The integral

The mass of an object depends not only on its size but also on how its material is distributed. For a uniform object, one density measurement can describe the whole thing. When density varies from place to place, different parts contribute different amounts to the total mass.

A rod reduces this problem to a single direction: we can move along its length, determine how much each short piece contributes, and combine those contributions. Consider a 2-meter metal rod that is thicker at one end than the other. How might its changing thickness affect its total mass?

The rod's linear density tells us how much mass is packed into each meter of length. For a rod with constant linear density, \[\text{mass}=\text{density}\times\text{length}.\]

For example, if the rod has linear density of \( 2 \) kg/m, its total mass would be kg.

For our rod, though, the density changes as we move from one end to the other. Let \( x \) be the distance in meters from the left end. At position \( x \), suppose the density is

\[\rho(x)=1+x\qquad(0\le x\le2),\]

in kilograms per meter. The Greek letter \( \rho \), pronounced “rho,” is often used for density. A single density can't give the exact mass of the whole rod, but we can use the density function to find an underestimate and an overestimate.

The minimum density along the rod is kg/m. If the entire \( 2 \) m rod had this minimum density, its mass would be kg. The actual density is never smaller, so this is a lower estimate for the rod's mass.

The same reasoning in the other direction gives an upper estimate. The maximum density is kg/m; if the entire rod had this maximum density, its mass would be kg. The actual density is never larger, so the rod can't have more than this much mass.

We now know that the rod's mass lies between \( 2 \) kg and \( 6 \) kg. This range is broad because each estimate uses one density for the entire rod, even though the density changes from one end to the other.

Instead, divide the rod into four pieces, each \( 0.5 \) m long. The density varies less within one short piece than it does across the whole rod, so the smallest and largest densities on each piece should give closer mass estimates. Since the density is increasing, those values occur at the piece's left and right endpoints.

Between \( x=0 \) and \( x=0.5 \), the length is m, while the density of that piece is at least kg/m and at most kg/m. This means that the mass of this first piece of the rod must be at least kg and at most kg.

We can repeat this calculation for the other three pieces of the rod:

Then the total mass of the rod is the sum of the four smaller masses. Adding the four lower contributions gives a lower estimate of kg; adding the four upper contributions gives an upper estimate of kg. That's a much tighter range and gives us a better idea of what the actual mass might be.

Graphing density against position turns each mass estimate into an area. A rectangle's width is the length of one piece in meters, and its height is an assigned density in kilograms per meter. Its area has units of kilograms, so it represents the estimated mass of that piece. Adding the four rectangle areas gives the mass estimates we just found.

We can get a better approximation by cutting the rod into eight narrower pieces, represented by eight narrower rectangles. Their stair-step edges stay closer to the density graph, leaving less space between the lower and upper approximations.

If we add the areas of these rectangles, each of which has a width of m, the lower and upper bounds for the rod's mass become \( 3.75 \) kg and \( 4.25 \) kg. The gap between the bounds has changed from 1 kg to kg. With more, thinner pieces, each assigned density represents its piece more closely, and the lower and upper estimates squeeze toward the same mass.

We can continue increasing the number of rectangles. As their widths shrink, both collections of rectangles come to resemble the same region beneath the density graph.

Number of pieces: \(n\)=

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At 100 pieces, the lower and upper estimates are \( 3.98 \) kg and \( 4.02 \) kg. The two staircases are almost impossible to tell apart from the region beneath the graph, although a small gap remains.

Smaller subdivisions give us better estimates of the rod's mass, but the calculations become more tedious, and no finite subdivision makes the density perfectly constant on every piece. We can always cut the rod again. The pattern suggests the same infinitesimal transition we used when we zoomed infinitely close to a curve: what if we continued subdividing the rod into infinitely many infinitesimal pieces? Each piece would then be thinner than any finite length we could measure, so its density could be treated as constant at that location.

The process of splitting something into infinitely many infinitesimal pieces and then adding them back up to get the whole is called integration.

Let's focus on one such piece at position \( x \). Its infinitesimal length is \( \dd{x} \), and its density is \( \rho(x) \). Since mass equals density times length, the piece has infinitesimal mass

\[\begin{aligned}\dd{m}&=\rho(x)\dd{x}\\ &=(1+x)\dd{x}\text.\end{aligned}\]

The symbol \( \dd{m} \) represents the mass of one infinitesimal piece.

To find the mass of the whole rod, we need to collect all of these infinitesimal masses, beginning at \( x=0 \) and continuing to \( x=2 \).

We write this new operation as

\[\begin{aligned}m&=\int_0^2\rho(x)\dd{x}\\ &=\int_0^2(1+x)\dd{x}\text.\end{aligned}\]

We call this expression an integral. The stretched-out letter S (\( \int \)) says that we're Summing infinitely many infinitesimal contributions. The \( 0 \) and \( 2 \) tell us where along the rod those contributions begin and end. Inside the integral, \( \rho(x)\dd{x} \) is the mass \( \dd{m} \) contributed by one infinitesimal piece.

The units multiply as we would expect: \( (\text{kg}/\text{m})\cdot\text{m}= \)Choose one kg kg/m square meters. Therefore, this integral represents the Choose one density at one point total mass length.

Back on the graph, the lower and upper staircases meet the density line when the pieces become infinitesimal. The whole region beneath the line represents the accumulated mass. For this linear density, that region is a trapezoid.

We can find the trapezoid's area using familiar geometry by splitting it along the dashed line. The part below the line is a rectangle of width \( 2 \) and height \( 1 \), while the part above it is a triangle with base \( 2 \) and height \( 2 \). Adding these areas gives us the value of our integral:

\(\displaystyle \int_0^2(1+x)\dd{x}=\)\(\displaystyle \text.\)

The horizontal units on our graph are meters and the vertical units are kilograms per meter, so this graphical area represents \( 4 \) kg of mass. It lies between every pair of lower and upper estimates we found.

Differentiation and integration are the two central tools of calculus. So far, we've used differentiation to understand change. In the next part of our study, we'll develop integration as a way to describe, calculate, and use accumulations.