Language learning | English » Area calculation using basic methods and integral calculus

Area calculation using basic methods and integral calculus BASIC AREA CALCULATION AREA OF POLYGONS: We regard a positive number associated to every polygon, which represents its (surface) area. The following properties must stand: • We need a unit. Let us agree that the area of a square with sides being one (length) unit is our one unit of area. • Identical plane figures must have equal areas. • If we divide a polygon into two polygons, then the sum of the areas of the new polygons must be equal to the area of the original polygon. (Thus, we can divide it into any finite number of polygons.) Area of a rectangle: (a difficult but rather nice proof, it is not obligatory) • The area of a square with sides n ( n ∈ Ζ + ) is n 2 , because we can divide it into n 2 square units, if we construct lines parallel to the sides, one unit in distance. • The area of a rectangle with sides n and m ( n, m ∈ Ζ + ) is nm, because we can divide it into nm square unit, following the

previous sequence of thought. 1 1 ( n ∈ Ζ + ) is 2 , because we can divide the square • The area of a square with sides n n 2 unit into n identical squares, as in the previous two cases. • The area of a rectangle with sides a and b ( a, b ∈ R + ). Let us construct segments 1 parallel to the sides, in distance. Let m and k be numbers ( m, k ∈ N + ) that n 1 1 1 1 m ≤ a < (m + 1) and k ≤ b < (k + 1) n n n n • From the property of area derives that the area of ABCD is bigger than the area of AB’C’D’, but smaller than AB”C”D”. If we denote the area of ABCD with t, we get 1 1 1 1 m k ≤ t < (m + 1) (k + 1) n n n n 1 1 1 1 • and from m and k, it is clear that m k ≤ ab < (m + 1) (k + 1) n n n n • From these two inequalities, we get m k 1 1 1 a b 1 a + b +1 . t − ab < (m + 1)(k + 1) 2 − mk 2 = 2 + 2 + 2 < + + = n n n n n n n n n t − ab has to be smaller than any positive number, because a + b + 1 is a constant, and we can choose n to

be arbitrarily great, thus t = ab. Thus, the area of a rectangle is T = a ⋅ b , where a and b are the sides. • Area of a parallelogram: Let us construct a perpendicular segment to one of the sides, such that it shall contain one of the vertices of the other side. Using this method, we have cut the parallelogram into pieces which fit together to form a rectangle. 1 Thus, the area of a parallelogram is T = a ⋅ ma , where a is a side, and m a is the altitude to a. It is trivial, that we can calculate the area of any regular quadrilateral with the help of the preceding two methods. Area of a triangle: If we complete a triangle to a parallelogram, we have doubled its area, therefore the area is half the area of the parallelogram. a ⋅ ma Thus, the area of a triangle is: T = , where a is a side, and m a is the altitude for side a. 2 Theorem: (with an easy proof) The area of a triangle can be calculated if two sides and the enclosed angle are known, such that: a ⋅ c ⋅ sin β

. T= 2 Proof: • Let us suppose that we know sides a and c and their enclosed angle β, in addition to a ⋅ ma that, the formula T = is known. 2 • • ma , thus we can conclude that ma = c ⋅ sin β , c irrespectively of β being an acute or an obtuse [because sin β = sin (π − β ) ] or even a right [because then sin β = 1 and the two sides serve as altitudes] angle. a ⋅ c ⋅ sin β Substituting this into the known formula we get: T = . 2 In the right angle triangle ∆ ABT sin β = Area of polygons: Every convex polygon can be divided into triangles, if we connect one vertex with every other vertex. Every concave polygon can be divided into a convex polygon with the lines of the sides lengthened. Then, we can follow the previous sequence of thought AREA OF THE CIRCLE: The area of the circle is T = r 2π . 2 INTEGRAL CALCULUS Integral calculus expresses the area under a curve, that is, the area under a specific function. DEFINITIONS: Let f (x) be a limited

function defined on [a; b] , the units: a = x0 < x1 < . < x n −1 < x n = b n S = ∑ mi ( xi − xi −1 ) , def: Lower sum: i =1 where mi is the lower limit of the function on interval [x i-1 , x i ]. n S = ∑ M i ( xi − xi −1 ) , def: Upper sum: i =1 where M i is the upper limit of the function on interval [x i-1 , x i ]. def: Definite integral: We regard function f defined on [a; b] integrable. As we refine the units of partitioning, the upper and lower sums tend towards the same number. If there exists only one number, which is not smaller than any of the lower sums, and not greater than any of the upper sums of function f¸ then this number is called the definite integral of function f on interval [a; b] . def: Integral function: If function f can be integrated on the interval [a; b] , then function F is the integral function of f: f : [a; b] R def: Primitive function: A primitive function of function f (x) (provided that it exists) is a function,

whose derivative is f (x) . def: Indefinite integral: The set of primitive functions of f (x) . THEOREMS: 1) Rules of indefinite integral: If functions f(x) and g(x) are integrable on interval [a; b] , then the following differential quotients exist in the same point, and the following relations are true: • a ∫ f ( x)dx = 0 a • b ∫ a a f ( x)dx = − ∫ f ( x)dx b 3 • • b b a a ∫ cf ( x)dx = c ∫ f ( x)dx b b a • b ∫ ( f ( x) ± g ( x))dx =∫ f ( x)dx ± ∫ g ( x)dx a a c b c a a b ∫ f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx 2) If function f (x) is continuous at every point of interval [a; b] , then F (x) , the integral function of f (x) is differentiable at every point of [a; b] , and F ′( x) = f ( x) 3) The primitive functions of a given function may only differ in a constant. 4) Newton-Leibniz theorem: if f (x) is a continuous function defined on interval [a; b] , and F (x) is a primitive function of it, then: b ∫ f ( x)dx

= F (b) − F (a) a 5) The primitive functions of some basic functions: • f ( x) = c F ( x) = cx • • f ( x) = x n where n ≠ −1 1 f ( x) = x • f ( x) = a x • f ( x) = e x F ( x) = • f ( x) = log a x n +1 x , n +1 • • • 1 (x ln x − x ) ln a f ( x) = ln x F ( x) = x ln x − x f ( x) = sin x F ( x) = − cos x f ( x) = cos x F ( x) = sin x • f ( x) = tan x F ( x) = − ln cos x • f ( x) = cot x F ( x) = ln sin x F ( x) = F ( x) = ln x 1 x a ln a F ( x) = e x F ( x) = 6) The volume of a body of revolution: b V = π ∫ ( f ( x) ) dx 2 a APPLICATIONS OF INTEGRAL CALCULUS 1. 2. 3. 4. 5. Finding the area under a function. Area, surface area and volume calculation. Calculations with bodies of revolution. Finding the centroid of figures. Finding the inertial momentum of bodies (of revolution) rotating around an axis (for example a solid plate). 6. Using it in sciences like mathematics, geography (land survey) and physics 4