# Definite integral. Mathematics FRDIS MENDELU

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1 Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1

2 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the region bounded by y = f(x), the x-xis nd the lines x = nd x = b? The re cn be pproximted with the sum of rectngles: We cut the intervl [, b] into subintervls. Ech of these subintervls forms the bse of rectngle, where the height of the rectngle is equl to the vlue of the function f evluted t n rbitrry point from the given subintervl. The pproximtion improves s the rectngles become nrrower nd the number of rectngles increses. We define the re of the region to be the limit of the rectngle re sums s the rectngles become smller nd smller nd the number of rectngles we use pproches infinity. Such limit cn be defined even for more generl functions nd we cll it definite integrl. The definite integrls cn be defined in mny different wys, we will define the Riemnn definite integrl. Construction of the Riemnn integrl Let f be function defined on n intervl [, b] nd suppose tht f is bounded on this intervl. The sequence of points D = {x, x 1, x,..., x n } such tht = x < x 1 < x < < x n = b is sid to be prtition of the intervl [, b]. The intervls [x, x 1 ], [x 1, x ],..., [x n 1, x n ] re clled subintervls of the prtition. The number ν(d) = mx{x i x i 1, i = 1,,..., n} is clled norm of the prtition D, i.e., the norm of the prtition is the length of the longest subintervl of the prtition. We choose n rbitrry number from ech of the subintervls ξ 1 [x, x 1 ], ξ [x 1, x ],..., ξ n [x n 1, x n ] nd we denote Ξ = {ξ 1, ξ,..., ξ n } the set of these numbers. Then the sum σ(f, D, Ξ) = n f(ξ i )(x i x i 1 ) i=1 is clled the integrl sum ssocited to the function f, the prtition D nd the choice of the numbers ξ i in Ξ. 1

3 Integrl sum: ξ 1 ξ ξ ξ 4 ξ 5 ξ 6 = x x 1 x x x 4 x 5 x 6 = b σ(f, D, Ξ) = f(ξ 1 )(x 1 x ) + f(ξ )(x x 1 ) + f(ξ )(x x ) +f(ξ 4 )(x 4 x ) + f(ξ 5 )(x 5 x 4 ) + f(ξ 6 )(x 6 x 5 ) 6 = f(ξ i )(x i x i 1 ) i=1 Refinement of the prtition: ξ 1 ξ ξ ξ n = x x 1 x x n 1 x n = b σ(f, D, Ξ) = f(ξ 1 )(x 1 x ) + f(ξ )(x x 1 ) + + f(ξ n )(x n x n 1 ) n = f(ξ i )(x i x i 1 ) i=1

4 Definition (Riemnn integrl) Let f be function defined nd bounded on n intervl [, b]. Let D 1, D, D,..., D n,... be sequence of prtitions of [, b] which stisfies lim n ν(d n) = nd Ξ 1, Ξ, Ξ,..., Ξ n,... be sequence of the corresponding choices of numbers ξ i from subintervls of the prtitions. The function f is sid to be integrble on [, b] (in sense of Riemnn) if there exists number I R with the property lim σ(f, D n, Ξ n ) = I n for every sequence of prtitions (with the bove given property) nd for rbitrry prticulr choice of the points ξ i in Ξ n. The number I is sid to be Riemnn integrl of the function f on [, b] nd it is denoted I = f(x) dx. The number is clled lower limit of the integrl nd the number b is clled n upper limit of the integrl. We hve to distinguish the definite integrls from the indefinite integrls: Indefinite integrl is set of functions. Definite integrl is limit (number). We will see, tht there is connection between the definite nd indefinite integrls, since definite integrls cn be evluted using indefinite integrls.

5 Theorem (Sufficient conditions for integrbility) Let f be function which stisfies t lest one of the following conditions: 1 f is continuous on [, b], f is monotone on [, b], f is bounded on [, b] nd contins t most finite number of discontinuities on this intervl. Then the function f is integrble (in sense of Riemnn) on [, b], i.e., f(x) dx exists. Properties of the Riemnn integrl Theorem (Additivity nd homogenity with respect to the integrnd) Let f nd g be functions which re integrble on [, b], c R. Then the functions f + g nd cf re lso integrble on [, b] nd it holds: [f(x) + g(x)] dx = cf(x) dx = c f(x) dx + f(x) dx g(x) dx Theorem (Additivity with respect to the domin of integrtion) Let f be function defined of [, b], nd let c (, b) be ny number. Then the function f is integrble on [, b] if nd only if it is integrble on both the intervls [, c] nd [c, b] nd it holds: f(x) dx = c f(x) dx + c f(x) dx 4

6 Theorem Let f nd g be functions integrble on [, b] such tht f(x) g(x) on this intervl. Then It follows from the lst theorem tht f(x) dx g(x) dx. if g(x), then g(x) dx, i.e., integrl of the nonnegtive function is nonnegtive. Evlution of the Riemnn integrl Theorem (Newton - Leibniz formul) Let f be function integrble on [, b] nd let F be n ntiderivtive of f on (, b) which is continuous on [, b]. Then f(x) dx = [F (x)] b = F (b) F (). The previous theorem sys tht to clculte the Riemnn integrl of f over [, b], ll we need to do is: 1 find n ntiderivtive F of f, clculte the number F (b) F (). 5

7 Exmple 1 x dx = [ x ] = = 19. π sin x dx = [ cos x ] π = cos π + cos = 1. 1 x dx = ( x) dx + 1 x dx = ] [ x [ x + ] 1 = = 5. Theorem (Integrtion by prts) Let u, v be functions hving continuous derivtives on [, b]. Then u(x)v (x) dx = [u(x)v(x)] b u (x)v(x) dx. Exmple 1 x ln x dx = u = ln x u = 1 x v = x v = x [ ] x = ln x x x dx [ ] x = 4 ln 1 ln 1 1 = ln 1 [ 4 1 ] 1 x dx = ln 1 = ln

8 Theorem (Substitution method) Let f be function continuous on [, b] nd let ϕ be function which hs continuous derivtive ϕ on [α, β]. Further suppose tht ϕ(x) b for x [α, β]. Then β α f[ϕ(x)]ϕ (x) dx = ϕ(β) ϕ(α) f(t) dt. The formul in the theorem cn be used from left to right (the 1 st substitution method) nd from right to left (the nd substitution method). It my hppen in some prticulr cses tht the lower limit the upper limit fter the substitution. For this reson we introduce the following extension: Extension The symbol f(x) dx cn be extended for b s follows: f(x) dx =, f(x) dx = f(x) dx. b We hve two possibilities when evluting the definite integrl with using the substitution method: 1 We use the previous theorem, i.e., we trnsform the limits of the integrl nd then we use the Newton-Leibniz formul with the new limits. (We do not substitute the originl vrible into the ntiderivtive obtined fter the integrtion.) We do not use the previous theorem. We evlute the indefinite integrl (i.e., we substitute the originl vrible fter integrtion) nd then we pply the Newton-Leibniz formul with the originl limits. 7

9 Exmple Evlute π sin x cos x dx. 1 We trnsform the limits: π sin x cos x dx = t = sin x dt = cos x dx t 1 = sin = t = sin π = 1 = 1 t dt = [ t ] 1 = 1. We do not trnsform the limits: sin x cos x dx = t = sin x dt = cos x dx = Tedy t dt = t = sin x + c. π sin x cos x dx = [ sin x ] π = sin π sin = 1. Applictions of the Riemnn integrl in geometry The re under curve nd between two curves Let f be nonnegtive nd continuous function on [, b]. The re S of the region in the plne bounded by y = f(x), the x-xis nd the lines x = nd x = b is: S = f(x) dx Let f, g be continuous functions nd suppose f(x) g(x) for x [, b]. The re S of the region in the plne bounded by y = f(x), y = g(x) nd the lines x = nd x = b is: S = [f(x) g(x)] dx (The signs of f nd g re rbitrry.) 8

10 Volume of the solid of revolution I Let f be nonnegtive nd continuous function on [, b]. The volume V of the solid generted by revolving the region bounded by y = f(x), the x-xis nd the lines x = nd x = b bout the x-xis is: V = π f (x) dx Volume of the solid of revolution II Let f, g be nonnegtive continuous functions nd suppose f(x) g(x) for x [, b]. The volume V of the solid generted by revolving the region bounded by y = f(x), y = g(x) nd the lines x = nd x = b bout the x-xis is: V = π [ f (x) g (x) ] dx 9

11 Mny other res nd volumes cn be clculted using the bove formuls since we cn cut the given region into severl pieces which stisfy the bove conditions. S = S 1 + S = c [f(x) h(x)] dx + c [g(x) h(x)] dx Exmple (Volume of the cone) Find the formul for volume of cone such tht the rdius of the bse is r nd the ltitude of the cone is v. Solution: If the following tringle revolves bout the x-xis, we obtin the cone: V = π v ( r v x ) dx = πr v v x dx = πr v [ x ] v = πr v v = πr v 1

12 Exmple (Volume of the bll) Find the formul for the volume of bll with the rdius r. Solution: The eqution of circle with rdius r nd the center t [, ] is x + y = r. The upper hlf-circle is the grph of the function y = r x, the lower hlf-circle is the grph of the function y = r x. If the hlf-circle revolves bout the x-xis, we obtin bll. If qurter of circle revolves bout the x-xis, we obtin hlf of the bll. V = π r r = π [r x x (r x ) dx = π ] r = π r (r r (r x ) dx ) = 4πr Appliction of the definite integrl in physics Consider region bounded by the grphs of the functions y = f(x), y = g(x), nd the lines x =, x = b, where g(x) f(x) on, b. Suppose tht the region hs constnt density ρ. Then: Mss of the region: m = ρ [f(x) g(x)] dx Moments of force with respect to x-xis, y-xis: S x = 1 ρ [ f (x) g (x) ] dx, S y = ρ x [f(x) g(x)] dx Center of mss: T = [ Sy m, S ] x m 11

13 Exmple (Center of mss) Find the center of mss of the tringle given by the vertices [, ], [, ], [, ]. Suppose tht the density ρ is constnt. Mss: m = ρ Moments of force: S x = 1 ρ S y = ρ [ ] x x dx = ρ Center of mss: T 1 = S y m =, 4 9 x dx = 9 ρ x x dx = ρ [ x = ρ 9 = ρ ] = 9 ρ 9 = ρ x dx = ρ [ x T = S x m = = T = [, ] ] = ρ 9 = 6ρ Approximtion of definite integrls Numericl methods for pproximting the definite integrl the following cses: f(x) dx re used in The ntiderivtive of the function f hs no elementry formul, hence the Newton-Leibniz formul cnnot be used ( sin x x, sin x, ex x, e x,... ). A formul for the function f is not known, we hve only set of mesured vlues. 1

14 The trpezoidl rule The trpezoidl rule for the evlution definite integrl is bsed on pproximting the region between curve nd the x-xis with trpezoids. = x x 1 x x x 4 = b Let f be function bounded on [, b]. To evlute f(x) dx : We cut the intervl [, b] into n subintervls [x, x 1 ], [x 1, x ],..., [x n 1, x n ], (x =, x n = b). Suppose tht the length of ech subintervl is h = b n. Denote y = f(x ), y 1 = f(x 1 ),..., y n = f(x n ). We pproximte the function f on [x i 1, x i ], (i = 1,,..., n) with the liner function pssing through [x i 1, y i 1 ], [x i, y i ]. This liner function is of the form Hence xi x i 1 f(x) dx y = y i 1 + y i y i 1 h xi x i 1 (x x i 1 ). [ y i 1 + y ] i y i 1 (x x i 1 ) h dx 1

15 xi x i 1 f(x) dx = xi [ y i 1 + y ] i y i 1 (x x i 1 ) dx x i 1 h [ y i 1 x + y ] xi i y i 1 (x x i 1 ) h x i 1 = y i 1 h + y i y i 1 h = h (y i 1 + y i ), h which (in cse when f is positive function ) is the well-known formul for evluting n re of the trpezoid with corners [x i 1, ], [x i 1, y i 1 ], [x i, ], [x i, y i ]. Hence, f(x) dx = x1 x xn f(x) dx + f(x) dx + + f(x) dx x x 1 x n 1 h (y + y 1 ) + h (y 1 + y ) + + h (y n 1 + y n ) = h (y + y 1 + y + + y n 1 + y n ). The trpezoidl rule Let f be function bounded on [, b], nd let = x < x 1 < < x n = b be prtition of [, b] such tht the length of ech subintervl of this prtition is h = b n. Then f(x) dx h (y + y 1 + y + + y n 1 + y n ), where y = f(x 1 ), y = f(x ),..., y n = f(x n ). (We suppose tht f is defined t the points x, x 1,..., x n.) Some other rules for pproximting the definite integrls cn be used, e.g., the so-clled Simpson s rule is bsed on pproximting curves with prbols insted of lines. 14

16 Using the computer lgebr systems Wolfrm Alph: Mthemticl Assistnt on Web (MAW): wood.mendelu.cz/mth/mw-html/index.php?lng=en&form=min Exmple Using the Wolfrm Alph find the integrl π sin x dx. Solution: integrte sin x dx from x= to pi 15

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