INTEGRAL CALCULUS

Integral calculus involves the inverse process to differentiation, called integration. Given a function f, we seek a function F with derivative F¢ = f; this is an integral or antiderivative of f, written F(x) = ∫ f(x)dx or simply F = ∫f dx (a notation we will explain later). Tables of derivatives can be used for integration: thus x2 has derivative 2x, so 2x has x2 as an integral. If F is any integral of f, the most general integral of f is F + c, where c is an arbitrary constant called the constant of integration; this is because a constant has derivative 0, so ( F + c)¢ = F¢ + c¢ = f + 0 = f. Thus ∫2xdx = x2 + c, for instance.

The calculation of the area under a curve is a classic example of the use of integral calculus. Here the area between the curve and the x-axis from x = a to x = b is approximately equal to the sum of a large number of thin rectangles like the one shown. This has an area approximately equal to f (x) times h. As h is made smaller, the rectangles become thinner and more numerous, and their total area approaches ever more closely to the true area under the curve. Integral calculus is concerned with deriving the exact value of the area from a knowledge of the equation describing the curve, y = f (x).

The basic rules for integrating compound functions resemble those for differentiation. The integral of a sum or difference is the sum or difference of their integrals, and likewise for multiplication by a constant. Thus x = ½.2x has integral ½x2, and similarly ∫xm dx = xm+1/(m + 1) for any m≠ -1. (We exclude m = -1 to avoid dividing by 0; the natural logarithm ln|x| is an integral of x-1 = 1/x for any x≠ 0.) Integration is generally harder than differentiation, but many of the more familiar functions can be integrated by these and other rules (see the table).

A classic application of integration is to calculate areas. Let A be the area of the region between the graph of a function y = f(x) and the x-axis, for a≤x≤b. For simplicity, assume that f(x) ³ 0 between a and b. For each x³a, let L(x) be the area of this region to the left of x, so we need to find A = L(b). First we differentiate L(x). If h is a small change in x, the region below the graph between x and x + h is approximately a rectangle of height f(x) and width h (see figure 3); the corresponding change k = L(x + h) - L(x) in area is therefore approximately f(x)h, so k/h is approximately f(x). As h→ 0 these approximations become more exact, so k/h→ f(x) and hence L¢(x) = f(x). Thus L is an integral of f, so if we know any integral F of f then L = F + c for some constant c. Now L(a) = 0 (since the region to the left of x vanishes when x = a), so c = -F(a) and hence L(x) = F(x) - F(a) for all x³a. In particular, A = L(b) = F(b) - F(a), written

This is the Fundamental Theorem of Calculus, valid whenever f is continuous between a and b, provided we assign negative areas to any regions below the x-axis, where f(x) < 0. (Continuity means that f(x) → f(x0) as x→x0, so f has an unbroken graph.) For example f(x) = x2 has integral F( x) = x3/3, so

and with this formula it is possible to work out a large number of useful quantities. For example, the volume of a cone of height h and radius r can be found by evaluating the expression ∫ً (rx / h)2 dx between the limits x = 0 and x = 1; this is because the radius at a distance x below the apex of the cone is rx / h and the cross-sectional area is ً (rx / h)2. The result is ً r2 h / 3.
Here  is a definite integral of f; this is a number, whereas the indefinite integral ∫f(x)dx is a function F(x) (more precisely, a set of functions F(x) + c). The symbol ∫ (a 17th-century S) suggests summation of areas f(x)dx of infinitely many rectangles of height f(x) and infinitesimal width dx; more precisely,  is the limit of a sum of finitely many rectangular areas, as their widths approach 0.

The derivative dy/dx = f¢(x) of a function y = f(x) can be differentiated again to obtain a second derivative, denoted by d2y/dx 2, f¢¢(x) or D2f. If x is time and y is distance travelled, for instance, so that dy/dx is velocity v, then d2y/dx2 = dv/dx is rate of change of velocity, that is, acceleration. By Newton's second law of motion, a body of constant mass m subject to a force F undergoes an acceleration a satisfying F = ma. For example, if the body falls under the gravitational force F = mg (where g is the gravitational field strength) then ma = F = mg implies a = g, so dv/dx = g. Integrating, we have v = gx + c where c is constant; putting x = 0 shows that c is the initial velocity. Integrating dy/dx = v = gx + c, we have y = ½gx2 + cx + b where b is constant; putting x = 0 shows that b is the initial value of y.

Higher derivatives f(n)(x) = dn y/dxn = Dnf of f( x) are found by successively differentiating n times. Taylor's Theorem states that if f(x) can be represented as a power series f(x) = a0 + a1x + a2x2 + ... + anx n + ... (where a0,a1, ... are constants), then an = f(n)(0)/n! where 0!=1 and n!= 1 × 2 × 3 × ... × n for all n³ 1. Most commonly used functions can be represented as power series; for instance if f(x) = ex then f(n)(x) = ex for all n, so f(n)(0) = e0 = 1 and hence: