Moments, Centers, and Probability

An average that respects where things are

An ordinary average treats every value equally. A physical balance cannot do that: mass farther from a pivot has a greater turning effect than the same mass close to the pivot. A continuous probability model has the same structure because values that occur more often should influence the expected value more strongly. In both settings, calculus builds a weighted average from many tiny contributions. The words moment, center of mass, and expected value describe different applications of one organizing idea.

Begin with discrete masses $m_1,m_2,\ldots,m_n$ located at positions $x_1,x_2,\ldots,x_n$ on a line. Their total mass is $M=\sum m_i$. The balance point is

$$\bar{x}=\frac{\sum x_i m_i}{\sum m_i}.$$

The numerator is a moment: each position is multiplied by the mass located there. Its units are mass times length. Dividing by total mass leaves length, as a position should. This unit check also explains why $sum x_i$ by itself is not enough unless all masses are equal. A balance point is not merely the midpoint of occupied positions.

From point masses to a density function

For a thin rod on the interval $a\le x\le b$, mass is distributed continuously. Let $\lambda(x)$ be its linear density, measured in $mathrm{kg}/\mathrm{m}$. A short piece of length $dx$ has mass

$$dm=\lambda(x)\,dx.$$

The total mass is therefore

$$M=\int_a^b\lambda(x)\,dx,$$

and its moment about the origin is

$$M_y=\int_a^b x\lambda(x)\,dx.$$

The subscript $y$ reflects the fact that horizontal location creates a tendency to rotate about a vertical, or $y$, axis. The center of mass is $\bar{x}=M_y/M$. The notation may look like a new collection of letters, but each factor has a job: $\lambda(x)dx$ gives a small mass, and multiplying by $x$ weights it by its distance from the reference point. The origin is a chosen reference, not a physical object that must be present in the drawing.

Suppose $lambda(x)=2+x$ kilograms per meter on $0\le x\le2$ meters. Density increases to the right, so the center should lie to the right of the midpoint $x=1\,\mathrm{m}$. The total mass is $M=\int_0^2(2+x)\,dx=6\,\mathrm{kg}$. The moment is $M_y=\int_0^2x(2+x)\,dx=\frac{20}{3}\,\mathrm{kg}\cdot\mathrm{m}$, giving $\bar{x}=\frac{10}{9}\,\mathrm{m}$. The estimate matches the physical prediction: extra mass on the right shifts the balance point rightward.

Centroids describe uniform regions

When a flat lamina has uniform density, its center of mass is called its centroid. For a region between $y=f(x)$ and $y=g(x)$ on $a\le x\le b$, where $f(x)\ge g(x)$, a thin vertical strip has area $dA=[f(x)-g(x)]dx$. The total area is

$$A=\int_a^b[f(x)-g(x)]\,dx.$$

Its horizontal moment contribution is $x\,dA$, so

$$\bar{x}=\frac{1}{A}\int_a^bx[f(x)-g(x)]\,dx.$$

To find $\bar{y}$, use the strip’s own midpoint height, $\frac12[f(x)+g(x)]$, before weighting it by its area. This produces

$$\bar{y}=\frac{1}{2A}\int_a^b\left[f(x)^2-g(x)^2\right]dx.$$

These formulas are compact summaries of a geometric argument, not arbitrary patterns. A common error is to use $f(x)-g(x)$ as a vertical coordinate. That expression is a height, while the strip’s actual vertical location is its midpoint. Draw one strip and mark its center before attempting the integral; the picture distinguishes location from size.

Probability uses normalized density

A continuous random variable has a probability density function $p(x)$. It is not itself a probability at one exact point; it is a density whose area over an interval gives probability. The normalization condition

$$\int_a^bp(x)\,dx=1$$

plays the role of total mass. The expected value is the weighted average

$$E[X]=\int_a^b x p(x)\,dx.$$

This is mathematically the same structure as ar{x}=M_y/M, with total probability already normalized to $1$. If $p(x)$ puts more area near larger values, the expected value shifts right; it need not equal the most likely value. Its units are the same as the units of $x$, because probability is dimensionless. This dimensional result is a useful way to remember what an expected value represents.

Retrieval check

A rod lies from $x=-1\,\mathrm{m}$ to $x=3\,\mathrm{m}$ with density $\lambda(x)=4-x$ kilograms per meter. Before evaluating, predict whether its center of mass lies left or right of the midpoint and explain your reasoning. Then set up the mass and moment integrals with correct units, and divide only after both have been found. This closes Unit III: every lesson has used an integral to add a small contribution whose meaning was established first.