🧱Riemann Sums & Approximating Area

Geometry can measure a rectangle, a triangle and a trapezoid. It cannot measure the region under a curve, because the top edge will not sit still.

So do the only thing available: cover the region with rectangles you can measure, and add them up.

Cut the interval from a to b into n strips of the same width. That width is the same every time and it is worth writing down first:

width of one strip = (b - a)/n

Then give each strip a height by picking one x inside it and reading f there. Each rectangle has area height x width, and the total is an estimate of the area under the curve. A total built this way is a Riemann sum.

Two things decide a Riemann sum completely: how many strips there are, and where inside each strip the height is read. Change either one and you get a different number for the same region.

Notice what the arithmetic looks like. Every rectangle is multiplied by the same width, so the width can be factored out:

Riemann sum = width x (sum of all the heights)

Add the heights first, multiply by the width once at the end. It is less work and there is only one place left to lose the width.

Three standard choices of where to read the height, on the interval from a to b with n strips of width w:

• ◆Left sum. Read f at the left edge of each strip: at a, a + w, a + 2w, up to a + (n - 1)w. The right-hand end b is never used.
• ◆Right sum. Read f at the right edge of each strip: at a + w, a + 2w, up to b. The left-hand end a is never used.
• ◆Midpoint sum. Read f in the middle of each strip: at a + w/2, a + 3w/2, and so on, half a width further along than the left edges.

Every one of them uses exactly n heights, one per strip. There are n + 1 edges but only n gaps between them, and it is the gaps that carry the rectangles. Counting n + 1 heights is the single most common slip here.

When f is monotone the picture settles which side of the truth the estimate falls on:

• ◆f only rising: the left edge is the shortest point of each strip and the right edge is the tallest, so the left sum under-estimates and the right sum over-estimates.
• ◆f only falling: the other way round. The left sum over-estimates and the right sum under-estimates.

If f rises on part of the interval and falls on the rest, neither argument applies. Some rectangles overshoot and others undershoot, and the shape alone cannot tell you which way the total leans.

A midpoint sum has no such rule, and in practice it is usually much closer than either endpoint sum: too much on one side of the middle roughly cancels too little on the other.

A Riemann sum can be read from a table as easily as from a rule. If speeds are logged every second and you want a midpoint sum over three 2-second pieces, the middles are the readings at 1, 3 and 5 seconds. The other readings are simply not used, and that is not a mistake.

A rectangle has a flat top, which is a poor match for a sloping curve. Join the two points of the curve at the edges of a strip with a straight chord instead, and each strip becomes a trapezoid.

One trapezoid of width w with heights h1 and h2 at its two edges has area ((h1 + h2)/2) x w. Add the trapezoids across the whole interval and every inside edge gets counted twice, once for the strip on its left and once for the strip on its right:

trapezoidal estimate = (w/2) x [f0 + 2f1 + 2f2 + ... + 2f(n-1) + fn]

Only the first and last values are counted once, because each of them belongs to a single trapezoid. Every value in between is doubled. Both the doubling and the division by 2 are easy to drop, and dropping either changes the answer by a lot.

There is a second way to see the same formula: the trapezoidal estimate is exactly the average of the left sum and the right sum. That is a fast way to check it.

Concavity decides which way a trapezoidal estimate leans, and the argument is about chords, not about endpoints:

• ◆f concave up (bending upward): a chord joining two points of the curve lies above the curve, so every trapezoid holds a little too much. The estimate is an over-estimate.
• ◆f concave down (bending downward): the chord lies below the curve, so every trapezoid misses a sliver. The estimate is an under-estimate.

The trapezoidal rule is exact only for a straight line, where the chord is the curve.

Every Riemann sum is an estimate. Use more strips and the estimate improves; the natural question is what it improves towards.

That limit is the definite integral:

the integral from a to b of f(x) dx = the limit, as n grows without bound, of the sum of f(sample point) x width

Written out with right-hand sample points, the width is (b - a)/n and the i-th sample point is a + i(b - a)/n. For the interval from 1 to 3 with f(x) = x², the width is 2/n, the i-th point is 1 + 2i/n, and the sum is

the sum from i = 1 to n of ((1 + 2i/n)²)(2/n)

which becomes the integral from 1 to 3 of x² dx.

Reading such a limit backwards is a standard exercise, and it needs only two observations:

• ◆The factor multiplying every term is the width, so it tells you b - a. A factor of 2/n means the interval has length 2, not that b is 2.
• ◆The number the sample point counts up from is a. In 1 + 2i/n that number is 1, so the interval is from 1 to 1 + 2 = 3.

Whatever the sample point is fed into is the integrand, unchanged. The width factor becomes the dx of the integral; it never becomes a coefficient in front of the function.

Some definite integrals need no antiderivative at all, because the region really is a shape geometry already knows.

• ◆A constant. The integral from a to b of c dx is the rectangle c(b - a).
• ◆A line. The region is a triangle or a trapezoid. Where the line crosses the axis inside the interval, split the region at the crossing and treat the two pieces separately.
• ◆A semicircle. y = sqrt(r^2 - x²) is the top half of the circle x² + y² = r². From -r to r the region is half that circle, so the integral is (pi r²)/2; from 0 to r it is a quarter, (pi r²)/4. Answers here are exact multiples of pi.

Two words that are often used as if they meant the same thing do not. Area is always positive. An integral is signed: a part of the region below the x-axis counts as negative. For a graph that dips below the axis the two questions have different answers, and the question always says which one it wants.

The properties of the definite integral let one integral be built out of others, again with no antiderivative in sight:

integral from a to a of f = 0 integral from b to a of f = -(integral from a to b of f) integral from a to b of f + integral from b to c of f = integral from a to c of f integral of k f = k x (integral of f) integral of (f + g) = integral of f + integral of g

So if the integral of f from 1 to 4 is 7 and the integral of f from 4 to 9 is -3, then the integral of f from 1 to 9 is 7 + (-3) = 4, and the integral of (3f(x) - 2) from 1 to 9 is 3(4) - 2(9 - 1) = 12 - 16 = -4. The constant term contributes its own rectangle, which is easy to forget.