For smoothly changing continuous functions {integrand}, with y-axis range and x-axis domain, you can calculate area between curve and x-axis {integral} {first integral} {integration, calculus}, from lower domain value {lower limit} to higher domain value {upper limit}. For example, you can calculate enclosed-surface area and volume.
summation
See Figure 1. Domain goes continuously from lower value x1 to higher value x2, while range goes continuously from lower value y1 to higher value y2. Length x2 - x1 can divide into number N of intervals with equal widths (x2 - x1)/N.
In Figure 1, dashed line divides x2 - x1 interval into two intervals, each with width (x2 - x1)/2, because N = 2.
Range at left of each small interval is f(x1 + (ni - 1) * (x2 - x1)/N), for ith interval. For i = 1, it is f(x1). For i = 2, it is f(x1 + (x2 - x1)/2). Range halfway between left and right of each small interval is f(x1 + (ni - 1/2) * (x2 - x1)/N). For i = 1, it is f(x1 + -(1/2) * (x2 - x1)/2). For i = 2, it is f(x1 + (3/2) * (x2 - x1)/2). Range at right of each small interval is f(x1 + ni * (x2 - x1)/N). For i = 1, it is f(x1 + (x2 - x1)/2). For i = 2, it is f(x1 + 2 * (x2 - x1)/2). Product of range and interval width is rectangular area. For example, using left of each small interval, (f(x1 + (ni - 1) * (x2 - x1)/N)) * (x2 - x1)/N, for ith interval. For i = 1, it is f(x1) * (x2 - x1)/2. For i = 2, it is (f(x1 + (x2 - x1)/2)) * (x2 - x1)/2. Sum of all interval areas approximates total area between curve and x-axis, between domain values. As interval number increases, widths decrease, and area sum approaches true area.
interval position
The three different ways of taking interval range do not matter, because total area is same. For example, using right of each small interval, (f(x1 + ni * (x2 - x1)/N)) * (x2 - x1)/N, for ith interval. For i = 1, it is (f(x1 + (x2 - x1))/2) * (x2 - x1)/2. For i = 2, it is f(x1) * (x2 - x1)/2. Total area is the same.
number of intervals
Number of intervals does not matter. Use function f(x) = x^2, as in parabola. Interval is x1 = 0 to x2 = b. Number of subintervals is N = 3. (x2 - x1)/N = b/3. If f(x) is at midpoint of each interval, sum from x1 = 0 to x2 = b of f(x1 + (ni - 1/2) * (x2 - x1)/N) * (x2 - x1)/N is ((b/6)^2 + (b/2)^2 + ((5 * b)/6)^2) * (b/3), which is (b^3)/3. If function f(x) = x^2, x1 = 0, x2 = b, and N = 6, sum from x1 = 0 to x2 = b is ((b/12)^2 + ((3 * b)/12)^2 + ((5 * b)/12)^2 + ((7 * b)/12)^2 + ((9 * b)/12)^2 + ((11 * b)/12)^2) * (b/6), which is (b^3)/3. Therefore, results for different numbers of intervals are the same.
line
If f(x) = x, function is line. See Figure 2. Sum from x1 = a to x2 = b with N = 1 of f(x1 + (ni - 1/2) * (x2 - x1)/N) * (x2 - x1)/N is ((b + a)/2) * (b - a), which is area of trapezoid of base b - a and heights a and b.
constant
If f(x) = C, function is constant. See Figure 3. Sum from x = b to x = a with N = 1 of f(x1 + (ni - 1/2) * (x2 - x1)/N) * (x2 - x1)/N is C * (b - a), which is area of rectangle with height C and length b - a.
definite integral
Given function f(x) and interval (b - a), you can calculate integral from x = a to x = b of f(x) * dx {definite integral}. Domain-value variable dx {dummy variable} {variable of integration} does not appear in definite-integral result, because domain values over interval replace it.
indefinite integral
Without using interval, formula or other method can calculate integral {indefinite integral} {anti-differential} {antiderivative} {antiderived function}. Antiderivatives are functions from which original function can derive by differentiation.
Because derivatives of constants equal zero, function antiderivatives differ by a constant {constant of integration}. Knowing original-function domain and range allows calculating constant.
Degree-greater-than-four integrals {hyperelliptic integral} can be rational elliptic-integral products.
Equations {integral equation} can represent an infinite number of ordinary differential equations.
For two-independent-variable functions, calculate integral {double integral} by holding first variable constant and integrating over second variable, then holding second variable constant and integrating over first variable, and then adding results. It does not matter which variable is first. Double-integral domain can be surface {closed region} inside closed curve.
To find surface area {area, integration}, take double integral over differential area (1 + (df(x,y) / dx)^2 + (df(x,y) / dy)^2)^0.5 * dx * dy.
Triple integral, over volume, of scalar product of del operator and vector function equals double integral, over surface, of scalar product of function and normal vector to surface {divergence theorem}.
If two solids have equal altitudes and all sections parallel to base have same ratio, volumes have same ratio {Cavalieri's theorem} {Cavalieri theorem}.
The gradient of scalar electric potential is vector electric field. A scalar function has a divergence of the gradient {Laplacian}: D^2f/Dx^2 + D^2f/Dy^2 + D^2f/Dz^2, where D is partial derivative. Potential relates to charge density as Laplacian of potential equals negative of charge density divided by electrostatic constant (Poisson's equation). If charge density is zero, Laplacian of potential equals zero (Laplace's equation).
Curved-surface area is not the limit of surface's plane-triangle areas {Schwarz's paradox} {Schwarz paradox}.
In a plane region, if function has no singularities, is single-valued, and solves the potential equation, double integral of ((du/dx)^2 + (du/dy)^2) * dx * dy over x and y has a minimum {Dirichlet principle} {Thomson principle}.
Integral between two curve points is integral {line integral} {curvilinear integral} from a to b of p(x, f(x)) * dx, where f(x) is curve function, and p is surface function.
For closed curve, line integral over line equals double integral over closed region {Green's theorem} {Green theorem}. Line integral over regular simply connected closed curve equals zero. Line integrals over any two regular curves between two region points are equal. For closed surface, double integral over surface equals triple integral over closed volume.
Double integral, over surface, of cross product of del operator and vector function, equals line integral, over boundary curve, of vector function {Stokes theorem}.
In complex plane, paths {closed contour} can loop around origin. There is line integral {contour integration} around the path. Looping once increases integral by 2 * pi * i. Looping counterclockwise once increases integral by -2 * pi * i.
Integration paths over complex functions do not matter {Cauchy integral theorem}. For complex function f(z), integral of f(z) = F(z) = (1 / (2 * pi)) * (integral from p = -pi to p = +pi of (x * f(x) / (x - z)) * dp), where z is complex number, and x = |z| * e^(i*p) {Cauchy integral formula}. Cauchy integral formula makes series {majorant series} and has residue {integral residue}.
To integrate constant times function {constant times function integration}, take integral of k * u(x) * dx = k * (integral of u(x) * dx), where k is constant, and u(x) is function.
Integral from a to b of f(x) * dx equals f(mean) * (b - a) {linear function, integration}. Integral of sum equals sum of integrals.
To integrate power function {power function, integration}: Increase exponent by one and divide by new exponent. Integral of x^p * dx = x^(p + 1) / (p + 1), if p != -1 (p <> -1), so integral of x^3 = x^4 / 4. Integral from x = 1 to x = b of (1/x) * dx equals ln(b).
Integral of e^x = e^x {exponential function, integration}. Integral of b^x = (1 / ln(b)) * b^x.
Integral of ln(e^x) = integral of x {logarithmic function, integral}.
For degree 1, 2, or 3 polynomials P(x), definite integral over interval (a,b) is ((b - a) / 6) * (P(a) + P(a + b) / 2 + P(b)) {midpoint rule}.
Integral of sin(x) = -cos(x) {trigonometric function, integration}. Integral of cos(x) = sin(x). Integral of tan(x) = - ln(|cos(x)|). Integral of cot(x) = ln(|sin(x)|). Integral of sec(x) = ln(|sec(x) + tan(x)|). Integral of csc(x) = ln(|csc(x) - cot(x)|). Integral of (sin(x))^2 = (x - sin(x) * cos(x)) / 2. Integral of (cos(x))^2 = (x + sin(x) * cos(x)) / 2. Integral of (tan(x))^2 = tan(x) - x. Integral of (cot(x))^2 = -cot(x) - x. Integral of (sec(x))^2 = tan(x). Integral of (csc(x))^2 = -cot(x).
If numerator polynomial has higher degree, divide polynomials to get quotient and remainder. Then integrate quotient, integrate remainder, and add results {ratios of polynomials integration}.
To integrate sum of functions {sum of functions integration}: Integral of |u(x) + v(x)| * dx = integral of u(x) * dx + integral of v(x) * dx. Integral of |u(x) - v(x)| * dx = integral of u(x) * dx - integral of v(x) * dx.
Definite integrals {improper integral} can have infinity as limit or integrate over an open interval. To integrate improper integrals, take integral limit as variable approaches infinity. If limit is infinity, split domain at value zero, integrate over both intervals, and add results.
To find curved-figure areas and volumes, add many small discrete triangular or trapezoidal areas {method of exhaustion} {exhaustion method}.
Magnitudes can be infinite numbers of small units {indivisibles method} {method of indivisibles}. Cavalieri invented this calculus forerunner [1629].
For degree-n functions, definite integral over (a,b) is ((b - a) / (3*n)) * (sum from k = 0 to k = n of c * f(a + k * (b - a) / n)), where n is even integer, c = 4 if k is odd, c = 2 if k is even, and c = 1 if k = 0 {Simpson's rule} {Simpson rule}. Error is less than or equal to M * (b - a)^5 / (180 * n^4), where M is less than or equal to fourth-derivative absolute value.
For functions sin^m(x), cos^m(x), and cos^m(x) * sin^m(x), where m is positive integer, such as sin(x) and sin^2(x), reduction formulas {Wallis's formula} {Wallis formula} can evaluate definite integrals from x = 0 to x = pi/2.
Integral of u*dv equals u*v minus integral of v*du {parts integration} {integration by parts}.
For two functions that depend on the same variable {composite function integration}, integral of u(x) * dv(x) = u(x) * v(x) - integral of v(x) * du(x).
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Date Modified: 2022.0225