Chapter 5. Numerical Integration


 Megan Robinson
 2 years ago
 Views:
Transcription
1 Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with similr nmes: Reverse of differentition Indefinite integrl f(x) dx = most generl ntiderivtive for f(x) Definite integrl This is relted to summtion (it is limit of sums of certin kind). The integrl sign ws originlly invented s modified S (for sum). There is no reson to expect connection between these two different things, but there is. See course S, the book by Anton or the ppendix for more detils. Wht we need is the ide of Riemnn sum. 5.2 The Definite Integrl. b b number nd does not involve x. Nottion is convenient. Grphicl interprettion: grph y = f(x), the xxis nd the verticl lines x =, x = b. f(x) dx (red integrl from to b of the function f). Answer is f(x) dx is the re of the region in the plne bounded by the
2 Mthemtics S3 (Timoney) (picture good for cse f(x) 0) Problems: need wy to compute the re of such shpe. More fundmentlly, need definition of wht you men by the re. 5.3 Nottion. We need nottion to write down formule for these pictures. Deling with b f(x) dx n subdivisions of intervl [, b] Number division point = x 0 < x < x 2 < < x n = b Heights of rectngles y j = f(x j), x j x j x j ny point. Are of jth rectngle = width height = (x j x j )y j Totl re = sum of these = y (x x 0 ) + y 2 (x 2 x ) + + y n (x n x n ) = f(x )(x x 0 ) + f(x 2)(x 2 x ) + + f(x n)(x n x n ) (clled Riemnn sums for the integrl)
3 Numericl Integrtion Definition. b f(x) dx = limit of these Riemnn sums s n nd mx width Theorem (Importnt Theorem). This limit mkes sense if f is continuous on the finite closed intervl [, b] (including end points). Proof of this is too hrd for us. 5.6 Nottion. We cn use Sigm nottion for sums to mke the formule look shorter. For numbers u, u 2,..., u n 5.7 Exmples. u i mens u + u u n i= 5 j 2 = = = 55 j= 25 k= 2k 2 + k = (2 + ) + ( ) + + (2(25) ) 5.8 Remrk. Usul choice for Riemnn sums: ll n subintervls eqully wide. Common width is then h = b n Then x j = + jh for 0 j n Riemnn sum becomes f(x j)(x j x j ) = j= f(x j)h = h f(x j) j= j= Limit of these is the integrl. Limit hrd to find directly s rule, but computer cn find the sum for lrge n. 5.9 Trpezoidl Rule. The trpezoidl rule is technique for finding definite integrls b f(x) dx numericlly. It is one step more clever thn using Riemnn sums. In Riemnn sums, wht we essentilly do is pproximte the grph y = f(x) by step grph nd integrte the step grph. In the Trpezoidl rule, we pproximte y = f(x) by continuous grph mde up of bits of lines.
4 Mthemtics S3 (Timoney) Use n equl divisions. h = b n. x i = + ih. Let y i = f(x i ) (for 0 i n). The trpezoidl rule formul cn be written b ( f(x) dx = h 2 y 0 + y + y 2 + y n + ) 2 y n. Proof (of this formul) uses re of ith trpezoid = h 2 (y i + y i ) 5.0 Exmple. Find 3 e x2 dx pproximtely using the Trpezoidl rule with n = 0. Solution: i x i y i Weight Weight y i / /2 Sum Sum times h is Remrk. A nturl question to sk t this point is: how ccurte is the Trpezoidl rule?
5 Numericl Integrtion 5 Theoreticlly we know tht s n, the trpezoidl rule pproximtion b f(x) dx, but tht does not help us to know how close we re to the limit if we use n = 00 or n = 000. The following theorem gives worst cse scenrio. 5.2 Theorem. Let T n denote the result of using the trpezoidl rule formul with n steps to pproximte b f(x) dx. Then b f(x) dx T n b 2 h2 M 2 where M 2 is the lrgest vlue of f (2) (x) = f (x) for x b. Note: Cn rewrite this in terms of n using h = (b )/n. We won t prove this, or sy nything bout how to show it is true. 5.3 Exmple. (i) In 3 ex2 dx with n = 0 how fr off cn the trpezoidl rule be? Solution: For this we need to know M 2 nd tht involves f (x) with f(x) = e x2. f (x) = 2xe x2 f (x) = 2e x2 + 2x(2x)e x2 = (2 + 4x 2 )e x2 It is firly cler then tht (in this cse) f (x) > 0 lwys (so tht f (x) = f (x)) nd the lrgest vlue for 0 x 3 is M 2 = f (3) = 38e 9 = The theorem tells us then tht the pproximtion T 0 differs from the ctul integrl by t most (or t worst) 3 e x2 dx T 0 b 2 h2 M 2 = 3 2 h2 (30797). The vlue of h is h = b n (0.04)30797 = = 3 0 = 0.2 nd so the error could be s lrge s (ii) How lrge should we chose n so tht the trpezoidl rule pproximtion T n to the sme integrl is certinly within 0.5 of the right vlue? Solution: It will certinly be enough to choose n so tht b 2 h2 M 2 = 3 ( ) 2 3 M 2 < n Rerrnging this, it sys we re sfe if n > So n = 64 will certinly do. 6 (4)M 2/0.5 = = 640 3
6 Mthemtics S3 (Timoney) 5.4 Simpsons Rule. Simpsons Rule is the next most sophisticted method fter the trpezoidl rule. With Riemnn sums we used pproximtion by step grphs (bits of constnt grphs one fter the other), with the trpezoidl rule we used bits of stright lines, nd now we use bits of qudrtic grphs y = x 2 + bx + c. The first problem is tht, while 2 points determine line, we need 3 points to pin down qudrtic grph. Then we lso need formul ( for the re under qudrtic grph (or the integrl of it) nlogous to the formul h 2 y 0 + ) 2 y we used for the re of trpezoid. 5.5 Lemm. If we hve 3 points in the plne with different xcoordintes, then there is exctly one qudrtic grph pssing through them. Proof. (just n ide of how it works) If the points re (x 0, y 0 ), (x, y ) nd (x 2, y 2 ) then one wy to find the qudrtic is to write it down vi the Lgrnge interpoltion formul q(x) = (x x )(x x 2 ) (x 0 x )(x 0 x 2 ) y 0 + (x x 2)(x x 0 ) (x x 2 )(x x 0 ) y + (x x 0)(x x ) (x 2 x 0 )(x 2 x ) y 2 Another pproch is to strt with generl q(x) = x 2 +bx+c nd use the 3 equtions q(x 0 ) = y 0, q(x ) = y, q(x 2 ) = y 2 to find, b, c. 5.6 Lemm. For 3 points eqully spced horizontlly (x 0, y 0 ) = (x h, y 0 ), (x, y ) nd (x 2, y 2 ) = (x + h, y 2 ) nd y = q(x) the qudrtic grph through the 3 points x +h ( q(x) dx = h 3 y y + ) 3 y 2 x h Proof. We ll skip it. You cn find it in Anton. It is just bit messy, not relly complicted. It turns out to mke life lot esier if you ssume x = 0 nd this you cn ssume by shifting the yxis. 5.7 Simpsons Rule. Ide: For b f(x) dx, choose n even nd preferbly lrge. Divide the intervl [, b] into n equl sections ech of width h = b n, division points x i = + ih (0 i n), corresponding vlues y i = f(x i ). Pir off intervls nd pproximte y = f(x) on ech pir by qudrtic grph. b f(x) dx = sum of integrls of these qudrtics. The formul we get out of this is b ( f(x) dx = h 3 y y y y y n y n + ) 3 y n 5.8 Exmple. Find Solution: 3 e x2 dx pproximtely using Simpsons rule with n = 0.
7 Numericl Integrtion 7 i x i y i Weight Weight y i / / / / / / / / / / /3 Sum Sum times h is (nd this is the pproximte vlue for the integrl given by Simpson s rule). 5.9 Remrk. Now we sk bout the ccurcy of this method. Is it ny better fter the somewht greter compliction? 5.20 Theorem. Let S n denote the result of using Simpsons rule formul with n steps to pproximte b f(x) dx. Then b f(x) dx S n b 80 h4 M 4 where M 4 is the lrgest vlue of f (4) (x) for x b. Note: Cn rewrite this in terms of n using h = (b )/n. 5.2 Exmple. (i) In 3 ex2 dx with n = 0 how fr off cn Simpsons rule be? Solution: (detils not ll here) You cn find tht for f(x) = e x2, the fourth derivtive is f (4) (x) = (6x x 2 + 2)e x2 nd M 4 = 740e 9 = The theorem tells us then tht the pproximtion S 0 differs from the ctul integrl by t most (or t worst) 3 e x2 dx S 0 b 80 h4 M 4 = 3 80 h4 ( ) The vlue of h is h = b = 3 n 0 90 (0.2)4 ( ) = = 0.2 nd so the error could be s lrge s (ii) How lrge should we chose n so tht Simpsons rule pproximtion S n to the bove integrl is certinly within 0.5 of the right vlue? Solution: It will certinly be enough to choose n so tht b 80 h4 M 4 = 3 ( ) 4 3 M 4 < n
8 Mthemtics S3 (Timoney) Rerrnging this, sys we re sfe if n > ( ) /4 90 (6)M 4/0.5 = So n = 48 will certinly do. (Note how much smller this is thn 64 (needed for the trpezoidl rule in the sme sitution.) 5.22 Remrk. The methods we hve discussed were bout finding definite integrls numericlly. We will look lter t methods for finding ntiderivtives in firly systemtic wy. By the fundmentl theorem, if we cn find ntiderivtives we cn find definite integrls b f(x) dx nlyticlly (tht mens we cn come up with n exct formul for the vlue s opposed to numericl pproximte vlue). Our exmple 3 ex2 dx is one tht we will not ever be ble to do nlyticlly Remrk. In the trpezoidl rule, one rrely uses Theorem 5.2 to gurntee the desired ccurcy of the estimte T n. A more prgmtic pproch is to work with T 2, T 2 2 = T 4, etc until we get to vlue of T 2 k which hs stbilised nd where 2 k is resonbly big. There is wy to void mking the sme clcultions over nd over gin nd still to follow this strtegy. The point is tht if we did this for the exmple 3 f(x) dx (sy with the f(x) = ex2 bove), we would be clculting T = 3 2 ( 2 f() + ) 2 f(3) T 2 = 3 2 ( 2 2 f() + f(2) + ) 2 f(3) T 4 = 3 2 ( 4 2 f() + f(.5) + f(2) + f(2.5) + ) 2 f(3) Since we hve to reuse f() nd f(3) ech time, we might be tempted to remember the nswers. Then we need f(2) every time fter T 2 nd so we might like to keep record of the nswer to tht, nd so on. This involves lot of recording (uses up computer memory if we do it on computer) nd there is wy to void so much storge. It turns out tht T 2 = 2 T f(2) T 4 = 2 T (f(.5) + f(2.5)) 4 Thus we cn clculte T 4 using only T 2 nd vlues of f(x) tht were not needed for T 2. This bit is not in the book by Anton, I think.
9 Numericl Integrtion 9 In generl, T 2n = 2 T n + b 2n f i= ( + i ( )) b nd we cn keep finding T, T 2, T 4 = T 2 2, T 8 = T 2 3,... without ny mssive storge requirement (nd without reclculting ny vlues of f(x) we hd to clculte previously). This gives n efficient prgmtic strtegy: Compute T, T 2, T 4 = T 2 2, T 8,... until we get to T n with some resonbly lrge n nd the vlue of T n is roughly equl to the vlue of T n/2. The chnces re tht both re then close to the true vlue of the integrl b f(x) dx. This is n experimentl pproch nd not quite method tht is gurnteed to be 00% right Remrk. As for the trpezoidl rule, one rrely uses Theorem 5.20 to gurntee the desired ccurcy of the estimte S n. A more prgmtic pproch is to work with S 2, S 2 2 = S 4, etc until we get to vlue of S 2 k which hs stbilised nd where 2 k is resonbly big. It turns out tht we cn mke use of the erlier efficient method of working out T, T 2, T 2 2 = T 4, etc together with the formul 2 2n This formul is esy to check. For exmple S 2n = 4 3 T 2n 3 T n T = b ( 2 f() + ) 2 f(b) T 2 = b ( ( 2 2 f() + f + b ) + 2 ) 2 f(b) S 2 = b ( 2 3 f() + 4 ( 3 f + b ) + 3 ) 2 f(b) 4 3 T 2 3 T = b ( 2 2 = S 2 3 f() f ( 3 f() + 3 f(b) 5.25 Exmple. Try out strtegy for our exmple 3 ex2 dx. 2 This bit is not in the book by Anton, I think. ( + b 2 )) ) f(b)
10 Mthemtics S3 (Timoney) Appendix 5A.26 The Indefinite Integrl. We will explin the ide of n indefinite integrl by first giving n exmple. Every time you differentite something you know the ntiderivtive for the result. For exmple d dx (2x4 + 2x 3 7x 2 + x 7) = 8x 3 + 6x 2 4x + nd so if we hppen to wnt to know something tht hs derivtive 8x 3 + 6x 2 4x + we know tht 2x 4 + 2x 3 7x 2 + x 7 is such function. We cll it n n ntiderivtive of 8x 3 +6x 2 4x+ something which when differentited gives us 8x 3 + 6x 2 4x +. Now wht bout other possible ntiderivtives for the sme thing? We cn spot quite esily tht the constnt 7 is not very importnt. Any constnt there insted of 7 would still hve derivtive 0 nd so we relise tht nything of the form 2x 4 + 2x 3 7x 2 + x + C is lso n ntiderivtive for 8x 3 + 6x 2 4x +. In fct this is ll becuse if two functions f(x) nd g(x) hve the sme derivtive, in this cse if f (x) = g (x) = 8x 3 + 6x 2 4x + then d (f(x) g(x)) = 0 dx for ll x. The only functions tht re defined on n intervl nd hve derivtive 0 re constnt functions. So f(x) g(x) = C = constnt nd so f(x) = g(x) + C. This mens tht if we find one ntiderivtive (in our exmple 2x 4 + 2x 3 7x 2 + x 7) for the given function (in our exmple 8x 3 + 6x 2 4x + ) then the most generl one is of the form 2x 4 + 2x 3 7x 2 + x 7 + C. Since C 7 is nother constnt we cn write 8x 3 + 6x 2 4x + dx = 2x 4 + 2x 3 7x 2 + x + C. We cn go bout things bit more systemticlly, by strting with the simplest rules for differentition nd turning them into rules for finding ntiderivtives. (i) We know d dx xn = nx n nd so nx n dx = x n + C. For exmple, 4x 3 dx = x 4 + C. We cn use little ingenuity to find tht n ntiderivtive for x n is n xn (s long s n 0) nd we cn tidy this up to get the rule x n dx = n + xn+ + C (n ) It is perhps interesting to see tht we cn t immeditely write down x dx = /x dx. In fct the ntiderivtive for /x involves the nturl logrithm function ln nd is therefore much more complicted thing tht /x. We will leve this for lter.
11 Numericl Integrtion A detil we should mention, is tht when n < 0, x n is not wellbehved t x = 0. So the domin is not n intervl but two intervls x < 0 nd x > 0. So, techniclly, the +C is not dequte in these cses to describe the most generl ntiderivtive. We could switch from one vlue of C to nother s we pss x = 0 nd still hve vlid ntiderivtive (when n < 0). However, people rrely go into this nd probbly you lmost never should encounter this in prctice. Becuse things blow up t x = 0 there should not relly be prcticl problem where x > 0 nd x < 0 re both vlid (when deling with x n nd n < 0). (ii) We know the derivtive of sum is the sum of the derivtives d du (u + v) = dx we cn see tht n ntiderivtive of sum is the sum of the ntiderivtives. Similrly for constnt multiples. + dv dx dx nd so Using these rules, we cn integrte ll polynomils. For exmple ( ) ( ) ( ) ( ) 5x 3 x 2 + 3x + 2 dx = 5 4 x4 3 x x2 + 2 x + C = 5 4 x4 3 x x2 + 2x + C (iii) From the rules for differentiting trigonometric functions d dx sin x = cos x, d dx cos x = sin x, d dx tn x = sec2 x we cn write down rules for integrting some cos x dx = sin x + C, sin x dx = cos x + C, sec 2 x dx = tn x + C With little guesswork we cn figure out some relted integrls like cos 3x dx We might think of sin 3x s possible ntiderivtive but d sin 3x = 3 cos 3x dx is 3 times wht we wnt. Since 3 is constnt, we cn divide cross by it nd we get cos 3x dx = sin 3x + C 3 However, there is no esy wy to do sec x dx (we will see wht the nswer to this is bit lter). We cn write sec x dx = cos x dx
12 Mthemtics S3 (Timoney) nd we know how to integrte cos x but tht does not help. The is no good quotient rule for ntiderivtives. Unlike differentition, where we cn lern smll number of rules tht re enough to differentite lmost ny function we cn esily write down, there re esylooking functions where ntiderivtives re quite hrd to find. We hve seen /x dx, more recently sec x dx nd the exmple cos(x 2 ) dx is one tht is essentilly impossible. It is not tht there is no nswer. There is n ntiderivtive but it is known tht there is no wy to write finite formul for the ntiderivtive of cos(x 2 ) using the fmilir functions (powers, roots, frctions, trig functions, ln, e x ). 5A.27 Exmple. There re very few exmples which cn be worked out directly from the limit of Riemnn sums definition. This is one exmple 4 2 x + 2 dx Solution: Tke n lrge, n eqully wide subintervls of [, b] = [2, 4] with widths h = (b )/n = (4 2)/n = 2/n. This gives subdivision points x i = + ih = 2 + 2i for i = 0,, 2,..., n. To n mke life esy for ourselves we tke x i = x i = 2 + (2i)/n. The Riemnn sum is then ( ) 2i 2 f(x i )(x i x i ) = n + 2 n i= i= ( ) = 2 2i n n + 2 i= i= ( ) ( ) = 4 i n 2 n i= i= = 4 ( ) n(n + ) + 2 n 2 2 n (2n) ( = 2 + ) + 4 n 6 s n Here we used the fct tht i = n i= turns out to be n(n+) 2. This cn be discovered by writing the sum down bckwrds nd dding the two formule verticlly: s = n s = n + n + n s = (n + ) + (n + ) + (n + ) + + (n + ) = n(n + )
13 Numericl Integrtion 3 In most exmples, we will not be ble to write the Riemnn sum s short formul in the wy we did here nd so we would not find the limit. 5A.28 Remrk. There is n mzing connection between definite integrls nd differentition, which we will now stte. It comes bout by considering not just one definite integrl b f(x) dx but whole infinite number of them. Not just 2 f(x) dx but x f(t) dt for 0 x A.29 Theorem (Fundmentl Theorem of Integrl Clculus). Assume tht y = f(x) is continuous for x b. Consider A(x) = x f(t) dt for x b. (A(x) is new function, built from f nd definite integrtion.) Then A(x) is n ntiderivtive for f (tht is A (x) = f(x) for x b). In summry: ( d x ) f(t) dt = f(x) ( x b, if f continuous) dx This is one prt of the Fundmentl theorem, or one wy to stte it. 5A.30 Corollry. There is n ntiderivtive for every continuous function f. 5A.3 Exmple. Find d ( x ( ) ) t 3 cos dt dx t 8 + Solution: By the theorem ( d x ( ) ) t 3 x 3 cos dt = cos dx t 8 + x 8 + 5A.32 Theorem (Other prt of fundmentl theorem). Assume tht y = f(x) is continuous for x b nd suppose g(x) is n ntiderivtive for f(x) (tht is g (x) = f(x) for x b). Then b f(x) dx = g(b) g()= [g(x)] b x=
14 Mthemtics S3 (Timoney) 5A.33 Exmple. Find 4 2 2x + dx Solution: Antiderivtive g(x) = x 2 + x (since g (x) = 2x + ) nd so 4 2 2x + dx = [g(x)] 4 x=2 = [x 2 + x] 4 x=2 = ( ) ( ) = 4 Proof. (of A (x) = f(x) prt of Fund. Thm.) Use first principles x+h x A (x) = lim h 0 A(x + h) A(x) h f(t) dt = f(x)h for h smll. Thus limit is f(x). = lim h 0 Proof. (of b f(x) dx = g(b) g() prt) We know A(x) = x f(t) dt is n ntiderivtive. Hence x+h x f(t) dt h d (A(x) g(x)) = f(x) f(x) = 0. dx Thus A(x) g(x) = c = const. Use x = to find constnt: A() g() = c. But A() = f(t) dt = 0. Thus g() = c nd A(x) g(x) = c = g() for ll x [, b]. Use this for x = b to get A(b) g(b) = g(), A(b) = g(b) g(), tht is b f(t) dt = g(b) g(). 5A.34 Correction. Correction to (or refinement of) grphicl interprettion of definite integrls. b f(x) dx = re under grph is good when f(x) 0 lwys. When f(x) < 0, tht prt of the re is counted with minus sign. Consider negtive terms in Riemnn sum f(x i )(x i x i ) if f(x i ) < 0. i=
15 Numericl Integrtion 5 b f(x) dx = (sum of res where f(x) > 0) (sum where f(x) < 0) 5A.35 Nottion. There is one more thing we hve skimmed over. When > b, there is convention tht llows us to write b f(x) dx (for exmple 2 x dx) even though the limits re upside down. By convention, the mening for this is b f(x) dx = b f(x) dx if > b 3 x2 + This convention comes up in the proof of the fundmentl theorem (the prt where A(x + h) A(x) = x+h f(t) dt, where we hve to be ble to del with h < 0 s well s h > 0). x In fct the fundmentl theorem in the form ( d x ) f(t) dt = f(x) dx is lso vlid for x < s long s f is continuous on n intervl tht includes both nd x. (The importnt thing is to hve ll points in between s well s nd x.) The convention lso fits with the other form of the fundmentl theorem. If g (x) = f(x), then remins true when > b. b 5A.36 Exmples. (i) Find d dx Solution: ( d 0 dx x f(x) dx = [g(x)] b x= = g(b) g() ( 0 x ) t t 6 + dt ) t t 6 + dt = d ( x dx 0 using the bove convention nd the Fundmentl theorem. ) t t 6 + dt = x x 6 +
16 Mthemtics S3 (Timoney) ( ) (ii) Find d x 3 +x cos(t 2 + 4) dt dx Solution: Tke y = x 3 +x cos(t 2 + 4) dt = Chin rule dy dx = dy du nd by the Fundmentl theorem u cos(t 2 + 4) dt where u = x 3 + x. By the du dx = dy du (3x2 + ) dy du = cos(u2 + 4) = cos((x 3 + x) 2 + 4) Putting these together, we get ( ) d x 3 +x cos(t 2 + 4) dt = (3x 2 + ) cos((x 3 + x) 2 + 4) dx ( ) Notice tht the nswer would hve been the sme for d x 3 +x cos(t 2 + 4) dt, s the dx 25 vlue of the lower limit of the integrl does not come into the nswer (s long s it is constnt nd s long s the integrnd is continuous wherever we go). Richrd M. Timoney Mrch 7, 2007
Math& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationAPPROXIMATE INTEGRATION
APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose ntiderivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More informationNumerical Analysis: Trapezoidal and Simpson s Rule
nd Simpson s Mthemticl question we re interested in numericlly nswering How to we evlute I = f (x) dx? Clculus tells us tht if F(x) is the ntiderivtive of function f (x) on the intervl [, b], then I =
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More informationSYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus
SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationZ b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...
Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More informationCMDA 4604: Intermediate Topics in Mathematical Modeling Lecture 19: Interpolation and Quadrature
CMDA 4604: Intermedite Topics in Mthemticl Modeling Lecture 19: Interpoltion nd Qudrture In this lecture we mke brief diversion into the res of interpoltion nd qudrture. Given function f C[, b], we sy
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationNUMERICAL INTEGRATION
NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls
More informationdifferent methods (left endpoint, right endpoint, midpoint, trapezoid, Simpson s).
Mth 1A with Professor Stnkov Worksheet, Discussion #41; Wednesdy, 12/6/217 GSI nme: Roy Zho Problems 1. Write the integrl 3 dx s limit of Riemnn sums. Write it using 2 intervls using the 1 x different
More informationIntegrals  Motivation
Integrls  Motivtion When we looked t function s rte of chnge If f(x) is liner, the nswer is esy slope If f(x) is nonliner, we hd to work hrd limits derivtive A relted question is the re under f(x) (but
More informationIf u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then f(g(x))g (x) dx = f(u) du
Integrtion by Substitution: The Fundmentl Theorem of Clculus demonstrted the importnce of being ble to find ntiderivtives. We now introduce some methods for finding ntiderivtives: If u = g(x) is differentible
More informationMath 131. Numerical Integration Larson Section 4.6
Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationMAT 168: Calculus II with Analytic Geometry. James V. Lambers
MAT 68: Clculus II with Anlytic Geometry Jmes V. Lmbers Februry 7, Contents Integrls 5. Introduction............................ 5.. Differentil Clculus nd Quotient Formuls...... 5.. Integrl Clculus nd
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationLecture 20: Numerical Integration III
cs4: introduction to numericl nlysis /8/0 Lecture 0: Numericl Integrtion III Instructor: Professor Amos Ron Scribes: Mrk Cowlishw, Yunpeng Li, Nthnel Fillmore For the lst few lectures we hve discussed
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationLECTURE. INTEGRATION AND ANTIDERIVATIVE.
ANALYSIS FOR HIGH SCHOOL TEACHERS LECTURE. INTEGRATION AND ANTIDERIVATIVE. ROTHSCHILD CAESARIA COURSE, 2015/6 1. Integrtion Historiclly, it ws the problem of computing res nd volumes, tht triggered development
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationLecture 1: Introduction to integration theory and bounded variation
Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 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
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationThe Evaluation Theorem
These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for inclss presenttion nd should not
More informationStuff You Need to Know From Calculus
Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More information(0.0)(0.1)+(0.3)(0.1)+(0.6)(0.1)+ +(2.7)(0.1) = 1.35
7 Integrtion º½ ÌÛÓ Ü ÑÔÐ Up to now we hve been concerned with extrcting informtion bout how function chnges from the function itself. Given knowledge bout n object s position, for exmple, we wnt to know
More information1 Part II: Numerical Integration
Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove xxis) ( bove f under xxis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More information6.5 Numerical Approximations of Definite Integrals
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 6.5 Numericl Approximtions of Definite Integrls Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil,
More informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re openended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnkoutnnswer problems! (There re plenty of those in the
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationRiemann Integrals and the Fundamental Theorem of Calculus
Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums
More informationNumerical Integration
Chpter 1 Numericl Integrtion Numericl differentition methods compute pproximtions to the derivtive of function from known vlues of the function. Numericl integrtion uses the sme informtion to compute numericl
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationx = b a n x 2 e x dx. cdx = c(b a), where c is any constant. a b
CHAPTER 5. INTEGRALS 61 where nd x = b n x i = 1 (x i 1 + x i ) = midpoint of [x i 1, x i ]. Problem 168 (Exercise 1, pge 377). Use the Midpoint Rule with the n = 4 to pproximte 5 1 x e x dx. Some quick
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationSection 4: Integration ECO4112F 2011
Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic
More informationThe Fundamental Theorem of Calculus
The Fundmentl Theorem of Clculus MATH 151 Clculus for Mngement J. Robert Buchnn Deprtment of Mthemtics Fll 2018 Objectives Define nd evlute definite integrls using the concept of re. Evlute definite integrls
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More information1 Probability Density Functions
Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our
More informationFirst midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009
Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationInfinite Geometric Series
Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More information10. AREAS BETWEEN CURVES
. AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in
More information