Applications of Integration: Center of Mass, Mean Value, and Root Mean Square

Center of Mass

Consider a mechanical system with n discrete masses located at positions along an axis.
Index each mass with i = 1 to n.
The i^{th} body has mass mi and is located at position xi with respect to the origin.
The location of the center of mass \bar{x} is calculated as:
\bar{x} = \frac{\sum{i=1}^{n} xi mi}{\sum{i=1}^{n} m_i}

Four masses are located at positions -2, -1, 1, and 2.
The masses are 2, 1, 2, and 3, respectively.
Calculate the center of mass:
\bar{x} = \frac{\sum{i=1}^{4} xi mi}{\sum{i=1}^{4} mi} \bar{x} = \frac{x1m1 + x2m2 + x3m3 + x4m4}{m1 + m2 + m3 + m_4}
\bar{x} = \frac{(-2)(2) + (-1)(1) + (1)(2) + (2)(3)}{2 + 1 + 2 + 3}
\bar{x} = \frac{-4 - 1 + 2 + 6}{8}
\bar{x} = \frac{3}{8}

For a one-dimensional object with linear density \rho(x), where \rho(x) gives the density at each point x from the origin.
Assume the density function is defined from x = a to x = b.
Divide the interval [a, b] into n equal subintervals of length \Delta x = \frac{b - a}{n}.
Approximate the mass in each subinterval as \rho(x_i) \Delta x.
Approximate the center of mass as:
\bar{x} \approx \frac{\sum{i=1}^{n} xi \rho(xi) \Delta x}{\sum{i=1}^{n} \rho(x_i) \Delta x}
The exact center of mass is found by taking the limit as n \to \infty:
\bar{x} = \frac{\int{a}^{b} x \rho(x) dx}{\int{a}^{b} \rho(x) dx}

Find the center of mass of an object with linear density \rho(x) = ax^2, where a > 0 and 0 \le x \le L.
Apply the formula:
\bar{x} = \frac{\int{0}^{L} x(ax^2) dx}{\int{0}^{L} ax^2 dx}
\bar{x} = \frac{a \int{0}^{L} x^3 dx}{a \int{0}^{L} x^2 dx}
\bar{x} = \frac{\int{0}^{L} x^3 dx}{\int{0}^{L} x^2 dx}
\bar{x} = \frac{\frac{x^4}{4} \Big|0^L}{\frac{x^3}{3} \Big|0^L}
\bar{x} = \frac{\frac{L^4}{4}}{\frac{L^3}{3}}
\bar{x} = \frac{3}{4}L

The mean value of a function f(x) over an interval [a, b] is needed to find a varying quantity.
Approximate the mean value by subdividing the interval into n equal parts and averaging the function values at each point xi: \frac{1}{n} \sum{i=1}^{n} f(x_i)
Since \Delta x = \frac{b - a}{n}, we have n = \frac{b - a}{\Delta x}.
The mean value can be approximated as:
\frac{1}{b-a} \sum{i=1}^{n} f(xi) \Delta x
The exact mean value \bar{f} is:
\bar{f} = \lim{n \to \infty} \frac{1}{b-a} \sum{i=1}^{n} f(xi) \Delta x = \frac{1}{b-a} \int{a}^{b} f(x) dx
Geometrically, the area under the curve of f(x) between a and b is equal to the area of a rectangle with height \bar{f} and width (b - a).

Find the mean value of f(x) = x^2 over the interval [0, 1]:
\bar{f} = \frac{1}{1 - 0} \int{0}^{1} x^2 dx \bar{f} = \int{0}^{1} x^2 dx
\bar{f} = \frac{x^3}{3} \Big|_0^1
\bar{f} = \frac{1}{3}

Find the mean value of f(x) = \sin(\pi x) over the interval [-1, 1]:
The function is odd-symmetric, meaning f(-x) = -f(x).
The area under the curve cancels out, so the mean value should be zero.
\bar{f} = \frac{1}{1 - (-1)} \int{-1}^{1} \sin(\pi x) dx \bar{f} = \frac{1}{2} \int{-1}^{1} \sin(\pi x) dx
\bar{f} = \frac{1}{2} \left[ -\frac{\cos(\pi x)}{\pi} \right]_{-1}^{1}
\bar{f} = \frac{1}{2\pi} [-\cos(\pi) + \cos(-\pi)]
\bar{f} = \frac{1}{2\pi} [-(-1) + (-1)] = 0

The mean value of a function is sometimes useless (e.g., when the function is anti-symmetric over the interval of integration).
The root mean square (RMS) value is introduced for a function f(x) defined on an interval [a, b] to obtain more information when the average is uninformative.
The RMS value is defined as the square root of the mean value of the square of the function:
RMS = \sqrt{\text{Mean Value of } f(x)^2}
RMS = \sqrt{\frac{1}{b-a} \int_{a}^{b} [f(x)]^2 dx}

Calculate the RMS value for the function f(x) = 2\cos(x) over the interval [0, 2\pi]:
RMS = \sqrt{\frac{1}{2\pi - 0} \int{0}^{2\pi} (2\cos(x))^2 dx} RMS = \sqrt{\frac{1}{2\pi} \int{0}^{2\pi} 4\cos^2(x) dx}
RMS = \sqrt{\frac{4}{2\pi} \int_{0}^{2\pi} \cos^2(x) dx}
Use the power reduction formula: \cos^2(x) = \frac{1 + \cos(2x)}{2}
RMS = \sqrt{\frac{4}{2\pi} \int{0}^{2\pi} \frac{1 + \cos(2x)}{2} dx} RMS = \sqrt{\frac{4}{4\pi} \int{0}^{2\pi} (1 + \cos(2x)) dx}
RMS = \sqrt{\frac{1}{\pi} \left[ x + \frac{\sin(2x)}{2} \right]_{0}^{2\pi}}
RMS = \sqrt{\frac{1}{\pi} (2\pi + 0 - 0 - 0)}
RMS = \sqrt{\frac{2\pi}{\pi}}
RMS = \sqrt{2}