Area Where No Formula Reaches Application of Integrals

Elementary geometry hands over a formula for the area of a rectangle, a triangle, a trapezium, even a full circle. Every one of those shapes has straight edges, or in the circle's case, one perfectly uniform curve with a known formula built specifically for it. But ask for the area trapped between the x-axis and the curve y = x squared, from x=1 to x=3, and none of those formulas apply at all; the boundary is curved, and no ready-made shape matches it. This is exactly the gap the definite integral, built in the previous chapter as a limit of a sum of thin rectangles, was designed to close. Slice the region under the curve into thin vertical strips of width dx, each one so narrow that the curve barely changes height across it, so each strip is very nearly a thin rectangle of height y = f(x) and width dx, area y dx. Summing every strip's area and letting the width shrink to zero is precisely the definite integral: Area = the integral from a to b of y dx. A shape with no formula of its own borrows the machinery of calculus instead.

The strip method assumes each rectangle's height, y, is a positive number. If part of a curve dips below the x-axis, y is negative there, and the raw integral over that stretch comes out negative too, even though area itself should always be reported as a positive quantity. Consider the area between the x-axis and y = cos(x), from x = 0 to x = pi (a full half-period of the cosine wave). Cosine is positive from 0 to pi/2, and negative from pi/2 to pi. Split the integral at the point where the curve crosses the axis: the integral from 0 to pi/2 of cos(x) dx = [sin x] from 0 to pi/2 = 1 - 0 = 1, a positive true area. The integral from pi/2 to pi of cos(x) dx = [sin x] from pi/2 to pi = 0 - 1 = -1, a raw negative value reflecting that the curve sits below the axis there. The actual area of that second region is the absolute value, 1, not -1. The total true area enclosed is 1 + 1 = 2, found by splitting at every axis-crossing and taking the absolute value of each piece separately, never by integrating straight across a sign change and hoping the positive and negative parts will report an honest total on their own (they would instead partially cancel, understating the real area).

Find the area bounded by the curve y = x squared, the x-axis, and the vertical lines x = 1 and x = 3. Since y = x squared is never negative, no sign-splitting is needed here. Area = the integral from 1 to 3 of x squared dx = [x cubed / 3] evaluated from 1 to 3 = (27/3) - (1/3) = 9 - (1/3) = 26/3 square units. The same region can be swept out using horizontal strips instead of vertical ones, integrating with respect to y rather than x, which sometimes turns out to be the easier direction depending on the shape of the boundary; for this particular region, rewriting x in terms of y as x = the square root of y, the same area can be recovered as the integral of x dy over the appropriate y-range, a genuinely useful alternative worth knowing exists, even when the vertical-strip approach used above is simpler for this specific curve.

The single most natural question after finding the area under one curve is the area trapped between two curves, and it uses only a small extension of the same strip idea. If one curve stays above another, y = f(x) above y = g(x), across some interval, the vertical strip at any point x no longer runs from the axis up to the curve; it runs from the lower curve up to the upper curve, giving a strip height of [f(x) - g(x)] rather than just f(x). The area between the two curves, from x = a to x = b, is therefore the integral from a to b of [f(x) - g(x)] dx. The values a and b are almost always the x-coordinates where the two curves actually cross, found by setting f(x) = g(x) and solving, since that is where one region naturally ends and the strip height would otherwise become negative or meaningless. Find the area between y = x + 3 (a straight line, playing the role of the upper curve) and y = x squared - 1 (a parabola, the lower curve). First find where they intersect: set x + 3 = x squared - 1, giving x squared - x - 4 = 0... actually simplifying, x squared - x - 4 = 0 does not factor neatly, so instead pick a cleaner pair on purpose: y = x + 3 and y = x squared + 1 intersect where x + 3 = x squared + 1, i.e. x squared - x - 2 = 0, which factors as (x-2)(x+1) = 0, giving x = -1 and x = 2. Between these two x-values, check which curve sits higher (at x=0, the line gives 3, the parabola gives 1, so the line is on top). Area = the integral from -1 to 2 of [(x+3) - (x squared+1)] dx = the integral from -1 to 2 of (-x squared + x + 2) dx = [-x cubed/3 + x squared/2 + 2x] evaluated from -1 to 2. At x=2: -8/3 + 2 + 4 = -8/3 + 6 = 10/3. At x=-1: 1/3 + 1/2 - 2 = 1/3 - 3/2 = -7/6. Subtracting: 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 4.5 square units.

A typical exam-style question rarely states the limits of integration outright; finding them is part of the problem. Find the area enclosed between the parabola y = x squared and the line y = 2x. First find the intersection points by setting the two expressions equal: x squared = 2x, so x squared - 2x = 0, x(x-2) = 0, giving x = 0 and x = 2. Next, decide which curve is on top between these two values by testing a point in between, say x = 1: the line gives y = 2, the parabola gives y = 1, so the line sits above the parabola across this whole interval. The enclosed area is then the integral from 0 to 2 of [2x - x squared] dx = [x squared - x cubed/3] evaluated from 0 to 2 = (4 - 8/3) - (0 - 0) = 4/3 square units. Every area-between-curves problem follows this same three-step shape: solve for the intersection points to find the limits, test a point in between to identify which curve is genuinely on top, then integrate the difference (top minus bottom) across those limits.

Long before calculus existed as a named subject, the Greek mathematician Eudoxus, and later Archimedes, estimated the areas of curved shapes, most famously a circle, by inscribing regular polygons with more and more sides inside them, watching the polygon's easily-computed area creep closer and closer to the circle's true area as the number of sides grew. This approach, called the method of exhaustion, is the direct ancestor of the strip-and-limit idea used throughout this chapter and the previous one; a circle approximated by a hundred-sided polygon is the same basic move as a curve approximated by a hundred thin rectangles. It took roughly two thousand years, and the combined work of Newton, Leibniz, and later Cauchy tightening the loose ends into a fully rigorous limiting process, before this ancient method of getting closer and closer to an answer became the precisely defined integral used in this chapter today.