The Function Integral and Area under a Curve

Sub-Interval Summation The integral is used to find the area of a function. It is an additive process. The two basic principles for this process are (1) subdividing the function by increasingly small intervals and (2) using the definite integral to offer a closer approximation of the function area. For (1), the Riemann sum equation can be used to denote a quotient of range to n, the number of arbitrary subintervals: range n -1 In this broad definition, ∆x summation can be thought of as descendent from method of exhaustion. Described by Archimedes, it is a concept of measurement by filling a shape or space (Apostol, 1967). The following sections describe integrals to 'fill' a space under a curve and are based on Jordon measure.

The Riemann Sum Equation The sum of subintervals exists in a two dimension Cartesian plane. For a continuous and nonnegative function f for the interval [a, b] divided into subintervals, the space becomes demarcated by a series of rectangles each with amplitude at f. The representative points on the function curve x1, x2, …,xn, equal the subinterval height. The area under f is defined by a definite integral: x a b f ( x ) = n n f x 1 Δx f x 2 Δx ⋯ f x n Δx Where Δx b -1 a n -1 . How the point of representation intersects with f and the sub-interval area effects how the Riemann sum will approach the function area. For cases where the sub-intervals intersect with the (nonnegative) function at the right- end point of the rectangle. This point can be arbirarily changed to the mid-point or left- end point of the rectangle. The effect of this control can be thought of as the Riemann sum approximation approaching the function area from above or below the function line.

The Area between Two Functions The fundamental theorem of calculus states that for function f, the definite integral with limit a to b, the anti-derivative F is scalable from a to b: x a b f x F b -1 F a Given: x 2 4 x 12 , find the area. Solution: 23 x 32 , [2,4] = 163 -1 4 2 12 3 -1 ≃ 3.4477 The theorem uses the definite integral to find the area of function f(x) between a and b. This technique can be expropriated to find the area between two functions.

Area Between Functions 'f' and 'g' For values of x greater than the point of intersect, area is calculated: (G - F).Given two functions f and g (Figure 1), where f(x) is greater than g(x), an area exists between the two. The following equation defines the area between points a and b: x a b f x -1 g x Figure 1 shows an instance of mirroring the integral to the function value in relation to the other at the point of intersection.

The Area of a Radial Set This section has been adapted from Apostol, Tom M. "Calculus" One-Variable Calculus, with an Introduction to Linear Algebra, vol. 1. Blaisdell Publishing. 1967. (pp.109-110):

Area Defined by a Radial Set The area between a and b where 0 ≤ b – a ≤ 2π, for non-negative function f, has an area defined by a radial set S of a step function s. The circular arc is the summation of n subintervals ( θ k -1 , θ k ) for which the width of s is constant. The radial set has an angle of θ k - θ k -1 . Therefore, the area of each sector is 1 - 2 ( θ k - θ k -1 ) s k 2 and the area for the set is defined: a[S] = 12 k 1 n s k 2 • ( θ k -1 θ k -1 ) = 12 k 1 n s k 2 • θ k -1 θ k -1