Math 116 Calculus II


 Barbara Mosley
 2 years ago
 Views:
Transcription
1 Mth 6 Clculus II
2 Contents 5 Exponentil nd Logrithmic functions 5. Review Exponentil functions Logrithmic functions Reltion between Exp nd Log functions Integrtion 5 6. Antiderivtives nd the Rules of integrtion Integrtion by Substitution Are nd the definite integrl The fundmentl theorem of Clculus Evluting definite integrls Are between two curves Applictions of the Definite integrl to Business nd Economics
3 Chpter 5 Exponentil nd Logrithmic functions 5. Review 5.. Exponentil functions Definition: e = f(x) = e x. Properties: domin: (, ). Rnge: (, + ). f() =. Continuous. incresing. Lws of exponents: e x e y = e x+y ; ex e y = ex y ;
4 (e x ) y = e xy ; Differentition: d dx ex = e x ; d dx ef(x) = e f(x) f (x); Exmple Clculte e, e, e,e, e using clcultor. Exmple Grph f(x) = e x, [ 5, 5] [, ] using clcultor. Exmple 3 Find the derivtive of f(x) = xe 3x. Exmple 4 Determine the intervls where f(x) = e x / is incresing nd where it is decresing. 5.. Logrithmic functions Definition: Log functions re introduced s inverse functions of exponentil functions. y = ln x x = e y. ln denotes the nturl log = ln becuse e =. Properties: domin (, + ). Rnge (, ). ln =, becuse e =. ln e =, becuse e = e. ln e x = x, e ln x = x. (ln nd e cn cncel ech other. ) continuous. incresing. Lws of logrithms: ln(xy) = ln x + ln y, (x >, y > ). 3
5 ln( x ) = ln x ln y(x >, y > ). y ln x y = y ln x. Differentition: d dx ln x = x d ln f(x) = dx f(x) f (x). Exmple 5: Clculte ln(.), ln(.), ln e, ln() by using clcultor. Exmple 6: Sketch f(x) = ln x using clcultor. Exmple 7: Find f (x) for f(x) = ln(x + ). Exmple 8: Find f (x) for y = 3 x using logrithmic functions Reltion between Exp nd Log functions y = ln x e y = x. e ln x = x. ln e x = x; Exmple 9: Solve 4 t = 4; Exmple : Solve e.t 4 = 6; Exmple : Solve 3 t = 4; tke nturl log t both sides, ln 3 t = ln 4, t ln 3 = ln 4, t = ln 4 ln 3 + 4
6 Chpter 6 Integrtion 6. Antiderivtives nd the Rules of integrtion Exmple: the distnce tht cr trvels from its initil position is known to be d(t) = 4t ( t 3) Find its velocity t ny time t, ( t 3). Soln: v(t) = d (t) = 8t. This process is clled differentition. Often the velocity of cr cn be red recorded from its speedometer. If we know v(t) = 8t Cn we find the distnce tht cr trvels from its initil position? This is equivlent to finding d(t) such tht d (t) = v(t) = 8t. Exmples of such function include generl forms: d(t) = 4t + C, C is constnt. d(t) = 4t, 4t +, 4t +, A prticulr function, d(t) = 4t or d(t) = 4t + is clled n ntiderivtive. The generl form of of ntiderivtives is clled the indefinite integrl. Tht is 5
7 Antiderivtives of 8t: 4t, 4t +, 4t +,... Indefinite integrl of 8t: 4t + C. The process of finding the distnce function from velocity is clled integrtion, i.e., differentition (unique) distnce velocity integrtion (notunique) Question: How cn we determine which ntiderivtive is the distnce we wnt? Answer: An extr condition initil condition d(t) = 4t + C. initilly::the distnce from initil position is. I.e., d() = = d() = 4 + C C =. d(t) = 4t. Summry: Definition: An ntiderivtive of function f(x) is function, g(x), such tht g (x) = f(x). Definition: The indefinite integrl of function f(x) is the generl form or the fmily of ntiderivtives of f(x), denoted by g(x) + C }{{}}{{} nntiderivtive rbitrry constnt f(x) }{{} integrnt d }{{} x. integrl vrible Generl integrtion rules: [f(x) ± g(x)]dx = cf(x)dx = c f(x)dx. f(x)dx ± g(x)dx. Integrtions rules: dx = x + C. 6
8 x n dx = xn+ + C, n + n. dx = ln x + C. x e x dx = e x + C. Exmple : Find f(x) such tht f (x) = x. Exmple 3: ( + x + x + e x )dx. Exmple 4: x (x4 x + )dx t t Exmple 5: dt. Exmple 6: (x + )(3x )dx. t Exmple 6: find the function f given tht the slope of the tngent line to the grph of the function t (, 3) (initil condition) is f (x) = e x + x. Answers: f(x) = e x + x + 3. Exmple 7 Find the solution of the initil vlue problems: f (x) = 3x 4x + 8 nd f() = 9 }{{} IC Answers: f(x) = x 3 x + 8x Integrtion by Substitution Exmple : Find x(x + 3) 4 dx. Solution:. define the substitution function: u = g(x) = x + 3; (6.) 7
9 . find the totl differentil: 3. Rewrite the integrl: 4. Mke the substitution: 5. Replce u by g(x): du = g (x)dx = xdx; x(x + 3) 4 dx = (x + 3) }{{} 4 (xdx }{{} ) u du = u 4 du = 5 u5 + C = 5 (x + 3) 5 + C; 6. Verifiction: [ ] 5 (x + 3) 5 = x(x + 3) 4. Exmple : Exmple 3: Exmple 4: e 3x dx. Let u = 3x. x x + dx. Let u = x +. dx. Let u = ln x. x(ln x) Exmple 5: Find f(x) given the slope of the tngent line f (x) = Exmple 6 (Appliction Problem): 3x t (, ). x 3 The rte of chnge of the unit price p of Apex Ldies boots is given by p (x) = 5x (6 + x ) 3/ where x is the quntity demnded dily in units of hundred. Find the demnd function for these boots if the quntity demnded dily is 3 pirs (x = 3) where the unit price is 5/pir. If F (x) = f(x)dx f(x + b)dx = F (x + b) proof: 8
10 . u = x + b. du = dx, or dx = du f(x + b)dx = F (x + b) f(u) du = f(u)du = F (u) Exmples: e x+3 dx. e x dx = e x Soln = ex+3 dx. 3x + 3 dx = x x dx = x / / = 4x Soln = 3 [4(3x + 3) ] = 4 3 (3x + 3) 6.3 Are nd the definite integrl We know how to clculte res of the following shpes: A = A = πr h b A = h( + b) How bout 9
11 Exmple Compute the re of the region bounded by y =.5x, x =, x = nd xxis. Method:Approximtion. Procedure:. divide [, ] into n = subintervl. Let x =, x =, x = = (length of subintervls) n A xf(x ) = x.5x =.5. f(x ) f(x ) x x
12 . divide [, ] into n = subintervls: x = n = =.5 x =, x = x + x =.5, x 3 = x + x = A = xf(x ) + xf(x ) = x(f(x ) + f(x )) = (.5x +.5x ) =.85 f(x 3 ) f(x ) f(x ) x x x 3 3. divide [, ] into n = 4 subintervls: x = n = 4 =.5 x =, x =.5, x 3 =.5, x 4 =.75, x 5 = A = xf(x ) + xf(x ) + xf(x 3 ) + xf(x 4 ) = x(f(x ) + f(x ) + f(x 3 ) + f(x 4 )) = (.5x +.5x +.5x 3 +.5x 4) =.9845
13 4. n = 8, x = 8 = 8 =.5 # x i f(x) =.5x sum= Approximtion = x Sum = = When n, the pproximtions go to the ctul re s shown in the following grph.
14 Generl Procedure: Approximte the re between y = f(x) =.5x nd the xxis, on [, b] = [, ]. Given n >, divide [, b] into n subintervls. Length of subintervls : x = b n.. Clculte points: x =, x = + x, x i = + (i ) x,..., x n+ = b 3. Compute pproximtion (using the left endpoint) A = xf(x ) + xf(x ) xf(x n ). }{{} n terms = x[f(x ) f(x n )] In summry, n 4 8 Approximtions = 7 6 This shows x[f(x ) f(x n )] n This limit is clled the definite integrl of f(x) =.5x from x = to x =, denoted by Definition: (definite integrl) b f(x)dx = lim x[f(x ) f(x n )] = lim n }{{} n Riemnn Sum 7 6 width n {}}{ f(x i ) x. }{{} height i=.5x dx. 3
15 Here n i= x = b, or x = b n Existence Theorem: If f is continuous on [, b], then the definite integrl integrble on [, b]. b f(x)dx exists, or f is Geometric interprettion: Generlly speking: b f(x)dx is not the re of the region bounded by y = f(x), x =, x = b nd xxis. It is the re bove the xxis minus the re below the xxis. Exmple The re of the region below =. However ( x + )dx = y = f(x) = x + Actully, if f(x) on [, b], then b b nd the xxis. if f(x) on [, b], then y = f(x), x =,x = b nd the xxis. f(x)dx is the re of the region bounded by y = f(x), x =,x = b f(x)dx is negtive of the re of the region bounded by ( x + )dx = b h = = ( x + )dx = b h = = ( x + )dx = ( x + )dx + ( x + )dx = 4
16 6.4 The fundmentl theorem of Clculus Theorem Let F (x) be ny ntiderivtive of f(x): F (x) = f(x). Then b f(x)dx = F (x) b F (b) F (). (definite integrl, which hs unique vlue) Exmple : Find the re of the region bounded by y = x, x =, x = nd the xxis. A = x dx = x3 3 = = 7 3 Exmple Find the re of the region bounded by y = x + 4, x =, x = 3 nd the xxis. Are 3 ( x + 4)dx Are = x + 4 =, x = 4, x = ± ( x + 4)dx 3 ( x + 4)dx Exmple 3 : Find the re of the region bounded by y = x + from x = to x =. A = (x + )dx = x3 3 + x = 3 3 ( )3 + ( ( )). 3 Exmple 4 Clculte 3 3 (3x + e x )dx. (3x + e x )dx = 3x e x 3 = e 3 e = 6 + e 3 e. Exmple 5: Find the re of region bounded by the grph of y = x, x =, x = nd the xxis. y = 3 y = x 4 5
17 Soln: A = [ ( x )]dx = x3 3 = 3 3 ( )3 3 = 3. Note tht re is lwys positive number. But the integrl of x over [, ] is negtive. x dx = 3. Ex 3: Find the re of the region bounded by y = x +, x =, x = nd xxis. Note tht the integrl ( x + )dx is positive in the blck region nd negtive in the red region. So the re is the sum bsolute vlue of these two prts, which leds to [ ] A = ( x + )dx + ( x + )dx = or A = ( x + )dx + ( x + )dx = Ex 4: Find the re enclosed by the grph of f(x) nd xxis on the intervl [, b]. () f(x) = (x )(x 3), [, 4] Find the roots of f(x) = : (x )(x 3) =, x =, 3 x =, 3 divides [, 4] into [, ], [, 3] nd [3, 4]. Find the re of ech intervl nd dd them up: 3 Are= f(x)dx + 4 f(x)dx + f(x)dx (b) f(x) = (x )(x 3), [, 4] Find the roots of f(x) = : x =, 3. x =, 3 divides [, 4] into [, 3] nd [3, 4]. 6 3
18 Are= 3 f(x)dx f(x)dx. (c) f(x) = (x )(x 3), [4, 5] Find the roots of f(x) = : x =, 3 x =, 3 divides [4, 5] into [4, 5]. 5 Are= f(x)dx Evluting definite integrls Integrtion Rules: b b cf(x)dx = c f(x)dx. b b b [f(x) ± g(x)]dx = f(x)dx + g(x)dx. f(x)dx =. b f(x)dx = f(x)dx. b b c b f(x)dx = f(x)dx + f(x)dx. Substitution Method Exmple : c xe x + dx. Let u = x +, du = xdx. then xe x + dx = u {}}{ e (x + ) }{{} xdx = du + + e u du = e u 5 = e 5 e. Since we re performing the integrtion with respect to new vrible u now, the rnge of integrtion hs to be chnged to reflect the fct tht the integrtion is being performed with respect to the new vrible u. Tht is sying, the upper limit nd lower limit hs to be chnged correspondingly. In this exmple, since u = x +, the upper limit is chnged to + = 5, nd the lower limit is chnged to + =. Generl Cse: if x u = u(x), b u(b) u() 7
19 Exmple : x (x 3 + ) 4 dx. Let u = x 3 +, du = 3x dx. x (x 3 + ) 4 dx = (x 3 + ) }{{} 4 x } {{ dx } = u 3 du = u = ( ) u4 du = 3 5. Averge of function: verge of f(x) on [, b] f(x ) + f(x ) f(x n ) n = b x[f(x ) f(x n )], x = b }{{} n b b Riemnn Sum f(x)dx. verge of f(x) on [, b] = b b f(x)dx. Exmple : A cr is trveling t velocity v(t) from time to b. Find the verge velocity of this cr during this time period. Soln: Totl Distnce= b v(t)dt verge velocity = Totl Distnce Totl time = b v(t)dt b Exmples xe x dx x x 3 + dx 4 x 9 + x dx 8
20 6.6 Are between two curves b f(x)dx is the re of the region bounded by y = f(x), x =, x = b nd xxis. If f(x), for ll x [, b]. Finding the re between two curves: If f(x) g(x) on [, b], then the re of the region bounded bove by y = f(x) nd below by y = g(x) on [, b] is given by b [f(x) g(x)]dx 5 4 y = f(x) 3 y = g(x) Approximtion: Height=f(x) g(x), width= x = b n. b Are n i= width {}}{ f(x) g(x) dx }{{} Height [f(x i ) g(x i )] b n = lim n n f(x i ) g(x i ) x i= n = 9
21 4 y = f(x) 3 y = g(x) n = 4 4 y = f(x) 3 y = g(x) n = 4 y = f(x) 3 y = g(x)
22 Exmple : Find the re of the region tht is completely enclosed by the grphs of y = x nd g(x) = x 4. 6 A 4 y = x A y = x 4 4 Soln: We first find out the intersection point of these two curves by solving Then A = 3 [(x ) (x 4)]dx. x = x 4, x x 3 =, (x 3)(x + ) =, x =, 3. Exmple : Find the re of the region completely enclosed by the grphs of the functions f(x) = x 3 3x + 3, nd g(x) = x + 3
23 6 f(x) = x 3 3x + 3 g(x) = x + 3 A 3 4 A A We get the xvlues of the points of intersection of f(x) nd g(x) by solving x 3 3x + 3 = x + 3 x 3 4x = (x 4)x = (x )(x + )x = x =,,. In the red region, since f(x) g(x), its re is its re is g(x) f(x)dx. So A = f(x) g(x)dx + f(x) g(x)dx. In the blck region, since g(x) f(x), g(x) f(x)dx = 8 Exmples () f(x) = x x +, g(x) = x f(x) = g(x), x x + = x, x( x + x) = x =, or x + x = x + = x, x + = x, x x =, (x )(x + ) =, x =, However, x = is not solution becuse +. x =,
24 A = x x + x dx (b) f(x) = x 3 3x, g(x) = x f(x) = g(x), x 3 x 3x =, x(x 3)(x + ) =, x =,, 3 3 Are= f(x) g(x)dx + f(x) g(x)dx 6.7 Applictions of the Definite integrl to Business nd Economics Consumers nd Producers Surplus: Consumers Surplus: Demnd { function reltes unit price p of commodity to the quntity x demnded of it. if p is high, x is smll if p is low, x is lrge p = D(x) indictes how much people is willing to py for the unit price t level quntity x. Suppose tht fixed unit mrket price p hs been estblished for the commodity nd the corresponding quntity demnded is x. Thus the ctul pyment by consumers = p x the mount consumers re willing to py = x D(x)dx. consumers Surplus = mount consumers re willing to py 3 = mount consumers ctully py x D(x)dx p x. Demnd function p = D(x) (willing to py) p 3 x 3
25 p consumers surplus p consumers ctully py p consumers willing to py 3 x 3 x 3 x Producers Surplus Supply function reltes the unit price of commodity p nd the quntity x produced by the supplier. Let p = S(x), if p is high, then x is lrge; if p is low, then x is smll. p = S(x) indictes how much the producers re willing to receive for the unit price t the level of quntity x. p p = S(x) 3 4 x Suppose tht fixed unit mrket price p hs been estblished for the commodity nd the corresponding quntity supplied is x. The ctul mount received by producers = p x (t the current mrket) The mount producers re willing to receive = x S(x)dx Producers surplus = mount ctully received mount tht producers re willing to received = p x x S(x)dx. 4
26 Actul received Willing to receive p p 3 x 3 x p Producers surplus 3 x Exmple The demnd function for commodity is p = D(x) =.x + 5 where p is the unit price in dollrs nd x is the quntity demnded in units of thousnd. The supply function for the commodity is p = S(x) =.6x +. + Determine the consumers nd producers surplus if the mrket price of the commodity is set t the equilibrium price. 5
27 D(x) S(x) p x x x x : supply < demnd, x x : supply > demnd, x x: supply = demnd, mrket price, stble. Soln: x is the solution of S(x) = D(x). Solve the eqution, we get x = 65, 3. Since x cn t be negtive, x = 3. And p = D( x) = 6. Consumers surplus = 3 Producers surplus = p x The future nd present vlue of n income strem D(x)dx p x = 8,. 3 S(x)dx =, 7. firm genertes strem of income over period of time. R(t) rte of income genertion t time t. E.g., R(t) = 3 dollr/hour. The relized income is reinvested (in the bnk) nd erns interest t fixed rte: r interest rte compounded continuously The period of time = T 6
28 If put in the bnk, nd interest is compounded continuously, Totl interest + Principle = Principl e rt Principl: the initil money you borrowed from or deposited to the bnk. t is the totl mount of time. Future vlue & present vlue P R(t)=rte of income reinvested nd erns interest rte r F (Now) T Wht is the totl mount F? (future vlue) Future vlue f = T T R(t)e r(t t) dt = e rt R(t)e rt dt. R(t) Y = R(t) t t + t R(t) t: Profit generted from time period (t, t + t). R(t) t e r(t t) : future vlue of profit generted from time period (t, t + ). }{{} Principl T lim R(t) t e T t = R(t)e r(t t) dt: totl income t 7
29 Present Vlue P R(t)=rte of income reinvested nd erns interest rte r F (Now) Present vlue P =? T future vlue F The present vlue P tht will yield the sme ccumulted vlue s the income strem itself when P is invested for the sme period of time t the sme rte of interest. T P e rt = F = e rt R(t)e rt dt P = T R(t)e rt dt. Exmple : The owner of locl cinem is studying pln for renovting nd improving the theter. the pln clls for n immedite outly of $ 5,. It hs been estimted tht the pln would result in net income strem generted t the rte of R(t) = $63, per yer If the previling interest rte for the next 5 yers is % nnully, determine the net income in the present vlue nd the futher vlue t the end of 5 yers. Soln: R(t) = $63,, r =., T = 5 5 Future vlue of the pln F = e. 5 63, e.t dt =, 63, 845 e. 5 =, 69, 6 future vlue of the cost 5, if put in the bnk B = 5e. 5 = 4, 8. A B =, 79, 96 (future vlue of profit) present vlue P = 5 63, e.t dt =, 63, 845 P cost =, 63, 845 5, =, 38, 845 8
30 Review: Chpter 5 & 6 e.7888 e x+y = e x e y e x y = ex e x e y (e x ) y = e xy d dx = d ex, dx ef(x) = e f(x) f (x) ln(xy) ( ) = ln x + ln y x ln = ln x ln y y ln x ln x y = y ln x d dx ln x = x d ln f(x) = dx f(x) f (x) Reltion:ln e x = x, e ln x = x. Chpter 6 Integrtion: see pge Problems: (Review of chpter 6, Pge ) 4, 3, 34, 38, 5. Review: Integrtion Concepts: differentition integrtion. ntiderivtives F (x) = f(x). indefinite integrl f(x)dx = F (x) + C. definite integrl Riemnn Sum b f(x)dx = F (b) F (). reltion between re nd definite integrl. Consumers surplus: CS = x D(x)dx p x. 9
31 Producers surplus: P S = p x Present Vlue: P = T x R(t)e rt dt. T Future Vlue: F = e rt R(t)e rt dt. Integrtion Techniques: S(x)dx. f(x) ± g(x)dx = cf(x)dx = c dx = x + C. f(x)dx. f(x)dx + x n dx = xn+ + C, n. n + dx = ln(x) + C. x e x dx = e x + C. e x dx = ex + C,. g(x)dx. f(x)dx =. b f(x)dx = f(x)dx. b b c f(x)dx = f(x)dx + Method of substitution. b fundmentl theorem of clculus: Applictions: c f(x)dx. b f(x)dx = F (x) b = F (b) F (). Compute res between curves. 3
32 compute CS, PS, FV,PV verge= b b f(x)dx. 3
x = 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 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 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 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 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 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 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 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 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 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 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 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 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 informationThe Fundamental Theorem of Calculus, Particle Motion, and Average Value
The Fundmentl Theorem of Clculus, Prticle Motion, nd Averge Vlue b Three Things to Alwys Keep In Mind: (1) v( dt p( b) p( ), where v( represents the velocity nd p( represents the position. b (2) v ( dt
More information5 Accumulated Change: The Definite Integral
5 Accumulted Chnge: The Definite Integrl 5.1 Distnce nd Accumulted Chnge * How To Mesure Distnce Trveled nd Visulize Distnce on the Velocity Grph Distnce = Velocity Time Exmple 1 Suppose tht you trvel
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 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 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 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 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 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 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 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 informationMath 190 Chapter 5 Lecture Notes. Professor Miguel Ornelas
Mth 19 Chpter 5 Lecture Notes Professor Miguel Ornels 1 M. Ornels Mth 19 Lecture Notes Section 5.1 Section 5.1 Ares nd Distnce Definition The re A of the region S tht lies under the grph of the continuous
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 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 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 informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
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 informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
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 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 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 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 informationMath 116 Final Exam April 26, 2013
Mth 6 Finl Exm April 26, 23 Nme: EXAM SOLUTIONS Instructor: Section:. Do not open this exm until you re told to do so. 2. This exm hs 5 pges including this cover. There re problems. Note tht the problems
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 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 informationTest 3 Review. Jiwen He. I will replace your lowest test score with the percentage grade from the final exam (provided it is higher).
Test 3 Review Jiwen He Test 3 Test 3: Dec. 46 in CASA Mteril  Through 6.3. No Homework (Thnksgiving) No homework this week! Hve GREAT Thnksgiving! Finl Exm Finl Exm: Dec. 1417 in CASA You Might Be Interested
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 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 informationCalculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties
Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ fuehr@mth.rwthchen.de Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009 Overview 1 Motivtion:
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 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 . 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 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 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 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 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 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 informationx = b a N. (131) The set of points used to subdivide the range [a, b] (see Fig. 13.1) is
Jnury 28, 2002 13. The Integrl The concept of integrtion, nd the motivtion for developing this concept, were described in the previous chpter. Now we must define the integrl, crefully nd completely. According
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 informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
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 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 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 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 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 information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
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 informationMath RE  Calculus II Area Page 1 of 12
Mth RE  Clculus II re Pge of re nd the Riemnn Sum Let f) be continuous function nd = f) f) > on closed intervl,b] s shown on the grph. The Riemnn Sum theor shows tht the re of R the region R hs re=
More informationMath Bootcamp 2012 Calculus Refresher
Mth Bootcmp 0 Clculus Refresher Exponents For ny rel number x, the powers of x re : x 0 =, x = x, x = x x, etc. Powers re lso clled exponents. Remrk: 0 0 is indeterminte. Frctionl exponents re lso clled
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 informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
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 informationF (x) dx = F (x)+c = u + C = du,
35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil
More informationUnit Six AP Calculus Unit 6 Review Definite Integrals. Name Period Date NONCALCULATOR SECTION
Unit Six AP Clculus Unit 6 Review Definite Integrls Nme Period Dte NONCALCULATOR SECTION Voculry: Directions Define ech word nd give n exmple. 1. Definite Integrl. Men Vlue Theorem (for definite integrls)
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 informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
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 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 informationOverview of Calculus
Overview of Clculus June 6, 2016 1 Limits Clculus begins with the notion of limit. In symbols, lim f(x) = L x c In wors, however close you emn tht the function f evlute t x, f(x), to be to the limit L
More information!0 f(x)dx + lim!0 f(x)dx. The latter is sometimes also referred to as improper integrals of the. k=1 k p converges for p>1 and diverges otherwise.
Chpter 7 Improper integrls 7. Introduction The gol of this chpter is to meningfully extend our theory of integrls to improper integrls. There re two types of soclled improper integrls: the first involves
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 informationUnit #10 De+inite Integration & The Fundamental Theorem Of Calculus
Unit # De+inite Integrtion & The Fundmentl Theorem Of Clculus. Find the re of the shded region ove nd explin the mening of your nswer. (squres re y units) ) The grph to the right is f(x) = x + 8x )Use
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 informationSample Problems for the Final of Math 121, Fall, 2005
Smple Problems for the Finl of Mth, Fll, 5 The following is collection of vrious types of smple problems covering sections.8,.,.5, nd.8 6.5 of the text which constitute only prt of the common Mth Finl.
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 informationCalculus III Review Sheet
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
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 Functions Defined in Terms of Integrals
November 5, 8 MAT86 Week 3 Justin Ko Functions Defined in Terms of Integrls Integrls llow us to define new functions in terms of the bsic functions introduced in Week. Given continuous function f(), consider
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 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 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 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 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 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 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 information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a nonconstant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
More informationES 111 Mathematical Methods in the Earth Sciences Lecture Outline 1  Thurs 28th Sept 17 Review of trigonometry and basic calculus
ES 111 Mthemticl Methods in the Erth Sciences Lecture Outline 1  Thurs 28th Sept 17 Review of trigonometry nd bsic clculus Trigonometry When is it useful? Everywhere! Anything involving coordinte systems
More informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationPrep Session Topic: Particle Motion
Student Notes Prep Session Topic: Prticle Motion Number Line for AB Prticle motion nd similr problems re on the AP Clculus exms lmost every yer. The prticle my be prticle, person, cr, etc. The position,
More informationSection 5.4 Fundamental Theorem of Calculus 2 Lectures. Dr. Abdulla Eid. College of Science. MATHS 101: Calculus 1
Section 5.4 Fundmentl Theorem of Clculus 2 Lectures College of Science MATHS : Clculus (University of Bhrin) Integrls / 24 Definite Integrl Recll: The integrl is used to find re under the curve over n
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 informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
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 informationDistance And Velocity
Unit #8  The Integrl Some problems nd solutions selected or dpted from HughesHllett Clculus. Distnce And Velocity. The grph below shows the velocity, v, of n object (in meters/sec). Estimte the totl
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 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 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 informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More information