1 The Definite Integral

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1 The Definite Integrl Definition. Let f be function defined on the intervl [, b] where<b.thenfis Riemnn integrble on [, b] ifthereisnumberl stisfying the following: For every ">, there is > such tht for ll smled rtitions (P, S), if P <,then nx f(s i ) i L <". i= Reclling tht R S P (f) =P n i= f(s i) i,wewrite The it L is denoted by P! RS P(f) = P! nd is clled the definite integrl of f on [, b]. nx f(s i ) i = L. i= f() d The net theorem gve us method for ctully evluting definite integrls. Theorem. Suose tht f is Riemnn integrble on [, b]. Let (P n, S n ) be sequence of smled rtitions of the intervl [, b] such tht n! P n =. Then n! RSn P n (f) = Theorem. Suose tht <b. If f is Riemnn integrble on [, b] nd le c<dle b, thenf is Riemnn integrble on [c, d]. In rticulr, f is Riemnn integrble on [, d]. The Men Vlue of Function Definition 4. Suose tht <bnd tht f is Riemnn integrble on [, b]. Then, we define the men vlue, or verge vlue, of f on [, b] tobe b Theorem 5 (Men Vlue Theorem for Integrtion). Suose tht f is continuous on the closed intervl [, b]. Then, there eists c in [, b] such tht f() d = f(c)(b ). Thus, if <b,thereeistsc in the intervl [, b] such tht f(c) equls the men vlue of f on [, b].

2 THE FUNDAMENTAL THEOREM OF CALCULUS The Fundmentl Theorem of Clculus Definition 6. Let F () be function. Let nd b be rel numbers. Then F () b = F (b) F (). We shll write F () b rther thn F () b. question: The erly mthemticins sked the following Question: Is there wy to determine the vlue of definite integrl without evluting the it of the Riemnn Sums? Leibniz nd Newton both nswered yes to the bove question by discovering the following theorem. Theorem 7 (Fundmentl Theorem of Clculus, Prt ). If function f() is continuous function on the intervl [, b] with n nti-derivtive given by F () [tht is, F () =f()], then f() d = F () b = F (b) F (). Problem 8. In the revious section we evluted the definite integrl R 4 d by evluting the it of Riemnn Sums nd we obtined the vlue. Now evlute the sme definite integrl by using the Fundmentl Theorem of Clculus (FTC). Solution. To evlute the definite integrl R 4 d using the FTC, we must find the nti-derivtive F () of the function f() =. We know tht F () = R d = nd so, Z 4 d = le 4 = 4 = (4 )= 6 (64 ) = =. Remrk. Given function f(), in order to evlute R b f() d using rt of the FTC, the function f() must be defined nd continuous t ech oint in [, b]. Problem 9. Evlute the definite integrls using the FTC: Z Z ( + ) d. d. cos() d. sin() d. sec() tn() d. Problem. Suose f() hs the nti-derivtive given by. Evlute Z Problem. Suose f() hs the nti-derivtive given by F (), where F () = nd F (5) =. Evlute Z 5

3 Remrk. Let f be Riemnn integrble on [, b]. For ech in the intervl [, b] weknowby Theorem tht R f(t) dt is the re under f over the intervl [, ]. Thus, we cn define new function F by F () = R f(t) dt, for ech in [, b]. Emle. Let f() be the function on the intervl [, ] whose grh ers in the following figure. For ech in the intervl [, ], we cn define new function F by F () = R f(t) dt, which is the re under the grh of y = f() from to Y is y = f() F() = the re under f u to X is Are under f from to The erly mthemticins lso sked the following question: Question: If function f is continuous, then does it hve n nti-derivtive? This question ws nswered in the ositive when the following theorem ws estblished. Theorem (Fundmentl Theorem of Clculus, Prt ). If function f() is continuous on the intervl [, b], then the function is such tht F () =f(). F () = Z f(t) dt Problem 4. Find the derivtives of the following functions of :. F () =. G() =. H() = 4. F () = Z Z Z +t dt. sin(5t) dt. Z sin() +t 4 dt. (cos(t)+t ) dt.

4 4 THE FUNDAMENTAL THEOREM OF CALCULUS Emle. Let f() be the function on the intervl [, ] defined by f() = + whose grh ers in the following figure. Then for ech in the intervl [, ], F () = R (t + ) dt is the re under the grh of y = f() from to F() = the re under f u to f() = + 4 Y is X is Are under f() = + from to () Using rectngles nd tringles, evlute F (). () Using rectngles nd tringles, evlute F (). () Using F () = R (t + ) dt nd FTC rt, evlute F (). (4) Using your evlution of F () in () or (), determine F (). (5) Using F () = R (t + ) dt nd FTC rt, determine F ().. The Substitution Rule for Definite Integrl Procedure (Substitution Rule). To evlute the definite integrl erform the following five stes: f(g()) g () d () Ste : Let u be n inside rt such tht du is the left over, obtining: u = g() du = g ()d Z Ste : Mke the substitution into (), obtining f(u) du. Ste : Evlute the new lower it g() nd the new uer it g(b). Ste 4: Use the bove the results to obtin: Ste 5: Then f(g()) g () d = Z g(b) g() Z g(b) g() f(u) du. f(u) du.

5 . The Substitution Rule for Definite Integrl 5 Problem. Evlute the definite integrl Solution. Let u be the inside rt Z u = g() = + ( +) ( + ) d. du =( + ) d = ( + ) d We need the constnt in order to crry out the substitution. Now evlute g() nd g(), obtining g() = nd g() =. So, Z ( +) ( + ) d = Problem. Evlute the definite integrl Solution. Note tht cos (t)sin(t) dt = = = Z Z le u ( +) ( + ) d u du = cos (t)sin(t) dt. le = 9. [cos(t)] sin(t) dt. Let u be the inside rt u = g(t) = cos(t) du = sin(t)dt We need the constnt in order to crry out the substitution. Now evlute g() nd g( ), obtining g() = nd g( )=. So, [cos(t)] sin(t) dt = = = Z le 4 u4 [cos(t)] ( u du = sin(t)) dt le 4 ( )4 4 ()4 = Problem. Evlute the following definite integrls using the substitution rule.... Z 5 Z +d. sin(5) d. + 4 d.