MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus. Lern how to compute the ntiderivtive of some sic functions. Lern how to use the sustitution method to compute the ntiderivtive of more complex functions. Lern how to solve re nd distnce trveled prolems y mens of ntiderivtives. Assignment 22 Assignment 23 Assignment 24 (Review) So fr we hve lerned out the ide of the integrl, nd wht is ment y computing the definite integrl of function f(x) over the intervl [,]. As in the cse of derivtives, we now study procedures for computing the definite integrl of function f(x) over the intervl [,] tht re esier thn computing limits of Riemnn sums. As with derivtives, however, the definition is importnt ecuse it is only through the definition tht we cn understnd why the integrl gives the nswers to prticulr prolems. Ide of the Fundmentl Theorem of Clculus: The esiest procedure for computing definite integrls is not y computing limit of Riemnn sum, ut y relting integrls to (nti)derivtives. This reltionship is so importnt in Clculus tht the theorem tht descries the reltionships is clled the Fundmentl Theorem of Clculus. Computing some ntiderivtives: In previous chpters we were given function f(x) nd we found the derivtive f (x). In this section, we will do the reverse. We will e given function f(x) tht is the derivtive of nother function F(x) nd will compute F(x). In other words find function F(x) such tht F (x) = f(x). F(x) is clled n ntiderivtive of f(x). For exmple (x 3 ) = 3x 2 so n ntiderivtive of f(x) = 3x 2 is F(x) = x 3. Note tht F(x) = x 3 +2 is lso n ntiderivtive of f(x) = 3x 2 ecuse (x 3 +2) = 3x 2. In generl, if F(x) is n ntiderivtive of f(x), then so is F(x)+C where C is ny constnt. This leds to the following nottion. Definition of the indefinite integrl: The indefinite integrl of f(x), denoted y f(x)dx without limits of integrtion, is the generl ntiderivtive of f(x). For exmple, it is esy to check tht 3t 2 dt = t 3 +c, where c is ny constnt. Recll tht the power rule for derivtives gives us (x n ) = nx n 1. We multiply y n nd sutrct 1 from the exponent. Since ntiderivtives re the reverse of derivtives, to compute n the ntiderivtive we first increse the power y 1, then divide y the new power. 15
The formuls elow cn e verified y differentiting the righthnd side of ech expression. Some sic indefinite integrls: 1. x n dx = 1 n+1 xn+1 +C n 1 2. e x dx = e x +C n = 1 in formul 1 leds to division y zero, ut for this specil cse we my use (ln(x)) = 1 x : 1 3. dx = ln x +C x Rules for indefinite integrls: ( ) ( A. cf(x)dx = c f(x)dx B. (f(x)±g(x))dx = f(x)dx ± Exmple 1: Evlute the indefinite integrl (t 3 +3t 2 +4t+9)dt. ) g(x) dx Exmple 2: Evlute the indefinite integrl 6 t dt. Wrning: We do not hve simple derivtive rules for products nd quotients, so we should not expect simple integrl rules for products nd quotients. Exmple 3: Evlute the indefinite integrl t 3 (t+2)dt. 16
Exmple 4: Evlute the indefinite integrl 2 +9 dx. x 2 We now hve some experience computing ntiderivtives. We will now see how ntiderivtives give us n elegnt method for finding res under curves. Exmple 5: Find formul for A(x) = 1 (4t + 2)dt, tht is, evlute the definite integrl of the function f(t) = 4t+2 over the intervl [1,x] inside [1,1]. (Hint: think of this definite integrl s n re.) Find the vlues A(5), A(1), A(1). Wht is the derivtive of A(x) with respect to x? y f(t) = 4t+2 6 A(x) 1 x 1 t Oservtions: There re two importnt things to notice out the function A(x) nlyzed in Exmple 1: A(1) = 1 1 (4t+2)dt = A (x) = d ( ) (4t+2)dt = 4x+2. dx 1 }{{} A(x) Notice wht the lst equlity sys: The instntneous rte of chnge of the re under the curve y = 4t+2 t t = x is simply equl to the vlue of the curve evluted t t = x. Why? A(x) mesures the re of some geometric figure. As x increses, the width of the figure increses, nd so the re increses. A (x) mesures the rte of increse of the figure. Now, s x increses, the right wll of the figure sweeps out dditionl re, so the rte t which the re increses should e equl to the height of the right wll. The following pges will mke this ide more precise. 17
Ide: Suppose tht for ny function f(t) it were true tht the re function A(x) = A() = f(t)dt = A (x) = d dx ( ) f(t)dt = f(x). f(t) dt stisfies Moreover, suppose tht F(x) is ny ndtiderivtive of f(x) (i.e., F (x) = f(x) = A (x).) By The Constnt Function Theorem (Chpter 6), there exists constnt vlue c such tht F(x) = A(x)+c. All these fcts put together help us esily evlute f(t) dt. Indeed, f(t)dt = A() = A() = A() A() = F() F() = [A()+c] [A()+c] The ove specultions re ctully true for ny continuous function on the intervl over which we re integrting. These results re stted in the following theorem, which is divided into two prts: The Fundmentl Theorem of Clculus: PART I: Let f(t) e continuous function on the intervl [,]. Then the function A(x), defined y the formul A(x) = f(t)dt for ll x in the intervl [,], is n ntiderivtive of f(x), tht is A (x) = d ( ) f(t)dt = f(x) dx for ll x in the intervl [,]. PART II: Let F(x) e ny ntiderivtive of f(x) on [,], so tht F (x) = f(x) for ll x in the intervl [,]. Then f(x)dx = F() F(). Specil nottions: The ove theorem tells us tht evluting definite integrl is two-step process: find ny ntiderivtive F(x) of the function f(x) nd then compute the difference F() F(). A nottion hs een devised to seprte the two steps of this process: F(x) stnds for the difference F() F(). Thus f(x)dx = F(x) = F() F(). 18
Some properties of definite integrls: 1. 3. 4. f(x)dx = 2. ( (f(x)±g(x))dx = f(x)dx+ c f(x)dx = ) ( f(x)dx ± c ) g(x) dx f(x)dx 5. 6. If m f(x) M on [,] then m( ) Geometric illustrtion of some of the ove properties: f(x)dx M( ) kf(x)dx = k f(x)dx = f(x)dx f(x)dx Property 3. sys tht if f nd g re positive vlued Property 4. sys tht if f(x) is positive vlued functions with f greter thn g. Then function then the re underneth the grph of f(x) y y etween nd plus f (f(x) g(x))dx the re underneth gives the re etween the grph of f(x) etween nd c equls the grphs of f nd g the re underneth g f(x)dx g(x) dx the grph of f(x) x x etween nd c. c Property 5. follows from Properties 4. nd 1. y letting c =. Alterntively, the Fundmentl Theorem of Clculus gives us f(x)dx = F(x) = F() F() = [F() F()] = F(x) = f(x)dx. Property 6. is illustrted in the picture tht ccompnies the proof of the Fundmentl Theorem of Clculus. An ide of the proof of the Fundmentl Theorem of Clculus: We lredy gve n explntion of why the second prt of the Fundmentl Theorem of Clculus follows from the first one. To prove the first prt we need to use the definition of the derivtive. More precisely, we must show tht A A(x+h) A(x) (x) = lim = f(x). h h For convenience, let us ssume tht f is positive vlued function. Given tht A(x) is defined y quotient is f(t)dt, the numertor of the ove difference M m y h A(x+h) A(x) = +h f(t)dt f(t)dt. Using properties 4. nd 5. of definite integrls, the ove difference equls +h x x x+h f(t)dt. As the function f is continuous over the intervl [x,x +h], the Extreme Vlue Theorem sys tht there re vlues c 1 nd c 2 in [x,x+h] wheref ttins theminimumndmximumvlues, sy mndm, respectively. Thus m f(t) M on [x,x+h]. As the length of the intervl [x,x+h] is h, y property 6. of definite integrls we hve tht t f(c 1 )h = mh +h x f(t)dt Mh = f(c 2 )h or, equivlently, f(c 1 ) +h x f(t)dt h f(c 2 ). Finlly, s f is continuous we hve tht lim h f(c 1 ) = f(x) = lim h f(c 2 ). This concludes the proof. 19
Exmples illustrting the First Fundmentl Theorem of Clculus: Exmple 6: Compute the derivtive of F(x) if F(x) = 2 (t 4 +t 3 +t+9)dt. Exmple 7: Compute the derivtive of g(s) if g(s) = s 5 8 u 2 +u+2 du. Exmple 8: Suppose f(x) = 1 t 2 7t+12.25dt. For which positive vlue of x does f (x) equl? Exmple 9: Find the vlue of x t which F(x) = intervl [3,1]. The vlue of x tht gives minimum of F(x) is. 3 (t 8 +t 6 +t 4 +t 2 +1)dt tkes its minimum on the 11
Exmple 1: Find the vlue of x t which G(x) = [ 5,1]. The vlue of x tht gives mximum of G(x) is. 5 ( t +2)dt tkes its mximum on the intervl Exmples illustrting the Second Fundmentl Theorem of Clculus: Exmple 11: Evlute the integrl 5 (t 2 +1)dt. Exmple 12: Evlute the integrl 5 7 ( ) 1 2 dt. t Exmple 13: Evlute the integrl 2 e x dx. 111
Exmple 14: Evlute the integrl 2 2 (t 5 +t 4 +t 3 +t 2 +t+1)dt. Exmple 15: Evlute the integrl 12 6 t dt. Exmple 16: Evlute the integrl 5 2 ( 3u 5 + 7 ) du. u Exmple 17: Suppose Evlute the integrl 5 f(x)dx. 2x, x 2; f(x) = 8 x, x > 2. 112
The sustitution rule for integrls: suintervl I nd f is continuous on I, then In cse of definite integrls the sustitution rule ecomes Exmple 18: Evlute the integrl If u = g(t) is differentile function whose rnge is f(g(t))g (t)dt = f(u)du. (t+9) 2 dt. f(g(t))g (t)dt = g() g() f(u)du. Exmple 19: Evlute the integrl 3t+7dt. Exmple 2: Evlute the integrl 1 (5t+4) 2 dt. Exmple 21: Evlute the integrl 5 2t+1dt. 113
Exmple 22: Evlute the integrl 1 5e 5x+1 dx Exmple 23: Evlute the integrl 3 2x x 2 +1 dx. Exmple 24: Compute the derivtive of F(x) if F(x) = (Hint: Write F(x) = g(f(x)) where f(x) = x 2 nd g(x) = 2 2tdt. 2tdt nd use the chin rule to find F (x).) Generl formul: If F(x) = f(x) H(t)dt then F (x) = d dx f(x) H(t)dt = H(f(x)) f (x). 114
Word prolems: If the velocity v(t) of n oject t time t is lwys positive, then the re underneth the grph of the velocity function nd lyingove thet-xis represents thetotl distnce trveled y theoject from t = to t =. Exmple 25: A trin trvels long trck nd its velocity (in miles per hour) is given y v(t) = 76t for the first hlf hour of trvel. Its velocity is constnt nd equl to v(t) = 76/2 fter the first hlf hour. Here time t is mesured in hours. How fr (in miles) does the trin trvel in the second hour of trvel? Exmple 26: A trin trvels long trck nd its velocity (in miles per hour) is given y v(t) = 76t for the first hlf hour of trvel. Its velocity is constnt nd equl to v(t) = 76/2 fter the first hlf hour. Here time t is mesured in hours. How fr (in miles) does the trin trvel in the first hour of trvel? Exmple 27: A rock is dropped from height of 21 feet. Its velocity in feet per second t time t fter it is dropped is given y v(t) = 32t where time t is mesured in seconds. How fr is the rock from the ground one second fter it is dropped? 115
Exmple 28: Suppose n oject is thrown down from cliff with n initil speed of 5 feet per sec, nd its speed in ft/sec fter t seconds is given y s(t) = 1t+5. If the oject lnds fter 7 seconds, how high (in ft) is the cliff? (Hint: how fr did the oject trvel?) Exmple 29: A cr is trveling due est. Its velocity (in miles per hour) t time t hours is given y v(t) = 2.5t 2 +1t+5. How fr did the cr trvel during the first five hours of the trip? Averge Vlues: The verge of finitely mny numers y 1,y 2,...,y n is y ve = y 1 +y 2 + +y n. Wht n if we re deling with infinitely mny vlues? More generlly, how cn we compute the verge of function f defined on n intervl? Averge of function: Theverge offunctionf onnintervl [,] y equls the integrl of f over the intervl divided y thelength of the intervl: Geometric mening: f(x)dx f ve =. x If f is positive vlued function, f ve is tht numer such tht the rectngle with se [,] nd height f ve hs the sme re s the region underneth the grph of f from to. Exmple 3: Wht is the verge of f(x) = x 2 over the intervl [,6]? f ve 116
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