Calculus III Review Sheet


 Caitlin Preston
 3 years ago
 Views:
Transcription
1 Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing or decresing on its entire domin. Exmple 1.1 e x is incresing on (, ) nd ln x is incresing on (0, ). f(x) = 1/x is decresing on (, 0) nd on (0, ). The sine function is incresing on [ π/, π/], [3π/, 5π/], etc. A function f(x) is proportionl to nother function g(x) if there is nonzero constnt k such tht f(x) = kg(x). Exmple 1. The force of grvity F between two msses M nd m is proportionl to the inverse squre function of the distnce between them, r, nmely Here k = GMm nd g(r) = 1/r. F = GMm r The grph of function f is concve up if it bends upwrd s we move from left to right, tht is if, when we pick two points (x 1, y 1 ) nd (x, y ) on the grph of f, the grph lies below or on the line segment joining those points on tht intevl. Similrly, the grph of f is concve down if it bends downwrd, or if the grph of f lies bove or on ny line segment joining two points on its grph. An inflection point is point (x, f(x)) on the grph of f t which the grph chnges concvity (or equivlently just the x vlue of tht point). Exmple 1.3 The cosine function is concve up on [ π, 3π ] nd concve down on [ π, π ], etc. It hs inflection points t odd multiples of π/. A function f is sid to be even if f( x) = f(x) nd odd if f( x) = f(x). Even functions re symmetric bout the yxis, while odd ones re symmetric bout the origin. Exmple 1.4 The cosine function is even, f(x) = x is even, nd f(x) = e 3x is even, while the sine function is odd, f(x) = x 3 is odd, nd f(x) = rctn(x) is odd. 1. Limits nd Continuity Suppose f is defined on n intervl round given rel number c, except perhps t x = c itself. The limit of f s x pproches given rel number c is rel number L stisfying the following condition: for ll ɛ > 0 there is δ > 0 such tht x c < δ nd x c 1
2 implies In this cse we write f(x) L < ɛ lim f(x) = L Remrk 1.5 In plin English, this mens tht the limit L exists provided the following holds: however smll I wnt to mke my ɛwindow on the yxis round L, I cn find smll enough δwindow on the xxis round c which is mpped by f entirely into the ɛwindow. Remrk 1.6 Note the logicl implictions: I hve freedom to choose ɛ. I cn choose it to be one in billion, whtever I wnt. But once chosen, I m typiclly restricted in my choice of δ. For exmple, if f(x) = x nd L = 100, then clerly c = 10, nd to show tht the limit s x pproches 10 is 100, I need to find δ for ny given ɛ. Suppose I ve chosen my ɛ, sy ɛ = 1/1000. Then you cn check tht good enough δ is δ = 1/1000. It s not the only one, since I cn pick it to be smller thn this, but I m restricted by how big I cn mke it. It cn t be much bigger thn tht. Remrk 1.7 Note lso tht the definition of limit includes bsolute vlues round x c. We write x c nd this implies the existence nd equlity of the left nd right limits, lim f(x) nd lim f(x) + respectively. Exmple 1.8 The function f(x) = x 1 x 1 hs left nd right limits s x pproches 1, but the two limits re not equl. Hence f hs no limit t x = 1. We cn nlogously define lim f(x) x ± The difference is we cn t use the δ prt of the definition. The ɛ prt is the sme, we cn choose ny ɛ > 0 such tht f(x) L < ɛ under some pproprite condition on x. It s just tht condition cn t be the existence of δ such tht x c < δ (becuse c = nd x =, which is never less thn δ), so insted we demnd the existence of n M > 0 such tht x M implies f(x) L < δ. A function f is continuous t point x = c if the limit lim f(x) = L exists nd moreover L = f(c), i.e. if lim f(x) = f(c) We sy f is continuous on n intervl [, b] or (, b), etc., if it is continuous t ll points in tht intervl. Exmple 1.9 All polynomils re continuous on (, ). All rtionl functions p(x)/q(x) re continuous except t zeros of q(x). For exmple, x +3 x 5 is continuous everywhere except t ± 5. The exponentil function e x is continuous on (, ), the log function ln x is continuous on (0, ). The sine nd consine functions re continuous on (, ). The bsolute vlue function f(x) = x is continuous everywhere on (, ).
3 Discontinuities, or points of discontinuity of f, re xvlues where f is not continuous. For exmple 0 is point of discontinuity of csc x nd of f(x) = 1/x. A removble discontinuity is one tht cn be plugged, e.g. x = is removble discontinuity of f(x) = x 4 x+. An essentil discontinuity is one tht cnnot be plugged, typiclly becuse f goes to ± ner tht point, for exmple x = 1 is n essentil discontinuity of f(x) = /(x 1). A jump discontinuity is discontinuity where f jumps finite mount ner it, for exmple x = 1 for f(x) = x 1 x Rte of Chnge nd the Derivtive The verge rte of chnge of function f between x = nd x = b is defined to be the slope of the line connecting the points (, f()) nd (b, f(b)) on the grph of f: verge rte of chnge = f(b) f() b If we define h = b, then b = + h, nd the verge rte of chnge between nd b = + h becomes the difference quotient: f( + h) f() h If the limit of the difference quotient s h pproches 0 exists, then we sy tht f is differentible t x = nd we cll this limit the derivtive of f t. It is rel number, denoted equivlently by f () df dy () f( + h) f() dx dx = lim x= h 0 h If f is differentible on n entire intervl, then vrying the point, in other words letting x vry, we get function, the derivtive function f (x). One of our tsks is to find vrious derivtive functions for wellknown functions. Exmple 1.10 Common derivtive functions re the following: 1. (x n ) = nx n 1 (power rule). (sin x) = cos x 3. (cos x) = sin x 4. (tn x) = sec x 5. (e x ) = e x 6. (ln x) = 1 x 7. ( x ) = (ln ) x 8. (rctn x) = 1 1+x 9. (rcsin x) = 1 1 x 3
4 Remrk 1.11 To compute more complicted derivtives, such s those of xe x, for exmple, we will need to know how to brek up the process of tking derivtive into severl steps involving only derivtives of things we know how to compute, such s (1)(6) bove. For exmple, we know how to tke the derivtive of x, x nd e x, nd we will develop methods to compute (xe x ) in terms of these esier ones. This is the content of the sum, product, quotient nd chin rules below. The derivtive f () t x = cn be interpreted s the slope of the tngent line to the grph of f t the point (, f()). Since we hve slope, f (), nd point, (, f()), we cn find the eqution of the tngent line using pointslope: y f() = f ()(x ) Adding f() to both sides gives the eqution for the tngent line, which we lso cll the liner pproximtion to f ner x = or locl lineriztion ner x = : y = f() + f ()(x ) This is the key ide behind the ide of derivtive, to pproximte complicted function f(x) with the simplest one possible, line, t lest loclly. The tngent line is n honesttogoodness pproximtion of f ner x =, becuse we know tht s x, f(x) f() x f (), i.e. from the definition of limit we get f(x) f() x f () f(x) f() f ()(x ) multiply both sides by (x ) f(x) f() + f ()(x ) dd f() to both sides Exmple 1.1 Let us pproximte the vlue of f(x) = e sin x ner x = π using locl lineriztion, sy t x = 3 which is ner x = π. Since f (x) = e sin x cos x, we hve f (π) = e sin π cos π = 1, nd becuse f(π) = e sin π = 1, we get f(x) f(π) + f (π)(x π) = 1 (x π) = 1 + π x Therefore, f(3) 1 + π 3 = π 1.14 The error in the tngent/liner pproximtion is the difference between the ctul vlue f(x) t x nd the pproximte vlue obtined from the tngent line t x ner x = : E(x) = f(x) [ f() + f ()(x ) ] It is theorem, not hrd to prove, tht E(x) lim x x = 0 It is lso fct, not proven in the book until the section on Tylor s theorem, tht E(x) f () (x ) 4
5 Exmple 1.13 The error in the previous exmple is E(x) = f(x) (1 + π x) = e sin x + x 1 π so, for exmple our pproximtion t x = 3 is off by bout E(3) = e sin π Not bd! And, to complete this exmple, note tht f (π) (3 π) = 1 (cos 3e sin 3 (sin 3)e sin 3 )(3 π) which is indeed close to E(3). 1.4 Optimiztion A criticl point of f is point x = where f () = 0, tht is criticl points re zeros of the derivtive of f. Hence, when looking for them you hve to solve the eqution f (x) = 0 for x. The yvlue of f t criticl point y = f() is clled criticl vlue. Note tht f() 0 in generl it s f () tht equls 0! A locl minimum of f is point x = such tht f(x) f() ner, nd locl mximum of f is point x = such tht f(x) f() ner. These re upgrded to globl minimum nd globl mximum if f(x) f() or f(x) f(), respectively, for ll x in the domin of f under considertion, not just those nerby. 1.5 The Riemnn Integrl Let [, b] be nonempty closed intervl in (, ) nd let f : [, b] (, ) be bounded nd continuous. Let P = ( = x 0 < x 1 < < x n = b) be prtition of [, b], tht is n incresing finite incresing sequence of points in the intervl. For given prtition P choose, for ech i = 0, 1,..., n, n rbitrry t i in [x i 1, x i ]. The Riemnn Sum of f for given prtition P is then defined s S P = n f(t i ) (x i x i 1 ) i=1 If we choose the x i such tht x n x n 1 = x n 1 x n = = x 1 x 0, tht is if x = b n then the Riemnn sum is written simply s nd x i = + i x n S P = f(t i ) x i=1 If we choose the left endpoint every time, t i = x i, then we get the left Riemnn sum, n 1 S L = f(x i ) x i=0 5
6 nd if we choose the right endpoint, t i = x i + 1, then we get the right Riemnn sum, S R = n f(x i ) x i=1 Similrly, we hve the upper nd lower Riemnn sums, U nd L, obtined by choosing t i in [x i, x i+1 ] such tht f(t i ) = mx xi t x i+1 f(t) nd f(t i ) = min xi t x i+1 f(t), respectively (since f is continous, we cn find such t i by the Extreme Vlue Theorem!). We cn lso choose t i to be the midpoint of [x i, x i+1 ], in which cse we obtin the midpoint Riemnn sum. Remrk 1.14 Note tht U does not in generl equl S L or S R, nd likewise with L. A good exmple is the upper hlf of the circle, f(x) = 1 x on [ 1, 1]. Try to compute S L, S R, U nd L. Exmple 1.15 Let us compute the Riemnn sum for f(x) = 1 x on [ 1, 1], nd so obtin severl pproximtions of the upper hlf of the unit circle. Let us choose n = 4 for convenience, which mens x 0 = 1, x 1 = 1, x = 0, x 3 = 1, nd x 4 = 1, nd so P = ( 1, 1, 0, 1, 1) is our prtition of [ 1, 1]. Let us compute the left Riemnn sum first: this mens choosing t i = x i, nd since x = 1 ( 1) 4 = 1, we hve S L = 4 1 f(x i ) x x=0 = f( 1) 1 + f( 1 ) 1 + f(0) 1 + f(1 ) 1 ( 1 = ( 1) + 1 ( 1/) ) 1 (1/) 1 ( ) 3 3 = = Similrly, by choosing t i = x i+1, we get the right sum, S R = 4 1 f(x i+1 ) x x=0 = f( 1/) 1 + f(0) 1 + f(1/) 1 + f(1) 1 ( 1 = ( 1/) (1/) + ) ( ) 3 3 = = The upper sum requires little bit of extr work. We hve to figure out where f is lrgest on 6
7 ech subintervl [x i, x i+1 ], nd this will vry depending on i. Let us look t picture: 1 x Thus, for exmple, on the intervl [x 0, x 1 ] = [ 1, 1 ], it is cler tht f is lrgest when x = 1, sowe must choose t 0 = 1. Proceeding nlogously, we get t 1 = 0, t = 0, nd t 3 = 1, nd so U = n 1 f(t i ) x i=0 = f( 1 ) 1 + f(0) 1 + f(0) 1 + f(1 ) 1 ( 1 = ( 1/) ) 1 (1/) 1 ( ) 3 3 = = Similrly, to compute the lower Riemnn sum, we see from the picture 1 x tht, for exmple on [x 0, x 1 ] = [ 1, 1 ], f is smllest t x = 1, so we will need t 0 = 1, nd proceeding nlogously we get tht t 1 = 1, t = 1 nd t 3 = 1. Thus, L = n 1 f(t i ) x i=0 = f( 1) 1 + f( 1 ) 1 + f(1 ) 1 + f(1) 1 ( 1 = ( 1) + 1 ( 1/) + 1 (1/) + ) ( ) 3 3 = = As finl observtion, we note tht L S L = S R U. 7
8 The next step is, of course, refining the prtition of [, b] nd summing over more nd more rectngles, thus, hopefully, getting more ccurte pproximtion to the ctul re. A function f is sid to be Riemnn integrble on [, b] if, no mtter how the choice of prtitions P of [, b] is mde, the limit of these Riemnn sums s we dd more nd more rectngles, lim n S P n f(t i ) (x i+1 x i ) i=1 exists s rel number. It is fct, proven lter in undergrdute nlysis, tht the choice of prtition doesn t mtter, so you my choose, if you wnt, the left Riemnn sum every time. In this cse, if the limit exists, f is sid to be Riemnn integrble on [, b] nd the vlue of this limit is clled the Riemnn integrl or definite Riemnn integrl of f on [, b], nd this limit is denoted n 1 f(x) dx = lim S P = lim f(t i )(x i+1 x i ) n n If we choose x = x i+1 x i to be constnt nd equl to b n, then we my write i=0 n 1 f(x) dx = lim f(t i ) x n i=0 8
9 Theorems Theorem.1 (Limit Lws) Let k be constnt rel number. Then, 1. lim (kf(x)) = k lim f(x). lim ( f(x) + g(x) ) = lim f(x) + lim g(x) ( ) ( 3. lim f(x)g(x) = lim f(x) )( lim g(x) ) f(x) 4. lim g(x) = lim f(x) lim g(x) if g(x) 0 5. lim k = k (we think of k s constnt function) 6. lim x = c (we think of x s the function f(x) = x) Theorem. (Continuity Lws) Let k be constnt rel number nd suppose f nd g re continuous on n intervl [, b] (or n open or hlfopen intervl, it doesn t mtter). Then, the following functions re continuous: 1. kf(x). f(x) + g(x) 3. f(x)g(x) 4. f(x) g(x) if g(x) 0 5. (f g)(x) Theorem.3 (Intermedite Vlue Theorem) If f is continuous on closed nd bounded intervl [, b] nd k is yvlue between f() nd f(b) (whether f() k f(b) or f(b) k f()), then there is t lest one number c between nd b such tht f(c) = k. Remrk.4 This mens tht the entire rnge of yvlues between f() nd f(b) is hit by f on the intervl [, b], possibly more thn once. Exmple.5 Since f(x) = x is continuous on [4, 100], nd since f(4) = nd f(100) = 10, ll numbers between nd 10 re hit. For exmple 5: there is number, nmely c = 5, in between 4 nd 100, such tht f(c) = 5. Exmple.6 Consider the polynomil p(x) = 5x 3 + πx + 1. It hs t lest one root between 1 nd 0, becuse p( 1) = 4 π nd p(0) = 1, nd since 0 lies in between these two y vlues nd p(x) is continuous on [ 1, 0], we know there is number c between 1 nd 0 such tht p(c) = 0. Tht s our root. 9
10 Exmple.7 Consider f(x) = 1/x. Then f( 1) = 1 nd f(1) = 1, but the Intermedite Vlue Theorem doesn t pply to give c between 1 nd 1 such tht f(c) = 0, becuse f is not continuous on [ 1, 1]. Indeed, f(x) never equls 0. Theorem.8 (Extreme Vlue Theorem) If f is continuous function on closed nd bounded intervl [, b], then f ttins (globl) minimum nd (globl) mximum on tht intervl, tht is mx x b f(x) nd min x b f(x) exist s rel numbers. This mens there re points x 1 nd x in the intervl [, b] such tht f(x 1 ) = mx x b f(x) nd f(x ) = min x b f(x) Exmple.9 The prbol f(x) = x + 6x 1, being polynomil, is continuous everywhere, so for exmple on [0, 4] it must chieve its minimum nd its mximum. In fct, s cn be seen from the grph: f(0) = min f(x) = 1 nd f(3) = mx f(x) = 8 0 x 4 0 x 4 Of course, we don t need the grph, becuse (using the first nd second derivtive tests) we cn ctully find the mximum nd minimum nlyticlly s follows: 1. First, find ll criticl points, which mens ompute the derivtive, set it equl to zero nd solve for x: f (x) = x + 6 = 0, so x = 3 is criticl point.. Use, e.g. the second derivtive test to clssify it: f (3) = < 0, so it s locl mx. 3. Check the vlue of f t the criticl point (which lies in the intervl [0, 4]) ginst the vlue of f t the endpoints 0 nd 4: f(0) = 1, f(3) = 8 nd f(4) = 7, so it looks like 0 is the globl min nd 3 is the globl mx here, i.e. f(0) = min 0 x 4 f(x) = 1 nd f(3) = mx 0 x 4 f(x) = 8. Exmple.10 The function f(x) = 1 x is continuous on (, 0) (0, ), but on (0, ) the Extreme Vlue Theorem doesn t pply, becuse (0, 1) is not closed intervl. As it turns out, f(x) hs neither (locl or globl) mx or min on (0, 1) (it just gets bigger nd bigger without bound s x pproches 0 nd it gets closer nd closer to 1/ s x pproches, but it never reches 1/, nd so it never reches smllest vlue. 10
11 Theorem.11 (Differentition Rules) If two functions f nd g re differentible t x = nd k is constnt, then the functions c (the constnt function), cf(x), f(x) + g(x), f(x)g(x), f(x)/g(x) (when g(x) 0), nd (f g)() (when f is differentible t b = g()) re ll differentible t x =, nd moreover stisfy the following formuls: (1) c = 0 () (cf) () = cf () (constnt fctor rule) } (linerity) (3) (f + g) () = f () + g () (sum rule) (4) (fg) () = f()g () + f ()g() (product rule) (5) ( ) f () = f ()g() f()g () g g () (quotient rule) (6) (f g) () = f ( g() ) f () (chin rule) Theorem.1 (Differentibility Implies Continuity) If f is differentible t point x =, then it is continuous t x =. Consequently, if f is differentible on n open intervl (, b), then it is continuous on tht intervl. Remrk.13 Of course the converse is not true. You cn hve continuous function which is not differentible. For exmple f(x) = x is continuous t x = 0 but not differentible there (the left nd right limits of the difference quotient s h 0 don t gree). Theorem.14 (Men Vlue Theorem) If f is continuous on [, b] nd differentible on (, b), then there is some c between nd b such tht f(b) f() = f (c)(b ) Exmple.15 The Men Vlue Theorem is used for mny things, but the simplest of its uses is in proving inequlities. For exmple, let us prove the inequlity sin x x on [0, ). The trick is to define new function, h(x), s the difference of the two functions under considertion, tht is let h(x) = x sin x 11
12 Then we need only show tht h(x) 0 to prove our ssertion. Now, since for ny x > 0 we know tht h is continous on [0, x] nd differentible on (0, x), the MVT pplies to give the existence of c between 0 nd x such tht h(x) h(0) = h (c)(x 0) Now, h(0) = 0 sin 0 = 0, nd moreover h (x) = 1 cos x 0, which, long with the fct tht x > 0, implies h (x)(x 0) 0, so we hve x sin x = h(x) 0 on [0, ), which is wht we set out to prove. Corollry.16 (Constnt Function Theorem) Let f be continuous on [, b] nd differentible on (, b). Then f (x) = 0 on ll of (, b) if nd only if f is constnt on the intervl. Corollry.17 Let f nd g be continuous on [, b] nd differentible on (, b). If f (x) = g (x) on (, b), then f nd g differ by constnt, i.e. for some constnt c. f(x) = g(x) + c Remrk.18 This is importnt for integrls nd ntiderivtives. Nmely, ll ntiderivtives differ by constnt, nd so the most generl ntiderivtive of function is written f(x)dx = F (x) + C where F (x) is some specific ntiderivtive nd C is n rbitrry constnt, tht constnt showing up in the lst corollry. Corollry.19 Let f be differentible on n open intervl (, b). (1) If f (x) 0 on (, b), then f is incresing on (, b). () If f (x) 0 on (, b), then f is decresing on (, b). Moreover, if the inequlities re strict, then f is strictly incresing or strictly decresing, respectively. Corollry.0 (First Derivtive Test) Let f be differentible on n open intervl contining criticl point x 0 of f, i.e. one for which f (x 0 ) = 0. (1) If f (x) > 0 for x < x 0 nd f (x) < 0 for x > x 0, then x 0 is locl mximum of f. () If f (x) < 0 for x < x 0 nd f (x) > 0 for x > x 0, then x 0 is locl minimum of f. 1
13 Corollry.1 (Second Derivtive Test) Let f be twice differentible on n open intervl contining criticl point x 0 of f, i.e. one for which f (x 0 ) = 0. (1) If f (x 0 ) > 0, then x 0 is locl minimum of f. () If f (x 0 ) < 0, then x 0 is locl mximum of f. (3) If f (x 0 ) = 0, then no conclusion cn be drwn. x 0 my be locl minimum or locl mximum or point of inflection. Theorem. (Inverse Fuction Theorem) Let f be continuous nd invertible on [, b] nd differentible t some point x 0 in [, b]. If f (x 0 ) 0, then f 1 is differentible t f(x 0 ) nd if we let y 0 = f(x 0 ), then (f 1 ) ( ) 1 y 0 = f (x 0 ) Theorem.3 (L Hôspitl s Rule) Let f nd g be differentible on (, b) nd let g (x) 0 on (, b). We llow = nd b =. If for some x 0 in [, b] we hve lim x x0 f(x) = lim x x0 g(x) = 0 or lim x x0 f(x) = lim x x0 g(x) = ±, nd if the limit f (x) lim x x 0 g (x) = L f(x) g(x) exists s rel number or s ±, then the limit lim x x0 nd f(x) lim x x 0 g(x) = lim f (x) x x 0 g (x) = L exists s rel number or ±, Theorem.4 If f nd g nd continuous on [, b] nd k is constnt, then (linerity) (f(x) + g(x)) dx = nd if f(x) g(x) on [, b], then kf(x) dx = k f(x) dx As consequence, if m f(x) M on [, b], then f(x) dx + f(x) dx g(x) dx g(x) dx m(b ) f(x) dx M(b ) 13
14 Theorem.5 Let f be Riemnn continuous on [, b]. Then for ll c in [, b] we hve f(x) dx = c f(x) dx + c f(x) dx Theorem.6 f f is continous on [, b], then f(x) dx = b f(x) dx Remrk.7 This is more mtter of convention thn theorem. Theorem.8 (Averge/Men Vlue Theorem for Integrls) [, b], then there is number c in [, b] such tht If f is continuous on f(c) = 1 b f(x)dx Theorem.9 (Fundmentl Theorem of Clculus I) differentible on (, b), nd if f is continuous on (, b), then f (x) dx = f(b) f() If f is continuous on [, b] nd Theorem.30 (Fundmentl Theorem of Clculus II) then the function F : [, b] R given by F (x) = x f(t) dt is continuous on [, b] nd differentible on (, b), with derivtive given by If f is continuous on [, b], F (x) = f(x) Remrk.31 This theorem, FTC II, gives the existence of ntiderivtives for ny Riemnn integrble function on [, b]. However, this result is only of theoreticl interest, s the form of the ntiderivtive, F (x) = x f(t) dt, isn t relly computtionlly helpful. However, it is useful for fmilirizing yourself with the nottion nd definitions. For exmple, if we wnt to tke the xderivtive of F (sin x), this is strightforwrd, just pply the chin rule: letting y = sin x, d df dy F (sin x) = = f(x) cos x dx dy dx But it s slightly hrder to see in the form d sin x dx f(t) dt. However, the two expressions re the sme! d sin x f(t) dt = df dy = f(x) cos x dx dy dx 14
15 Theorem.3 (Integrtion by Prts) If f nd g re continuous on [, b] nd differentible on (, b), nd if f nd g re Riemnn integrble on [, b], then f(x)g (x) dx = [ ] b f(x)g(x) x= f (x)g(x) dx (.1) Theorem.33 (Chnge of Vrible) [, b] nd g is continuous on [α, β], then If f : [, b] [α, β] is continously differentible on g ( f(x) ) f (x) dx = f(b) f() g(y) dy (.) 15
Review 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 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 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 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 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 informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
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 information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationMath& 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 information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
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 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 informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
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 informationMathematics 19A; Fall 2001; V. Ginzburg Practice Final Solutions
Mthemtics 9A; Fll 200; V Ginzburg Prctice Finl Solutions For ech of the ten questions below, stte whether the ssertion is true or flse ) Let fx) be continuous t x Then x fx) f) Answer: T b) Let f be differentible
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 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 informationCalculus in R. Chapter Di erentiation
Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di
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 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 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 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 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 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 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 informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
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 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. Tody we provide the connection
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 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 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 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 informationBig idea in Calculus: approximation
Big ide in Clculus: pproximtion Derivtive: f (x) = df dx f f(x +h) f(x) =, x h rte of chnge is pproximtely the rtio of chnges in the function vlue nd in the vrible in very short time Liner pproximtion:
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 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 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 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 informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationIMPORTANT THEOREMS CHEAT SHEET
IMPORTANT THEOREMS CHEAT SHEET BY DOUGLAS DANE Howdy, I m Bronson s dog Dougls. Bronson is still complining bout the textbook so I thought if I kept list of the importnt results for you, he might stop.
More informationChapter 6. Riemann Integral
Introduction to Riemnn integrl Chpter 6. Riemnn Integrl WonKwng Prk Deprtment of Mthemtics, The College of Nturl Sciences Kookmin University Second semester, 2015 1 / 41 Introduction to Riemnn integrl
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 informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
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 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 informationWeek 10: Riemann integral and its properties
Clculus nd Liner Algebr for Biomedicl Engineering Week 10: Riemnn integrl nd its properties H. Führ, Lehrstuhl A für Mthemtik, RWTH Achen, WS 07 Motivtion: Computing flow from flow rtes 1 We observe the
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
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 information1 Techniques of Integration
November 8, 8 MAT86 Week Justin Ko Techniques of Integrtion. Integrtion By Substitution (Chnge of Vribles) We cn think of integrtion by substitution s the counterprt of the chin rule for differentition.
More informationCalculus II: Integrations and Series
Clculus II: Integrtions nd Series August 7, 200 Integrls Suppose we hve generl function y = f(x) For simplicity, let f(x) > 0 nd f(x) continuous Denote F (x) = re under the grph of f in the intervl [,x]
More informationDERIVATIVES NOTES HARRIS MATH CAMP Introduction
f DERIVATIVES NOTES HARRIS MATH CAMP 208. Introduction Reding: Section 2. The derivtive of function t point is the slope of the tngent line to the function t tht point. Wht does this men, nd how do we
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 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 information0.1 Chapters 1: Limits and continuity
1 REVIEW SHEET FOR CALCULUS 140 Some of the topics hve smple problems from previous finls indicted next to the hedings. 0.1 Chpters 1: Limits nd continuity Theorem 0.1.1 Sndwich Theorem(F 96 # 20, F 97
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 informationMATH SS124 Sec 39 Concepts summary with examples
This note is mde for students in MTH124 Section 39 to review most(not ll) topics I think we covered in this semester, nd there s exmples fter these concepts, go over this note nd try to solve those exmples
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 informationReview on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.
Mth 20B Integrl Clculus Lecture Review on Integrtion (Secs. 5.  5.3) Remrks on the course. Slide Review: Sec. 5.5.3 Origins of Clculus. Riemnn Sums. New functions from old ones. A mthemticl description
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 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 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 informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
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 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 informationMathematics for economists
Mthemtics for economists Peter Jochumzen September 26, 2016 Contents 1 Logic 3 2 Set theory 4 3 Rel number system: Axioms 4 4 Rel number system: Definitions 5 5 Rel numbers: Results 5 6 Rel numbers: Powers
More informationFinal Exam  Review MATH Spring 2017
Finl Exm  Review MATH 5  Spring 7 Chpter, 3, nd Sections 5.5.5, 5.7 Finl Exm: Tuesdy 5/9, :37:pm The following is list of importnt concepts from the sections which were not covered by Midterm Exm or.
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 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 informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
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 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 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 informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
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 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 informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
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 information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
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 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 information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
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 informationSection Areas and Distances. Example 1: Suppose a car travels at a constant 50 miles per hour for 2 hours. What is the total distance traveled?
Section 5.  Ares nd Distnces Exmple : Suppose cr trvels t constnt 5 miles per hour for 2 hours. Wht is the totl distnce trveled? Exmple 2: Suppose cr trvels 75 miles per hour for the first hour, 7 miles
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 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 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 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 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 information1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x
I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)
More informationPrinciples of Real Analysis I Fall VI. Riemann Integration
21355 Principles of Rel Anlysis I Fll 2004 A. Definitions VI. Riemnn Integrtion Let, b R with < b be given. By prtition of [, b] we men finite set P [, b] with, b P. The set of ll prtitions of [, b] will
More informationMath 118: Honours Calculus II Winter, 2005 List of Theorems. L(P, f) U(Q, f). f exists for each ǫ > 0 there exists a partition P of [a, b] such that
Mth 118: Honours Clculus II Winter, 2005 List of Theorems Lemm 5.1 (Prtition Refinement): If P nd Q re prtitions of [, b] such tht Q P, then L(P, f) L(Q, f) U(Q, f) U(P, f). Lemm 5.2 (Upper Sums Bound
More information( ) Same as above but m = f x = f x  symmetric to yaxis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.
AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find
More informationMath 116 Calculus II
Mth 6 Clculus II Contents 5 Exponentil nd Logrithmic functions 5. Review........................................... 5.. Exponentil functions............................... 5.. Logrithmic functions...............................
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 informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationFINALTERM EXAMINATION 2011 Calculus &. Analytical GeometryI
FINALTERM EXAMINATION 011 Clculus &. Anlyticl GeometryI Question No: 1 { Mrks: 1 )  Plese choose one If f is twice differentible function t sttionry point x 0 x 0 nd f ''(x 0 ) > 0 then f hs reltive...
More information