[
M = \left( \frac{\pi}{3} - \cos\left(\frac{\pi}{3}\right) \right) - \left( 0 - \cos(0) \right)
\]
\[
M = \left( \frac{\pi}{3} - \frac{1}{2} \right) - (-1)
\]
\[
M = \frac{\pi}{3} - \frac{1}{2} + 1
\]
\[
M = \frac{\pi}{3} + \frac{1}{2}
\]