Accumulation & Area Under a Curve

Add up enough thin strips and the total is exactly the area — and usually a distance, a cost or a volume.

A car holds a steady 30 mph for 2 hours. How far did it go?

Sixty miles, obviously. And on a speed-time graph that is a rectangle: 30 tall, 2 wide, area 60.

a speed graph, and the distance hiding under it speed time each strip is speed × time, which is a distance — add them up

Now let the speed change. There is no single rectangle any more — but the graph is still there.

Chop the trip into short stretches. Over each one the speed barely changes, so each strip is almost a rectangle.

Add the strips and you have the distance, near enough. Make them thinner and "near enough" becomes exact. Area under a curve is accumulation, and the units tell you what is accumulating.

Thinner strips, closer to the true area 4 strips 8 strips 16 strips the gaps shrink faster than the strips do

Rate times time is a rectangle

When a rate is constant, the total is rate × time — and on the graph that is an area.

30 mph for 2 hours is a rectangle 30 tall and 2 wide, and 60 is both its area and the distance.

That coincidence is not a coincidence. It is the whole idea, seen in its simplest case.

Read the units

Height in mph times width in hours gives miles. Multiplying the axis units tells you what the area means — and it is rarely area.

Strips when the rate changes

Cut the interval into n equal strips of width b-an and treat the rate as constant on each.

Take the height from the left edge or the right edge — either gives an estimate. If the function is monotonic, the two sums bracket the integral; for a function that rises and falls, that bound need not hold.

For a rising function the left sum is too small and the right sum too large.

Two estimates are better than one

For a monotonic function, if the left and right sums are 18 and 24, the true integral is between them. The gap tells you how far off you might be.

Why thinner is better

The error lives in the little triangles between each rectangle and the curve.

For a straight-line graph, halving the width quarters each little triangular error, but doubles the number of strips. The total error is halved, not quartered.

Let the width go to zero and the error vanishes, leaving the exact area.

The same move as last unit

Derivatives shrank a gap to zero to get a slope; integrals shrink a width to zero to get a total. One limit, two uses.

Notation, and what it means

∫ab f(x)dx is the exact accumulation of f from a to b.

The ∫ is a stretched S for "sum", f(x) is the height of a strip and dx is its vanishing width.

Read as a sentence: add up height times width, all the way across.

Area below the axis counts negative

A velocity graph dipping below zero means motion in the negative direction. Speed is nonnegative; integrating speed, or the absolute value of velocity, gives distance. The integral subtracts that stretch, giving displacement rather than distance travelled.

Exact areas you can already find

If the graph is made of straight lines, no calculus is needed — split it into triangles and rectangles.

A speed rising steadily from 0 to 40 over 5 seconds traces a triangle: 12(5)(40) = 100 metres.

Geometry handles every straight-edged case exactly, and it is worth checking a Riemann sum against one.

A free accuracy check

Run your strip method on a straight-line graph where you already know the answer. If it does not converge to the right number, the method is being applied wrongly.

Worked examples

A left sum

Estimate the area under f(x)=x2 from 0 to 2 using 4 left rectangles.

  1. Width is 2-04 = 0.5; left edges are 0, 0.5, 1, 1.5. Four strips, heights from the left.
  2. Heights: 0, 0.25, 1, 2.25. Sum = 3.5. Square each left edge.
  3. Area ≈ 0.5 × 3.5 = 1.75. The true value is 83 ≈ 2.67, so a left sum underestimates here.

1.75

Distance from a speed graph

A car speeds up steadily from 0 to 40 m/s over 5 seconds. How far does it travel?

  1. The speed-time graph is a triangle with base 5 and height 40. Steady change means a straight line.
  2. Area = 12 (5)(40). No calculus needed for a straight edge.
  3. = 100 metres. Same as average speed 20 times 5 seconds.

100 m

Bracketing the answer

For an increasing function on [1,3], a left sum gives 12 and a right sum 18. What can you say?

  1. For an increasing function the left sum is too small. Each rectangle sits under the curve.
  2. The right sum is too large, for the mirror reason. Each rectangle pokes above.
  3. So the exact area is between 12 and 18. A midpoint estimate of 15 would be a reasonable guess.

Between 12 and 18

Words to know

Riemann sum
An estimate of area made from rectangles under a curve.
Left sum
Rectangles whose heights come from the left edge of each strip.
Right sum
Rectangles whose heights come from the right edge.
Definite integral
The exact area, written ∫ab f(x)dx.
Accumulation
The total built up by a rate acting over an interval.

Mistakes to watch for

Using the wrong edge heights. A left sum for four strips on [0,2] takes heights at 0, 0.5, 1, 1.5 — not at 2. Including the right endpoint turns it into a different estimate.

Forgetting to multiply by the strip width. Adding the heights gives 3.5, but the area is 0.5 × 3.5 = 1.75. The sum of heights is not an area until it is given a width.

Reading the area as an area. Under a speed-time graph the units are mph times hours, which is miles. Multiply the axis units to find out what has actually been accumulated.

Ignoring negative regions. A curve below the axis contributes a negative amount, so an integral can be zero for a graph with plenty of visible area.

Assuming the left sum is always too small. That holds for an increasing function; for a decreasing one it is the other way round. Check which way the curve runs before deciding.

Archimedes found the area of a parabolic segment this way around 240 BC, two thousand years before calculus had a name.

The integral sign ∫ is a stretched letter S, for "summa". Leibniz chose it in 1675 and it has not changed since.

The fundamental theorem of calculus says that accumulating and rating are inverse operations — the two halves of this course are the same subject seen from either end.

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