My English lesson Calculation of areas y O a b x BE CAREFUL! B Th Weierstrass theorem states The that a continuous function in a closed interval has a minimum and a maximum (see Unit 3). Our objective in this section is to determine the area of the surface subtended by the graph of a function y = f(x) defined and continuous in a closed interval [a ; b]. Let us start by considering the case in which such a function is always positive in [a ; b] and proceed, as in the case analyzed in Unit 4 regarding the case of a parabola, by dividing the interval [a ; b] into subintervals of equal amplitudes. Thus: b a Q by dividing the interval [a ; b] into 2 parts each of them will have amplitude _; 2 b a Q by dividing the interval [a ; b] into n parts, each of them has amplitude h = _. n The Weierstrass theorem assures us that, in each of the n closed intervals [a ; a + h], [a + h ; a + 2h], the function has, both a minimum mini(f) and a maximum maxi(f) where by i we have denoted the generic interval. These two values represent the heights of one of the rectangles that are respectively below and above the graph of the function so each of the products min1(f) h ; min2(f) h ; ... ; mini(f) h ; ... ; minn(f) h is the area of one of the rectangles that are below the graph of the function. The sum: n y s n = mi n i(f) h i=1 min, ( f ) gives, then, the area of the figure (beside in colour) inscribed in the surface identified by the x-axis and the graph of the function, in the interval [a ; b]. O a Similarly, the products: h bx max1(f) h ; max2(f) h ; ... ; maxi(f) h ; ... ; maxn(f) h are the areas of the rectangles that are above the graph of the function and the sum: n S n = max i(f) h i=1 gives the area of the figure (beside in colour) circumscribed by that bounded by the x-axis and the graph of the function, in the interval [a ; b]. maxi ( f ) y O a h bx In considering the following values of n N, we construct the sn succession of the areas of the inscribed figures and the Sn succession of the areas of the circumscribed figures. The sn succession is nondecreasing and upper bounded while Sn is nonincreasing and lower bounded; since the function is continuous, as n increases, the amplitude h decreases and the two areas get closer and closer because the gap between the maximum and the minimum of the function tends to zero. In every single part of the interval: lim s n = lim S n n BE CAREFUL! B Th sn succession is formed as The follows: s1 = inscribed area of the base rectangle [a ; b]; s2 = inscribed area of the 2 rectangles having as bases 2 equal parts of [a ; b]; ... sn = inscribed area of the n rectangles having as bases n equal parts of [a ; b]. In a similar way the Sn sequence is formed, with the circumscribed areas. 412 n The common value of this limit is by definition the area of the surface identified by the x-axis and the graph of the continuous function y = f(x), in the closed interval [a ; b]. example O Let us determine, both geometrically and analytically (through the definition of definite integral), the area subtended by the 1 graph of the function y = _ x in the interval 2 [1 ; 4]. 1 The function y = _ x has as its graph the 2 straight-line passing through the origin in the figure beside. Geometrically 1 15 areaABCD = areaOBC areaOAD = 4 _ = _ 4 4 C y 1 O D A B x

My English lesson