4.2 Definite Integral

Transcription

1 4. Defiite Itegrl Bri E. Veitch 4. Defiite Itegrl Whe we compute the re uder curve, we obtied limit of the form lim f(x i ) x. x=1 This sme limit shows up whe we cosider fidig the distce give the velocity, d it turs our tht this limit shows up i plety of other situtios. As it will pper for the ext, ohhh, forever, we give it specil defiitio: Defiitio 4.. If f is fuctio defied over the itervl [, b], we divide [, b] ito subitervls of equl width x = b. We let x =, x, x,..., x 1, x = b be the edpoits of these subitervls d we let x 1, x,..., x be y smple poits i the subitervls, so x i [x i 1, x i ]. The, the defiite itegrl of f from to b is f(x)dx f(x i ) x, x = b x i = + i x for right-hd edpoits provided tht this limit exists d gives the sme vlue for ll possible choices of smple poits. If this limit does exist, we sy tht f is itegrble over [, b]. For ottio, f(x)dx, 31

2 4. Defiite Itegrl Bri E. Veitch we sy tht is the itegrl sig, is the lower limit d b is the upper limit of itegrtio. The dx idictes tht the idepedet vrible is x ll other vribles my be treted s costts. Also, the itegrl f(x)dx is umber it does ot deped o x, d there is othig specil bout x. For the fuctio f, we hve f(x)dx = f(t)dt = f(y)dy, for y vrible s plceholder. Lstly, we sy tht the sum f(x i ) x is clled Riem sum. If our fuctio of iterest, f(x), is lwys o-egtive, the we c tret the Riem sum s sum of res of rectgles. Sice the defiite itegrl is the limit of Riem sum, defiite itegrl is the limit of the sum of re of rectgles, d thus the re uder the curve f(x) from x = to x = b. If we cosider fuctio f tht does tke o egtive vlues, such s fuctio like the oe below, 313

3 4. Defiite Itegrl Bri E. Veitch we cot just tke the heights of ll the pproximtig rectgles some of these heights would be egtive, d tht would t mke whole lot of sese to hve egtive re. The wy we c get roud this is to cosider the et re we fid the re of the curve bove the x-xis. The, seprtely, fid the re of the curve uder the x-xis. We the dd these two res together. So if A 1 is the re bove the x-xis but uder f(x) computed by bse times height, x f(x i ) d A is the re uder the x-xis but bove f(x), lso computed by bse times height, x f(x i ), we hve Are of shded regio = A 1 A, sice A is egtive. Exmple 4.6. Evlute 5 1. Let s tke look t the grph (3 x) dx d the re of the shded regio. 314

4 4. Defiite Itegrl Bri E. Veitch Do you see how some of the shded regio is bove the x-xis d some of it is below. Whe we evlute 5 (3 x) dx, it would cosider the re below egtive. We c do it two differet wys t this poit. Let s do it without clculus.. Do you see how the two shded regios re trigles? Guess wht? We kow the formul for the re of trigle, A = 1 bh. Are of top trigle: A = 1 (3)(3) = 9 Are of bottom trigle: A = 1 ()() = So our itegrl is 5 (3 x) dx = 9 = 5 =.5 3. Now if you wt re of the shded regio (pretedig like the re below the x-xis is lso positive), the wht we relly wt to evlute is 5 3 x dx 315

5 4. Defiite Itegrl Bri E. Veitch So the re of the shded regio is 5 3 x dx = 9 + = 13 = 6.5 Theorem 4.1. If f is cotiuous o [, b], or if f hs oly fiite umber of jump discotiuities, the f is itegrble o [, b] meig tht the defiite itegrl f(x)dx exists. This theorem is NOT esy to prove, by y stretch of the word esy. We do ot do it here, but it does tell us somethig. It sys tht there re fuctios which re ot itegrble. I order to simplify the clcultios, we c choose specific smple poits, d the right most edpoits re s good s y other. So, we hve hlf defiitio, hlf theorem: Defi-thereom: If f is itegrble o [, b], the f(x)dx f(x i ) x where x = b 316

6 4. Defiite Itegrl Bri E. Veitch d x i = + i x. I order to evlute these itegrls, we hve to be ble to work with some very commoly see sums the followig three equtios will be isely vluble i doig this: 1 = ( = 1) i = i ( + 1)( + 1) = 6 ( ) ( + 1) i 3 = The remiig rules tht will help evlute sums re very similr to rules tht let us evlute limits: c =c c i =c ( i + b i ) = ( i b i ) = i i + i b i b i Exmple 4.7. Let s fid the Riem sum for f(x) = 3 x by tkig right-hd edpoits over the itervl [,5]. By the wy, we kow the swer should be 5. I ll tke you through the procedure. Do this wheever you re sked to fid evlute the Riem Sum. 1. Fid x: x = b = 5 = 5 317

7 4. Defiite Itegrl Bri E. Veitch. Fid x i : 3. Fid f(x i ): x i = + i x = + i 5 = 5i f(x i ) = 3 x i ( ) 5i = 3 4. Fid A i : A i = f(x i ) x ( = 3 5i ) 5 = 15 5i 5. Fid A i: A i = 15 5i 1 5 = 15 = 15 5 = 15 i 5( + 1) ( ) ( + 1) 6. Our fil step is tke the limit s. 5( + 1) True Are 15 = 15 5 =.5 which is exctly wht we got from the before. 318

8 4. Defiite Itegrl Bri E. Veitch Exmple 4.8. Fid the Riem sum for f(x) = x 3 4x by tkig smple poits to be right edpoits, where =, b = d = 4. With = 4, the itervl width is 4 = 1, d the right edpoits re x 1 =.5, x = 1, x 3 = 1.5 d x 4 =. The Riem sum is R 4 = 4 f(x i ) x = x (f(.5) + f(1) + f(1.5) + f()) = 1 (( ) ( ) ( 4) = 5 ) + (16 8) Note tht sice this fuctio does dip egtive, this vlue is NOT the pproximtio for the re uder the curve. However, it does represet the differece i positive d egtive res of the pproximtig rectgles of the curve. But, this is just pproximtio. Now, let s evlute With subitervls, we hve x 3 4xdx. x = b =. d we hve x =, x 1 = /, x = 4/, d i geerl, x i = i/. We use right edpoits 319

9 4. Defiite Itegrl Bri E. Veitch d our theorem to let us evlute x 3 4xdx f(x i ) x ( ) i f ( ( ) 3 i 4 ( 16i 3 3 8i ) ( ) ) i ( ) 16 i 3 8 i 3 ( ( ) 16 ( + 1) 8 ( ) ) ( + 1) 3 ( ) 4( + 1) 4( + 1) ( + 1) =8 lim / 1/ =8 lim 1 + 1/ 1 1 =8 (1 1) = Clerly, this itegrl cot be iterpreted s the re uder curve, sice this curve clerly does ot hve re. However, it c be iterpreted s differece betwee the positive d egtive re. If we grph the fuctio, we should see tht the positive d egtive res re ideticl: 3

10 4. Defiite Itegrl Bri E. Veitch This is NOT how we will be computig these soo. But just like s with derivtives, we hve to go through it the log wy first before we get to the shortcuts. Exmple 4.9. Set up expressio for We let f(x) = x 7, = 3 d b = 7. Thus, 3 x = 7 3 x 7 dx s limit of sums. Do ot evlute. = 4. We hve x 1 = , x = 3 + 4,..., d we hve geeric term By our theorem, we hve 3 x 7 dx x i = 3 + 4i. f(x i ) x f 4 ( 3 + 4i ( 3 + 4i This is ot very esy. Soo eough we ll get to the shortcuts tht llow us to evlute this quickly. By the wy, ) ) x 7 dx =

11 4. Defiite Itegrl Bri E. Veitch Exmple 4.1. Evlute by iterpretig the itegrl s re. 4 x dx Sice f(x) = 4 x for x i [, ], we c iterpret the itegrl s the re uder y = 4 x from to. We c rewrite this s y = 4 x, which gives x + y = 4, which shows tht the grph of f is prt of circle with rdius 4 =. Sice our prt of the circle is stuck i Qudrt I, we hve qurter circle. Thus, we tke qurter of the re: A = 1 4 ( π r ) = 1 4 ( π ) = π. Exmple Evlute 3 3x 5 dx. 3

12 4. Defiite Itegrl Bri E. Veitch We re beefited by grphig this first: I order to fid the itegrl, ote tht we get positive vlues for the trigle bove the x-xis d egtive vlues from the trigle below the x-xis. We fid the res of these two trigles, d subtrct. First, fid the x-itercept of 3x 5 =, which gives x = 5 3. The upper trigle the hs bse equl to 3 5 = 4, with height of f(3) = 3(3) 5 = 4. Thus, 3 3 the upper trigle hs re A T = = 8 3. Similrly, the lower trigle, which hs bse 5 3 d height f() = 5 hs re A B = = 3 6. Thus, the itegrl is 33

13 4. Defiite Itegrl Bri E. Veitch 3 3x 5dx = A T A B = = 9 6 = Midpoit Rule These re ll good itegrls, but for some we still eed to pproximte, ythig with curved side, s mtter of fct, which is ot prt of circle. We so fr hve chose just left d right edpoits, d we sw i Sectio 4.1 how those gve us rther bd pproximtios. Better, perhps, to use the midpoit of the itervls! Defiitio 4.3 (The Midpoit Rule:). f(x)dx f( x i ) x = x (f( x 1 ) + f( x ) f( x )), where d x = b x i = 1 (x i 1 + x i ), which is the midpoit of the itervl [x i 1, x i ]. Exmple 4.1. Use the midpoit rule with = 4 to pproximte 1 1 x dx. We strt with x = 1 4 = 1 4. The righthd edpoits of the 4 itervls re x 1 = 1/4, x = 1/, x 3 = 3/4, x 4 = 1. To fid the midpoits, dd cosecutive pirs d divide by : 34

14 4. Defiite Itegrl Bri E. Veitch Thus, the midpoit rule gives x 1 = + 1/4 = 1 8 1/4 + 1/ x = = 3 8 1/ + 3/4 x 3 = = 5 8 x 4 = 3/4 + 1 = dx = x (f(1/8) + f(3/8) + f(5/8) + f(7/8)) x = 1 ( 1 4 1/ / /8 + 1 ) 7/8 = 1 ( 1 4 1/ / /4 + 1 ) 7/4 = 1 ( ) 7 = Now, sice the fuctio is lwys positive, we c thik of this s the re uder the curve. However, we still hve o ide how good of pproximtio this is. I this cse, it s horrible sice the re is. Exmple Let s evlute x dx usig the midpoit d 6 rectgles. Recll we did this exmple usig left d right hd edpoits. It took us 1 rectgles to get estimte of.747. We proved the exct re is

15 4. Defiite Itegrl Bri E. Veitch 1. x = 6 = 1 3. Fid x i x 1 = = 1 3 x = = 3 x 3 = = 3 3 x 4 = = 4 3 x 5 = = 5 3 x 6 = = 6 3 so the midpoits re x 1 = 1 6 x = 3 6 x 3 = 5 6 x 4 = 7 6 x 5 = 9 6 x 6 =

16 4. Defiite Itegrl Bri E. Veitch 3. The pproximte re is A 1 3 [f(x 1) + f(x ) + f(x 3 ) + f(x 4 ) + f(x 5 ) + f(x 6 )] [ (1 A 1 ) ( ) ( ) ( ) ( ) ( ) ] A.648 So oly usig 6 rectgles we got much closer to the true re th we did with 1 rectgles! I believe you eed somewhere roud 19 righthd rectgles for the sme pproximtio s 6 midpoit rectgles! 4.. Properties of the Defiite Itegrl Whe we defied the itegrl f(x)dx, we hve so fr defied this whe < b. However, if b <, the Riem Sum defiitio still works, but x chges from b b to = (b ). Thus, f(x)dx = b f(x)dx. Further, if = b, the x = b b =, d s such b f(x) dx =. 37

17 4. Defiite Itegrl Bri E. Veitch Further, to help evlute defiite itegrl, we hve the followig four properties, s log s f d g re cotiuous fuctios: 1. c dx = c(b ), for y costt c (f(x) + g(x)) dx = f(x) dx + cf(x) dx = c (f(x) g(x)) dx = g(x) dx f(x) dx for y costt c f(x) dx g(x) dx Provig these is quite simple tsk. Property 1 follows from the fct tht we re itegrtig costt height, d our figure would just be rectgle re of tht is height c times width b. Properties, 3 d 4 re ll prove similrly, d we prove property 3 here: cf(x)dx (cf(x i )) x c =c lim =c 38 f(x i ) x f(x i ) x f(x)dx

18 4. Defiite Itegrl Bri E. Veitch Thus, costt, but ONLY costt c be pulled out i frot of itegrl sig. Exmple Use the bove properties to fid 3 6x dx. Usig the differece property, we hve By property 1, we hve Also, 3 6x dx = 3 dx 3 dx = 3(7 ) = 15. 6x dx = 6 x dx, 6x dx but to evlute x dx, we eed to tret this s Riem Sum, with x = 7 d x i = + 5i = 5 39

19 4. Defiite Itegrl Bri E. Veitch x dx ( + 5i ) i + 5i [ ] i i ( 4 + ( + 1) + 5 ) ( + 1)( + 1) 6 ( ) 4 + 1( + 1) + =5 lim ( = ) 3 = ( + 1)( + 1) Thus, we hve ( ) x dx = 15 6 = = A fifth property ivolves combiig the bouds o two itegrls over the sme fuctio: c f(x)dx + f(x)dx = c which c esily be see by the picture below: f(x)dx, 33

20 4. Defiite Itegrl Bri E. Veitch After ll, if we wt to fid the re uder f(x) from to b, we c split it somewhere i the middle d dd those two res. Exmple Suppose tht 1 f(x) dx = 13 d f(x) dx = 3. Fid 1 7 f(x)ldx. The swer here lies i simple pplictio from the followig equtio: Therefore, 1 f(x) dx = 13 = 3 + f(x) dx f(x) dx f(x) dx 1 7 f(x) dx =

21 4. Defiite Itegrl Bri E. Veitch This property is true regrdless if < c < b or ot. It does t mtter t ll. Ay other cofigurtio would simply be rerrgemet of the vribles. However, the followig three properties re oly true for < b: 6) If f(x) for x b, the f(x) dx 7) If f(x) g(x) for x b, the f(x) dx g(x) dx 8) If m f(x) M for x b, the m(b ) f(x) dx M(b ). Property 6 implies positive fuctio gives positive res. Property 7 sys tht bigger (higher) fuctio will hve lrger re. Duh. Property 8 sttes tht is fuctio is trpped betwee two horizotl lies, d s such, the re uder the curve is trpped betwee the res of two rectgles. Noe of these relly eed to be prove, but we we c demostrte pictorilly: Exmple Estimte the vlue of Tke look t the grph, e x dx. 33

22 4. Defiite Itegrl Bri E. Veitch From the grph, you c tell tht it is DECREASING fuctio. You c lso fid this out by the first derivtive test. Ufortutely, we do ot ler the derivtive of expoetil fuctios util ext semester. You c see from the grph, our fuctio hs the smllest y vlue t x =, which is e 4. It hs its highest y-vlue t x =, which is e = 1. Therefore, Thus, by property 8, we hve e 4 ( ) which gives e 4 < e x < e = 1. e 4 e x dx 1( ) e x dx. 333