18.01 Single Variable Calculus Fall 2006

Transcription

1 MIT OpenCourseWre 8. Single Vrible Clculus Fll 6 For informtion bout citing these mterils or our Terms of Use, visit:

2 Lecture 8 8. Fll 6 Lecture 8: Definite Integrls Integrls re used to clculte cumultive totls, verges, res. Are under curve: (See Figure.). Divide region into rectngles. Add up re of rectngles 3. Tke limit s rectngles become thin b b (i) (ii) Figure : (i) Are under curve; (ii) sum of res under rectngles Emple. f() =, =, b rbitrry. Divide into n intervls Length b/n = bse of rectngle. Heights: ( ) b b st : =, height = n n ( ) b b nd : =, height = n n Sum of res of rectngles: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) b b b b b 3b b nb b = ( n ) n n n n n n n n n 3

3 Lecture 8 8. Fll 6 = b Figure : Are under f () = bove [, b]. We will now estimte the sum using some 3-dimensionl geometry. Consider the stircse pyrmid s pictured in Figure 3. n = 4 n Figure 3: Stircse pyrmid: left(top view) nd right (side view) st level: n n bottom, represents volume n. nd level: (n ) (n ), represents volumne (n ) ), etc. Hence, the totl volume of the stircse pyrmid is n + (n ) + +. Net, the volume of the pyrmid is greter thn the volume of the inner prism: n > (bse)(height) = n n = n nd less thn the volume of the outer prism: n < (n + ) (n + ) = (n + ) 3 3 3

4 Lecture 8 8. Fll 6 In ll, Therefore, b 3 nd the re under from to b is. 3 n n (n + ) 3 = 3 < < 3 n 3 n 3 3 n 3 b 3 lim ( n ) = b 3, n n 3 3 Emple. f() = ; re under bove [, b]. Resoning similr to Emple, but esier, gives sum of res: b n ( n) b (s n ) This is the re of the tringle in Figure 4. b b Figure 4: Are under f () = bove [, b]. Pttern: ( ) d b 3 = b db 3 ( ) d b = b db The re A(b) under f() should stisfy A (b) = f(b). 3

5 Lecture 8 8. Fll 6 Generl Picture y=f() c i b Figure 5: One rectngle from Riemnn Sum Divide into n equl pieces of length = Δ = b n Pick ny c i in the intervl; use f(c i ) s the height of the rectngle Sum of res: f(c )Δ + f(c )Δ + + f(c n )Δ n In summtion nottion: f(c i )Δ clled Riemnn sum. i= Definition: b n lim f(c i )Δ = f()d clled definite integrl n i= This definite integrl represents the re under the curve y = f() bove [, b]. Emple 3. (Integrls pplied to quntity besides re.) Student borrows from prents. P = principl in dollrs, t = time in yers, r = interest rte (e.g., 6 % is r =.6/yer). After time t, you owe P ( + rt) = P + P rt The integrl cn be used to represent the totl mount borrowed s follows. Consider function f(t), the borrowing function in dollrs per yer. For instnce, if you borrow $ /month, then f(t) =, /yer. Allow f to vry over time. Sy Δt = / yer = month. t i = i/ i =,,. 4

6 Lecture 8 8. Fll 6 f(t i ) is the borrowing rte during the i th month so the mount borrowed is f(t i )Δt. The totl is f(t i )Δt. i= In the limit s Δt, we hve f(t)dt which represents the totl borrowed in one yer in dollrs per yer. The integrl cn lso be used to represent the totl mount owed. The mount owed depends on the interest rte. You owe f(t i )( + r( t i ))Δt for the mount borrowed t time t i. The totl owed for borrowing t the end of the yer is f(t)( + r( t))dt 5

7 Lecture 9 8. Fll 6 Lecture 9: First Fundmentl Theorem of Clculus Fundmentl Theorem of Clculus (FTC ) If f() is continuous nd F () = f(), then b f()d = F (b) F () Nottion: F () b = F () =b = F (b) F () = 3 b 3 b b 3 3 Emple. F () =, F () = ; d = = Emple. Are under one hump of sin (See Figure.) π sin d = cos π = cos π ( cos ) = ( ) ( ) = ϖ Figure : Grph of f () = sin for π. Emple d = 6 = 6 = 6

8 Lecture 9 8. Fll 6 Intuitive Interprettion of FTC: d (t) is position; v(t) = (t) = is the speed or rte of chnge of. dt b v(t)dt = (b) () (FTC ) R.H.S. is how fr (t) went from time t = to time t = b (difference between two odometer redings). L.H.S. represents speedometer redings. n v( i)δ i= t t pproimtes the sum of distnces trveled over times Δt The pproimtion bove is ccurte if v( t) is close to v( t i ) on the i intervl. The interprettion of ( t) s n odometer reding is no longer vlid if v chnges sign. Imgine round trip so tht (b) () =. Then the positive nd negtive velocities v(t) cncel ech other, wheres n odometer would mesure the totl distnce not the net distnce trveled. Emple 4. π sin d = cos π = cos π ( cos ) =. The integrl represents the sum of res under the curve, bove the -is minus the res below the -is. (See Figure.) th + - ϖ Figure : Grph of f() = sin for π.

9 Lecture 9 8. Fll 6 Integrls hve n importnt dditive property (See Figure 3.) b c c f()d + f()d = f()d b b c New Definition: Figure 3: Illustrtion of the dditive property of integrls b f()d = b f()d This definition is used so tht the fundmentl theorem is vlid no mtter if < b or b <. It lso mkes it so tht the dditive property works for, b, c in ny order, not just the one pictured in Figure 3. 3

10 Lecture 9 8. Fll 6 Estimtion: b b If f() g(), then f()d g()d (only if < b) Emple 5. Estimtion of e Since e for, Thus e, or e. d e d e d = e = e e = e Emple 6. We showed erlier tht + e. It follows tht ( + )d e d = e 3 5 Hence, e,or, e. ( ) 3 ( + )d = + = Chnge of Vrible: If f() = g(u()), then we write du = u ()d nd g(u)du = g(u())u ()d = f()u ()d (indefinite integrls) For definite integrls: u f()u ()d = g(u)du where u = u( ), u = u( ) u ( ) 4 Emple d Let u = 3 +. Then du = 3 d = d = du ; 3 =, = = u = 3 + = 3, u = 3 + =, nd ( ) 4 4 du u d = u = =

11 Lecture 8. Fll 6 Lecture : Second Fundmentl Theorem Recll: First Fundmentl Theorem of Clculus (FTC ) If f is continuous nd F = f, then b f()d = F (b) F () We cn lso write tht s b =b f()d = f()d Do ll continuous functions hve ntiderivtives? Yes. However... Wht bout function like this? e d =?? The new function is defined s n integrl: F () = Yes, this ntiderivtive eists. No, it s not function we ve met before: it s new function. It will hve the property tht F () = e. If F () = e t dt / sin Other new functions include ntiderivtives of e, e,, sin( ), cos( ),... Second Fundmentl Theorem of Clculus (FTC ) = f(t)dt nd f is continuous, then F () = f() Geometric Proof of FTC : Use the re interprettion: F () equls the re under the curve between nd. ΔF = F ( + Δ) F () ΔF (bse)(height) (Δ)f() (See Figure.) ΔF Δ f() ΔF Hence lim Δ Δ = f() But, by the definition of the derivtive: ΔF lim Δ Δ = F ()

12 Lecture 8. Fll 6 y F() F + Figure : Geometric Proof of FTC. Therefore, F () = f() Another wy to prove FTC is s follows: [ ] +Δ ΔF = f(t)dt f(t)dt Δ Δ +Δ = Δ f(t)dt (which is the verge vlue of f on the intervl t + Δ.) As the length Δ of the intervl tends to, this verge tends to f(). Proof of FTC (using FTC ) Strt with F = f (we ssume tht f is continuous). Net, define G() = f(t)dt. By FTC, G () = f(). Therefore, (F G) = F G = f f =. Thus, F G = constnt. (Recll we used the Men Vlue Theorem to show this). Hence, F () = G() + c. Finlly since G() =, b f(t)dt = G(b) = G(b) G() = [F (b) c] [F () c] = F (b) F () which is FTC. Remrk. In the preceding proof G ws definite integrl nd F could be ny ntiderivtive. Let us illustrte with the emple f() = sin. Tking = in the proof of FTC, G() = cos t dt = sin t = sin nd G() =.

13 Lecture 8. Fll 6 If, for emple, F () = sin +. Then F () = cos nd b sin d = F (b) F () = (sin b + ) (sin + ) = sin b sin Every function of the form F () = G() + c works in FTC. Emples of new functions The error function, which is often used in sttistics nd probbility, is defined s erf() = e t dt π nd lim erf() = (See Figure ) Figure : Grph of the error function. Another new function of this type, clled the logrithmic integrl, is defined s dt Li() = ln t This function gives the pproimte number of prime numbers less thn. A common encryption technique involves encoding sensitive informtion like your bnk ccount number so tht it cn be sent over n insecure communiction chnnel. The messge cn only be decoded using secret prime number. To know how sfe the secret is, cryptogrpher needs to know roughly how mny -digit primes there re. You cn find out by estimting the following integrl: We know tht dt ln t ln = ln() (.3) = 46 nd ln = ln() 46 3

14 Lecture 8. Fll 6 We will pproimte to one significnt figure: ln t 5 for t. With ll of tht in mind, the number of -digit primes is roughly dt dt ( ) 9 ln t 5 = There re LOTS of -digit primes. The odds of some hcker finding the -digit prime required to brek into your bnk ccount number re very very slim. Another set of new functions re the Fresnel functions, which rise in optics: C() = cos(t )dt S() = sin(t )dt Bessel functions often rise in problems with circulr symmetry: π J () = π cos( sin θ)dθ On the homework, you re sked to find C (). Tht s esy! C () = cos( ) dt We will use FTC to discuss the function L() = t from first principles net lecture. The middle equlity in this pproimtion is very bsic nd useful fct b c d = c(b ) Think of this s finding the re of rectngle with bse (b ) nd height c. In the computtion bove, =, b =, c = 5 4

15 Lecture 8. Fll 6 Lecture : Applictions to Logrithms nd Geometry Appliction of FTC to Logrithms The integrl definition of functions like C(), S() of Fresnel mkes them nerly s esy to use s elementry functions. It is possible to drw their grphs nd tbulte vlues. You re sked to crry out n emple or two of this on your problem set. To get used to using definite integrls nd FTC, we will discuss in detil the simplest integrl tht gives rise to reltively new function, nmely the logrithm. Recll tht n+ n d = + c n + ecept when n =. It follows tht the ntiderivtive of / is not power, but something else. So let us define function L() by dt L() = t (This function turns out to be the logrithm. But recll tht our pproch to the logrithm ws firly involved. We first nlyzed, nd then defined the number e, nd finlly defined the logrithm s the inverse function to e. The direct pproch using this integrl formul will be esier.) All the bsic properties of L() follow directly from its definition. Note tht L() is defined for < <. (We will not etend the definition pst = becuse /t is infinite t t =.) Net, the fundmentl theorem of clculus (FTC) implies L () = Also, becuse we hve strted the integrtion with lower limit, we see tht dt L() = = t Thus L is incresing nd crosses the -is t = : L() < for < < nd L() > for >. Differentiting second time, L () = / It follows tht L is concve down. The key property of L() (showing tht it is, indeed, logrithm) is tht it converts multipliction into ddition: Clim. L(b) = L() + L(b) Proof: By definition of L(b) nd L(), b dt dt b dt b dt L(b) = = + = L() + t t t t

16 Lecture 8. Fll 6 b dt To hndle, mke the substitution t = u. Then t Therefore, This confirms L(b) = L() + L(b). dt = du; < t < b = < u < b b dt u=b du b du = = = L(b) t u= u u Two more properties, the end vlues, complete the generl picture of the grph. Clim. L() s. Proof: It suffices to show tht L( n ) s n, becuse the fct tht L is incresing fills in ll the vlues in between the powers of. L( n ) = L( n ) = L() + L( n ) = L() + L() + L( n ) = L() + L() + + L() (n times) Consequently, L( n ) = nl() s n. (In more fmilir nottion, ln n = n ln.) Clim 3. L() s +. ( ) Proof: = L() = L = L() + L(/) = L() = L(/). As +, / +, so Clim implies L(/). Hence L() = L(/), s + Thus L(), defined on < < increses from to, crossing the -is t =. It is concve down nd its grph cn be drwn s in Fig.. This provides n lterntive to our previous pproch to the eponentil nd log functions. Strting from L(), we cn define the log function by ln = L(), define e s the number such tht L(e) =, define e s the inverse function of L(), nd define = e L().. (,) to + to Figure : Grph of y = ln().

17 Lecture 8. Fll 6 Appliction of FTCs to Geometry (Volumes nd Ares). Ares between two curves f() y g() d b Figure : Finding the re between two curves. Refer to Figure. Find the crossing points nd b. The re, A, between the curves is A = b (f() g()) d Emple. Find the re in the region between = y nd y =. = y (4, ) (, ) y = (, ) (, -) Figure 3: The intersection of = y nd y =. 3

18 Lecture 8. Fll 6 First, grph these functions nd find the crossing points (see Figure 3). y + = = y y y = (y )(y + ) = Crossing points t y =,. Plug these bck in to find the ssocited vlues, = nd = 4. Thus the curves meet t (, ) nd (4, ) (see Figure 3). There re two wys of finding the re between these two curves, hrd wy nd n esy wy. Hrd Wy: Verticl Slices If we slice the region between the two curves verticlly, we need to consider two different regions. d = y (4, ) (, ) y = (, ) (, -) Figure 4: The intersection of = y nd y =. Where >, the region s lower bound is the stright line. For <, however, the region s lower bound is the lower hlf of the sidewys prbol. We find the re, A, between the two curves by integrting the difference between the top curve nd the bottom curve in ech region: { } 4 { } A = ( ) d + ( ) d = (y top y bottom ) d Esy Wy: Horizontl Slices Here, insted of subtrcting the bottom curve from the top curve, we subtrct the right curve from the left one. y= ( ) [ ] y A = ( left right ) dy = (y + ) y d = + y + y = +4 ( + ) = y= 4

19 Lecture 8. Fll 6 = y (4, ) (, ) dy y = ; ( = y +) (, ) (, -). Volumes of solids of revolution Figure 5: The intersection of = y nd y =. Rotte f() bout the -is, coming out of the pge, to get: y rotte n -y plne section by π rdins f() d z Figure 6: A solid of revolution: the purple slice is rotted by π/4 nd π/. We wnt to figure out the volume of slice of tht solid. We cn pproimte ech slice s disk with width d, rdius y, nd cross-sectionl re of πy. The volume of one slice is then: dv = πy d (for solid of revolution round the -is) Integrte with respect to to find the totl volume of the solid of revolution. 5

20 Lecture 8. Fll 6 Emple. Find the volume of bll of rdius. y d Figure 7: A bll of rdius The eqution for the upper hlf of the circle is y =. If we spin the upper prt of the curve bout the -is, we get bll of rdius. Notice tht rnges from to +. Putting ll this together, we find = ( ) ( ) π 3 4 V = πy d = π( )d = π = π 3 π 3 = π = One cn often eploit symmetry to further simplify these types of problems. In the problem bove, for emple, notice tht the curve is symmetric bout the y-is. Therefore, ( ) 3 V = π( )d = π( )d = π 3 (The svings is tht zero is n esier lower limit to work with thn.) We get the sme nswer: ( ) 3 ( ) = π 3 π π V = 3 = 4 π

21 Lecture 8. Fll 6 Lecture : Volumes by Disks nd Shells Disks nd Shells We will illustrte the methods of finding volume through n emple. Emple. A witch s culdron y Figure : y = rotted round the y-is. Method : Disks y thickness of dy Figure : Volume by Disks for the Witch s Culdron problem. The re of the disk in Figure is π. The disk hs thickness dy nd volume dv = π dy. The volume V of the culdron is V = π dy (substitute y = ) y π V = πy dy = π =

22 Lecture 8. Fll 6 If = meter, then V = π gives V = π m 3 = π ( cm) 3 = π 6 cm 3 6 liters ( huge culdron) Wrning bout units. If = cm, then V = π () = π 4 cm 3 = π 6 liters But cm = m. Why is this nswer different? The resolution of this prdo is hiding in the eqution. y = At the top, = = = cm. So the second culdron looks like Figure 3. By contrst, when cm cm Figure 3: The skinny culdron. = m, the top is ten times wider: = or = m. Our eqution, y =, is not scle-invrint. The shpe described depends on the units used. Method : Shells This relly should be clled the cylinder method. y Figure 4: = rdius of cylinder. Thickness of cylinder = d. Height of cylinder = y =.

23 Lecture 8. Fll 6 The thin shell/cylinder hs height, circumference π, nd thickness d. dv = ( )(π)d = V = ( )(π)d = π ( 3 )d = ( 4 ) ( ) ( ) π = π = π 4 = π = 4 4 (sme s before) Emple. The boiling culdron Now, let s fill this culdron with wter, nd light fire under it to get the wter to boil (t o C). Let s sy it s cold dy: the temperture of the ir outside the culdron is o C. How much energy does it tke to boil this wter, i.e. to rise the wter s temperture from o C to o C? Assume the 7 o C y o C Figure 5: The boiling culdron (y = = meter.) temperture decreses linerly between the top nd the bottom (y = ) of the culdron: T = 3y (degrees Celsius) Use the method of disks, becuse the wter s temperture is constnt over ech horizontl disk. The totl het required is How mny clories is tht? H = T (π )dy (units re (degree)(cubic meters)) = ( 3y)(πy)dy = π (y 3y )dy = π(5y y 3 ) = 4π (deg.)m 3 ( ) 3 cl cm # of clories = cm 3 deg (4π) = (4π)( 6 ) cl = 5 3 kcl m There re bout 5 kcls in cndy br, so there re bout ( ) # of clories = cndy br 3 5 cndy brs So, it tkes bout 5 cndy brs worth of energy to boil the wter. 3

24 Lecture 8. Fll 6 R velocity Figure 6: Flow is fster in the center of the pipe. It slows sticks t the edges (i.e. the inner surfce of the pipe.) Emple 3. Pipe flow Poiseuille ws the first person to study fluid flow in pipes (rteries, cpillries). He figured out the velocity profile for fluid flowing in pipes is: v = c(r r ) distnce v = speed = time cr v v=c(r -r ) R r Figure 7: The velocity of fluid flow vs. distnce from the center of pipe of rdius R. The flow through the nnulus (.k. ring) is (re of ring)(flow rte) re of ring = πrdr (See Fig. 8: circumference πr, thickness dr) v is nlogous to the height of the shell. 4

25 Lecture 8. Fll 6 r dr Figure 8: Cross-section of the pipe. R totl flow through pipe = v(πrdr) = c (R r )πrdr R ( R (R r r 4 ) R = πc r r 3 )dr = πc 4 π flow through pipe = cr 4 Notice tht the flow is proportionl to R 4. This mens there s big dvntge to hving thick pipes. Emple 4. Drt bord You im for the center of the bord, but your im s not lwys perfect. Your number of hits, N, t rdius r is proportionl to e r. N = ce r This looks like: R y = ce -r r Figure 9: This grph shows how likely you re to hit the drt bord t some distnce r from its center. The number of hits within given ring with r < r < r is r c e r (πrdr) We will emine this problem more in the net lecture. r 5

26 Lecture 3 8. Fll 6 Lecture 3: Work, Averge Vlue, Probbility Appliction of Integrtion to Averge Vlue You lredy know how to tke the verge of set of discrete numbers: or 3 Now, we wnt to find the verge of continuum. y 4. y=f(). 4 b Figure : Discrete pproimtion to y = f () on b. where nd nd The limit of the Riemnn Sums is Averge y + y y n n = < < n = b y = f( ), y = f( ),... y n = f( n ) n(δ) = b Δ = b n lim (y + + y n ) b = n n Divide by b to get the continuous verge b b f() d y + + y n lim = f() d n n b

27 Lecture 3 8. Fll 6 y= - re = ϖ/ Figure : Averge height of the semicircle. Emple. Find the verge of y = on the intervl. (See Figure ) ( π ) π Averge height = d = = 4 Emple. The verge of constnt is the sme constnt b b 53 d = 53 Emple 3. Find the verge height y on semicircle, with respect to rclength. (Use dθ not d. See Figure 3) equl weighting in θ different weighting in Figure 3: Different weighted verges.

28 Lecture 3 8. Fll 6 y = sin θ π π Averge = sin θ dθ = ( cos θ) = ( cos π ( cos )) = π π π π Emple 4. Find the verge temperture of wter in the witches culdron from lst lecture. (See Figure 4). m m Figure 4: y =, rotted bout the y-is. First, recll how to find the volume of the solid of revolution by disks. πy π V = (π ) dy = πy dy = = Recll tht T (y) = 3y nd (T () = o ; T () = 7 o ). The verge temperture per unit volume is computed by giving n importnce or weighting w(y) = πy to the disk t height y. T (y)w(y) dy w(y) dy The numertor is T πy dy = π ( 3y)ydy = π(5y y 3 ) = 4π Thus the verge temperture is: 4π = 8 o C π/ Compre this with the verge tken with respect to height y: T dy = ( 3y)dy = (y 5y ) = 85 o C T is liner. Lrgest T = o C, smllest T = 7 o C, nd the verge of the two is 7 + = 85 3

29 Lecture 3 8. Fll 6 The nswer 85 o is consistent with the ordinry verge. The weighted verge (integrtion with respect to πy dy) is lower (8 o ) becuse there is more wter t cooler tempertures in the upper prts of the culdron. Drt bord, revisited Lst time, we sid tht the ccurcy of your im t drt bord follows norml distribution : ce r Now, let s pretend someone sy, your little brother foolishly decides to stnd close to the drt bord. Wht is the chnce tht he ll get hit by stry drt? drt bord r₁ r₁ 3r₁ little brother Figure 5: Shded section is r i < r < 3r between 3 nd 5 o clock. To mke our clcultions esier, let s pproimte your brother s sector (the shded region in Fig. 5). Your brother doesn t quite stnd in front of the drt bord. Let us sy he stnds t distnce r from the center where r < r < 3r nd r is the rdius of the drt bord. Note tht your brother doesn t surround the drt bord. Let us sy he covers the region between 3 o clock nd 5 o clock, or of ring. 6 Remember tht prt probbility = whole 4

30 Lecture 3 8. Fll 6 r dr width dr, circumference πr weighting ce -r Figure 6: Integrting over rings. ( ) The ring hs weight ce r (πr)(dr) (see Figure 6). The probbility of drt hitting your brother is: 3r r ce r 6 πr dr ce r πr dr Denomintor: Recll tht = 5 3 is our pproimtion to the portion of the circumference where the little 6 brother stnds. (Note: e r = e ( r ) not (e r ) ) b ( ) e r b d re r dr = = e b + e e r = re r dr (Note tht e R s R.) R e R e r rdr = e r = + e = Figure 7: Norml Distribution. Probbility = 3r ce r πr dr 3r e r r dr 3r 6 r = 6 r = e r r dr = e r 3r ce r πr dr e r r dr 3 6 r r 5

31 Lecture 3 8. Fll 6 Probbility = e 9r + e 4r 6 Let s ssume tht the person throwing the drts hits the drtbord r r bout hlf the time. (Bsed on personl eperience with 7-yer-olds, this is relistic.) r P ( r r ) = = e r rdr = e r + = e r = e r = ( )9 ( )9 e 9r = e r = ( )4 ( ) 4 = e r = = 6 e 4r So, the probbility tht stry drt will strike your little brother is ( ) ( ) 6 6 In other words, there s bout % chnce he ll get hit with ech drt thrown. 6

32 Lecture 3 8. Fll 6 Compute Q = e d Volume by Slices: An Importnt Emple Figure 8: Q = Are under curve e ( ). This is one of the most importnt integrls in ll of clculus. It is especilly importnt in probbility nd sttistics. It s n improper integrl, but don t let those s scre you. In this integrl, they re ctully esier to work with thn finite numbers would be. To find Q, we will first find volume of revolution, nmely, V = volume under e r (r = + y ) We find this volume by the method of shells, which leds to the sme integrl s in the lst problem. The shell or cylinder under e r t rdius r hs circumference πr, thickness dr; (see Figure 9). Therefore dv = e r πrdr. In the rnge r R, R R = πe R e r πr dr = πe r + π When R, e R, V = e r πr dr = π (sme s in the drts problem) 7

33 Lecture 3 8. Fll 6 width dr r Figure 9: Are of nnulus or ring, (πr)dr. Net, we will find V by second method, the method of slices. Slice the solid long plne where y is fied. (See Figure ). Cll A(y) the cross-sectionl re. Since the thickness is dy (see Figure ), V = A(y) dy z A(y) y Figure : Slice A(y). 8

34 Lecture 3 8. Fll 6 y dy bove level of y in cross-section of re A(y) top view Figure : Top view of A(y) slice. To compute A(y), note tht it is n integrl (with respect to d) A(y) = e r d = e y d = e y e d = e y Q Here, we hve used r = + y nd e y = e e y nd the fct tht y is constnt in the A(y) slice (see Figure ). In other words, ce d = c e d with c = e y y fied ce - - Figure : Side view of A(y) slice. 9

35 Lecture 3 8. Fll 6 It follows tht V = A(y) dy = e y Q dy = Q e y dy = Q Indeed, Q = e d = e y dy becuse the nme of the vrible does not mtter. To conclude the clcultion red the eqution bckwrds: π = V = Q = Q = π We cn rewrite Q = π s e d = π An equivlent rescled version of this formul (replcing with / σ)is used: /σ e d = πσ /σ This formul is centrl to probbility nd sttistics. The probbility distribution e on πσ < < is known s the norml distribution, nd σ > is its stndrd devition.

36 Lecture 4 8. Fll 6 Lecture 4: Numericl Integrtion Numericl Integrtion We use numericl integrtion to find the definite integrls of epressions tht look like: b ( big mess) We lso resort to numericl integrtion when n integrl hs no elementry ntiderivtive. For instnce, there is no formul for 3 cos(t )dt or e d Numericl integrtion yields numbers rther thn nlyticl epressions. We ll tlk bout three techniques for numericl integrtion: Riemnn sums, the trpezoidl rule, nd Simpson s rule.. Riemnn Sum b Figure : Riemnn sum with left endpoints: (y + y y n )Δ Here, i i = Δ (or, i = i + Δ) = < < <... < n = b y = f( ), y = f( ),... y n = f( n )

37 Lecture 4 8. Fll 6. Trpezoidl Rule The trpezoidl rule divides up the re under the function into trpezoids, rther thn rectngles. The re of trpezoid is the height times the verge of the prllel bses: ( ) ( ) bse + bse y 3 + y 4 Are = height = Δ (See Figure ) y 4 y 3 ( ) y Figure : 3 + y 4 Are = Δ b Figure 3: Trpezoidl rule = sum of res of trpezoids. ( ) Totl Trpezoidl Are = Δ y + y + y + y + y + y y n + y n ( ) = Δ y + y + y y n + y n

38 Lecture 4 8. Fll 6 Note: The trpezoidl rule gives more symmetric tretment of the two ends ( nd b) thn Riemnn sum does the verge of left nd right Riemnn sums. 3. Simpson s Rule This pproch often yields much more ccurte results thn the trpezoidl rule does. Here, we mtch qudrtics (i.e. prbols), insted of stright or slnted lines, to the grph. This pproch requires n even number of intervls. y y y ₀ ₁ ₂ Figure 4: Are under prbol. ( ) y + 4y + y Are under prbol = (bse)(weighted verge height) = (Δ) 6 Simpson s rule for n intervls (n must be even!) ( ) Are = (Δ) [(y + 4y + y ) + (y + 4y 3 + y 4 ) + (y 4 + 4y 5 + y 6 ) + + (y n + 4y n + y n )] 6 Notice the following pttern in the coefficients:

39 Lecture 4 8. Fll 6 st chunk nd chunk 3 4 Figure 5: Are given by Simpson s rule for four intervls Simpson s rule: b Δ f() d 3 (y + 4y + y + 4y 3 + y y n 3 + y n + 4y n + y n ) The pttern of coefficients in prentheses is: 4 = sum = sum = sum 8 To double check plug in f() = (n even!). Δ Δ ( ( n ) ( n )) 3 ( ) = = nδ (n even) 3 4

40 Lecture 4 8. Fll 6 Emple. Evlute integrtion. d using two methods (trpezoidl nd Simpson s) of numericl + Figure 6: Are under (+ bove [, ]. ) By the trpezoidl rule: ( ) ( ( )) ( ) 4 4 Δ y + y + y = () + + = + + = By Simpson s rule: Ect nswer: ( ( )) Δ / 4 (y + 4y + y ) = = π π d = tn = tn tn = = /( + ) Roughly speking, the error, Simpson s Ect, hs order of mgnitude (Δ)

41 Em 3 Review 8. Fll 6 Lecture 5: Em 3 Review Integrtion. Evlute definite integrls. Substitution, first fundmentl theorem of clculus (FTC ), (nd hints?). FTC : d f(t) dt = f(t) d If F () = f(t) dt, find the grph of F, estimte F, nd chnge vribles. 3. Riemnn sums; trpezoidl nd Simpson s rules. 4. Ares, volumes. 5. Other cumultive sums: verge vlue, probbility, work, etc. There re two types of volume problems:. solids of revolution. other (do by slices) In these problems, there will be something you cn drw in D, to be ble to see wht s going on in tht one plne. In solid of revolution problems, the solid is formed by revolution round the -is or the y-is. You will hve to decide how to chop up the solid: into shells or disks. Put nother wy, you must decide whether to integrte with d or dy. After mking tht choice, the rest of the procedure is systemticlly determined. For emple, consider shpe rotted round the y-is. Shells: height y y, circumference π, thickness d Disks (wshers): re π (or π π ), thickness dy; integrte dy. Work Work = Force Distnce We need to use n integrl if the force is vrible.

42 Em 3 Review 8. Fll 6 Emple : Pendulum. See Figure Consider pendulum of length L, with mss m t ngle θ. The verticl force of grvity is mg (g = grvittionl coefficient on Erth s surfce) θ L mg mss m Figure : Pendulum. In Figure, we find the component of grvittionl force cting long the pendulum s pth F = mg sin θ. θ θ mg Figure : F = mg sin θ (force tngent to pth of motion).

43 Em 3 Review 8. Fll 6 Is it possible to build perpetul motion mchine? Let s think bout simple pendulum, nd how much work grvity performs in pulling the pendulum from θ to the bottom of the pendulum s rc. Notice tht F vries. Tht s why we hve to use n integrl for this problem. θ θ W = (Force) (Distnce) = (mg sin θ)(l dθ) W = Lmg cos θ θ = Lmg(cos θ ) = mg [L( cos θ )] In Figure 3, we see tht the work performed by grvity moving the pendulum down distnce L( cos θ) is the sme s if it went stright down. L θ L(-cosθ) Figure 3: Effect of grvity on pendulum. In other words, the mount of work required depends only on how fr down the pendulum goes. It doesn t mtter wht pth it tkes to get there. So, there s no free (energy) lunch, no perpetul motion mchine. 3