Math 116 Calculus II


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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
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