Integrals at Work: Area, Volume & Averages

The one idea behind all of them: accumulation

You already met the definite integral as the limit of a Riemann sum: chop a quantity into many thin pieces, add the pieces up, then let the pieces shrink toward zero width. "Area under a curve" was just the first example, because each rectangle's contribution f(x)*dx happens to look like an area. But nothing forces dx to measure width on a graph. The real move is always the same — slice, multiply, and sum — and the integral is what that sum becomes in the limit.

So the recipe for every application in this guide is: decide what one thin slice contributes, write that contribution as (something) times dx, and integrate. If a slice contributes a length, you get an area; if it contributes an area, you get a volume; if it contributes a bit of distance, you get total distance. The integral sign is just a tally machine for infinitely many infinitesimal pieces.

Area between curves, and volumes by spinning

Start with the area between two curves. Say a top curve f(x) sits above a bottom curve g(x) on the interval from a to b. A thin vertical strip at position x has height f(x) - g(x) (top minus bottom) and width dx, so it contributes (f(x) - g(x)) * dx. Add the strips and shrink them, and you get the area as one integral. Subtracting g first is what makes this work even when both curves dip below the x-axis.

area between curves =  integral from a to b of ( f(x) - g(x) ) dx
                                            \__top minus bottom__/

slice contribution:  height * width  =  ( f(x) - g(x) ) * dx

Now spin a region around the x-axis and you sweep out a 3-D solid of revolution — think of a vase shaped on a lathe. Slice it into thin coins (disks) perpendicular to the axis. A coin at position x is a circle of radius f(x), so its face area is pi * f(x)^2, and its tiny thickness is dx; one coin contributes pi * f(x)^2 * dx. Stack the coins with an integral and you have the volume. The contribution changed from a length to an area, but the slice-multiply-sum move is identical.

The average value of a function

How do you average something that takes infinitely many values, like temperature over a whole day? For a finite list you add the numbers and divide by how many there are. For a function on [a, b] you do the continuous version: add up all the values with an integral, then divide by the length of the interval b - a. That gives the average value of f.

average value of f on [a, b]  =  ( 1 / (b - a) ) * integral from a to b of f(x) dx

compare to a finite average:  ( sum of values ) / ( how many values )

There is a satisfying picture for this. Multiply both sides by (b - a) and the formula says: average * (b - a) = integral from a to b of f. In other words, a flat rectangle of width b - a and height equal to the average holds exactly the same area as the wiggly region under f. The average is the constant height that would have given you the same total — it levels the bumps out.

The unifying theme: integrate a rate to get a total

Here is the idea that ties the whole subject up. A derivative turns a total into a rate: position becomes velocity. An integral runs that backwards — it turns a rate back into a total. If you know how fast something is accumulating at every instant, integrating that rate over a time span gives the total amount accumulated. This is just the Fundamental Theorem read as a sentence about the real world.

1. Velocity is the rate of change of position, so over a tiny time dt you move about v(t) * dt. Add up all those tiny moves: total displacement = integral from t1 to t2 of v(t) dt.
2. Example: a car drives at a steady v = 60 km/h from t = 0 to t = 2 hours. The integral of 60 dt from 0 to 2 is 60*2 - 60*0 = 120 km — exactly the distance you'd expect. The integral just confirms the everyday answer.
3. Now let the speed vary, say v(t) = 6t. An antiderivative is 3t^2, so distance from 0 to 2 is 3*2^2 - 3*0^2 = 12 km. No clean arithmetic shortcut exists for changing speed — the integral is what handles it.