5/9/17. Lesson 51 - FTC PART 2. Review FTC, PART 1. statement as the Integral Evaluation Theorem as it tells us HOW to evaluate the definite integral

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1 Lesson - FTC PART 2 Review! We hve seen definition/formul for definite integrl s n b A() = lim f ( i )Δ = f ()d = F() = F(b) F() n i=! where F () = f() (or F() is the ntiderivtive of f() b! And hve seen n interprettion of the definite integrl s net/totl chnge 2 FTC, PART! Some tetbooks/resources refer to this sttement s the Integrl Evlution Theorem s it tells us HOW to evlute the definite integrl b b f ()d = F() = F(b) F() by finding ntiderivtives 3

2 A Grphic Representtion! We need to visulize our equtions & terms to keep our ides cler in our minds! Our function in question will be y = f(t)! The re under the curve of y = f(t) we hve clled A(), but it will now be clled g()! So here is the grph: A Grphic Representtion! So, s we hve done in pst eercises, the vlue of the upper limit (b) is being vried its vlue is chnging from nywhere between nd b (i.e. < < b)! Therefore, the re is constntly chnging s result of the vlue of! in other terms, the re is function of the vlue of! hence A() or rther now g()! So here is the grph: A Grphic Representtion of the Are Function! Let s work with the re under the curve f(t) = t 2, strting from = so our integrl will be A() =?? t 2 dt! Now, let s set the upper limit to = nd we get A() = t 2 dt = 3 6 2

3 A Grphic Representtion of the Are Function! Now, let s set the upper limit to = 2 nd we get 2 A() = t 2 dt = 8 3! Now, let s set the upper limit to = 3 nd we get 3 A() = t 2 dt = A Grphic Representtion of the Are Function! Now, let s set the upper limit to = nd we get A() = t 2 dt = 6 3! Now, let s set the upper limit to = nd we get A() = t 2 dt = A Grphic Representtion from Before! Now tht we hve seen few clcultions nd some dt points, let s see wht we relly hve??? 2 3 Are /3 8/3 27/3 6/3 2/3! We hve the function A() =

4 FTC, Prt 2! So wht does g() = f (t)dt REALLY men??! Try this s simple contrst! we hve looked s integrtion s process tht genertes fied numbers (tht correspond to net res under curves)! Now, with different notion! the vlue of the upper limit is vrible, we re now considering g() = f (t)dt s FUNCTION, not single number! Wht does the function represent! the ntiderivtive of y = f(t) FTC, Prt 2! So here s the shift in our understnding! the epression g() = f (t)dt! IS AN EXPRESSION for the ntiderivtive of y = f(t) which grphiclly could be understood s cumultive AREA function A Clrifying Emple (I Hope)! Let s go bck to y = f(t) = t 2 between = nd b =! So we hve worked out the integrl t 2 dt = 3 t 3 = 2 3! So now s the vlue of b chnges, we get t 2 dt = 3 t 3 = = 3 3! Where the ntiderivtive (/3 3 ) cn be used to evlute for the re under the curve 2

5 A Clrifying Emple (I Hope)! So we need to recognize the equivlence of the following 2 sttements: A() = 3 3 nd A() = t 2 dt! In one sttement, the ntiderivtive is given s simple, eplicit function, in the other, the ntiderivtive function is defined s n integrl of f(t)! To become consistent in our nottions, I will switch the sttements to g() = 3 3 nd g() = t 2 dt 3 A Clrifying Emple (I Hope)! Let s work with the function y(t) = t 2! So, gin, if y(t) = t -2 represents derivtive of some function, let s find tht other function! in other words, let s find the ntiderivtive of y(t)! We know the ntiderivtive to be Y(t) = -t - + c! Note tht we could hve sked the SAME thing s n indefinite integrl! Y(t) = t 2 dt = t + c A Clrifying Emple (I Hope)! Let s work with the grph of the ntiderivtive of on DESMOS y(t) = t 2

6 A Clrifying Emple (I Hope)! Let s work with the grph of the ntiderivtive of y(t) = on DESMOS t 2! Notice tht the ntiderivtive eqution hs been presented in different mnner! s n integrl!! 6 Y(t) = t 2 dt = t + A Clrifying Emple (I Hope)! Let s work with the grph of the ntiderivtive of y(t) = t 2 on DESMOS in the form of Y(t) = t 2 dt = t + c c 7 FTC, Prt 2! If f is continuous on [,b] then the fcn hs derivtive t every point in [,b] nd d d F() = d d! So wht does this relly men Every continuous function f() HAS n ntiderivtive (which simply hppens to be epressed s n integrl s: f (t)dt = f () F() = f (t)dt ) rther thn eplicitly in terms of elementry functions! f (t)dt 8 6

7 Using the FTC, Prt 2! Find the derivtive of the following functions g() = g() = g() = sint dt! For ll rel numbers of, define F() = Evlute 3 F ʹ nd ʹ F 2 2 π sint dt sint dt nd interpret sin( πt)dt 9 Using the FTC, Prt 2! Find the derivtive of the following functions (Composites) g() = sint dt 3 2 g() = sint dt g() = t 3 + dt sin g() = t 3 + dt 2 Using the FTC, Prt 2! Find dy/d if:! Find function, y = f() with derivtive of y = y = 2 cost dt 3t sint dt 2 y = dt 2 + e t 2 dy d = tn tht stisfies the condition f(3) = 2 7

8 Using the FTC! Given the function f be s shown nd let g() =! () Evlute g(), g(2), g(), g(7) nd g(9) nd g()! (b) On wht intervls is g incresing?! (c) Where does g hve mimum vlue? f (t)dt! (d) Sketch rough grph of g() 22 Using the FTC, Prt 2! Grph the function defined by F() = e t 2 dt! Address the following in your solution:! (i) determine nd discuss F ()! (ii) determine nd discuss F ()! (iii) find symmetry of F()! (iv) estimte some points using trpezoid sums 23 Using the FTC, Prt 2! We cn use the methods of differentil clculus to nlyze these functions!!! Find the intervl on which the curve! is concve up! Find the intervl on which the curve is incresing y( ) = + t + t dt 2 y( ) = + t + t dt 2 2 8

9 Using the FTC, Prt 2! Given tht F() = + t dt 2! () Find ll criticl points of F! (b) Determine the intervl on which F increse nd F decreses! (c) Determine the intervls of concvity nd inflection points of F! (d) Sketch grph of F 2 FTC, Prt 2! Wht is the dvntge of defining n ntiderivtive s n integrl? => we cn then simply use our numericl integrtion methods (RRAM, LRAM etc) to estimte vlues! =v&pid=sites&srcid=zgvmyxvsdgrvbwfpbnhcgnh bgnbhvzzxhhbxfzxnw9uc3nedoowyzte N2VkMjkNmZi 26 9