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Find the center of mass of the region bounded by the following functions. y=ln x,quad x=1,quad x=2,quad y=0

99. Find the center of mass of the region bounded by the following functions.\newliney=ln⁡x,x=1,x=2,y=0 y=\ln x, \quad x=1, \quad x=2, \quad y=0

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Q. 99. Find the center of mass of the region bounded by the following functions.\newliney=ln⁡x,x=1,x=2,y=0 y=\ln x, \quad x=1, \quad x=2, \quad y=0
  1. Calculate xˉ\bar{x}: To find the center of mass, we need to calculate the coordinates (xˉ\bar{x}, yˉ\bar{y}). First, let's find xˉ\bar{x}, which is the average x-value weighted by area.
  2. Find xˉ\bar{x} formula: The formula for xˉ\bar{x} is the integral of xx times the density function over the area, divided by the total mass (area). Since density is constant, it cancels out. So, xˉ=(1/Area)×∫abx(y) dy\bar{x} = (1/\text{Area}) \times \int_{a}^{b} x(y) \, dy, where y=ln⁡xy=\ln x and the bounds are x=1x=1 and x=2x=2.
  3. Calculate area: We need to find the area first. The area is given by the integral from 11 to 22 of ln⁡x dx\ln x \, dx.
  4. Calculate Area: Calculating the integral, we get Area = ∫12ln⁡x dx=[xln⁡x−x]12\int_{1}^{2} \ln x \, dx = [x \ln x - x]_{1}^{2}.
  5. Calculate xˉ\bar{x}: Plugging in the bounds, Area = (2ln⁡2−2)−(1ln⁡1−1)=2ln⁡2−2−(0−1)=2ln⁡2−1(2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 2 - (0 - 1) = 2 \ln 2 - 1.
  6. Calculate integral: Now, we calculate xˉ=1Area∫12xln⁡x dx\bar{x} = \frac{1}{\text{Area}} \int_{1}^{2} x \ln x \, dx.
  7. Evaluate integral: The integral of xln⁡x dxx \ln x \, dx is x22ln⁡x−x24\frac{x^2}{2} \ln x - \frac{x^2}{4}. So, we need to evaluate this from 11 to 22.
  8. Calculate xˉ\bar{x}: Plugging in the bounds, we get [222ln⁡2−224]−[122ln⁡1−124]=[2ln⁡2−1]−[0−14]=2ln⁡2−1+14[\frac{2^2}{2} \ln 2 - \frac{2^2}{4}] - [\frac{1^2}{2} \ln 1 - \frac{1^2}{4}] = [2 \ln 2 - 1] - [0 - \frac{1}{4}] = 2 \ln 2 - 1 + \frac{1}{4}.
  9. Simplify xˉ\bar{x}: Now, we plug the area and the integral result into the formula for xˉ\bar{x}. xˉ=(12ln⁡2−1)⋅(2ln⁡2−1+14)\bar{x} = \left(\frac{1}{2 \ln 2 - 1}\right) \cdot \left(2 \ln 2 - 1 + \frac{1}{4}\right).
  10. Identify mistake: Simplifying, we get xˉ=2ln⁡2−1+142ln⁡2−1\bar{x} = \frac{2 \ln 2 - 1 + \frac{1}{4}}{2 \ln 2 - 1}. Wait, there's a mistake here. We didn't multiply xx by ln⁡x\ln x in the integral for xˉ\bar{x}.

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