7.2 The Definite Integral


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1 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 F is ny ntiderivtive of f. This result is usully expressed in terms of n integrl, clled the definite integrl of f on [, b], nd denoted by: (Red s the definite integrl of f, from to b.) In this section, study of the definite integrl begins. the ctul definition of is bit complicted The ctul definition of the definite integrl is bit complicted, due minly to the fct tht (not surprisingly!) it is defined in terms of limit. The precise definition of is presented in the next section. This definition revels the following fcts (which you cn tke on fith for the moment, nd strt understnding now): For continuous function f, is NUMBER; nd if f hppens to be nonnegtive on [, b], then this number hs very nice interprettion; it gives the re under the grph of f on [, b]. Since F (b) F () gives this sme re, where F is ny ntiderivtive of f, we cn in this cse write: = F (b) F () lower limit of integrtion; upper limit of integrtion The definite integrl nd the indefinite integrl hve similr ppernces. The only difference is tht the definite integrl hs numbers nd b dorning the integrl sign. These two new components hve nmes: is clled the lower limit of integrtion b is clled the upper limit of integrtion compring the definite nd indefinite integrls Since the ppernce of the two integrls is so similr, you should be sking yourself the following questions: Why the similr ppernce? How re these integrls the sme? How re they different? Here re some nswers to these questions. 8
2 copyright Dr. Crol JV Fisher Burns 9 How re the integrls different? How re the integrls the sme? EXERCISE The definite integrl is NUMBER. If f is nonnegtive on [, b], then this number hs nice interprettion s the re under the grph of f on [, b]. However, the indefinite integrl is n INFINITE CLASS OF FUNCTIONS; ll the ntiderivtives of f. So, in one sense, the two integrls re very, very different. In nother sense, however, they re very much the sme. It will be seen tht if JUST ONE ntiderivtive of f is known, then the definite integrl cn be computed. This fct hs lredy been estblished in specil cse when f is continuous nd nonnegtive on [, b] nd we will see tht it ctully holds for ny continuous function f. This is precisely the content of the Fundmentl Theorem of Integrl Clculus (to be presented momentrily); nd is the justifiction for the similrity in the ppernce of the two integrls. TRUE or FALSE:. The definite integrl is function. 2. The number 5 x2 dx gives the re under the grph of x 2 on the intervl [, 5].. For continuous function f, =.. This text hs not yet presented the ctul definition of. We ve run cross sitution before where (precise) definition ws hrd to work with, nd fortuntely we could often get wy with NOT working with the definition. Remember the definition of the derivtive of function? f (x) := lim h f(x + h) f(x) h Evluting this limit ws nuisnce, even for firly simple functions f. Fortuntely, this definition rrely needs to be used ny more, becuse the definition ws USED to develop tools tht llow us to therefter BYPASS the definition; tools such s the simple power rule, chin rule, product rule, nd quotient rule. Similrly, the Fundmentl Theorem of Integrl Clculus gives convenient tool for computing the definite integrl of f, whenever we cn get our hnds on n ntiderivtive of f. Here s precise sttement of the fundmentl theorem: Fundmentl Theorem of Integrl Clculus Let f be continuous on [, b]. If F is ny ntiderivtive for f on [, b], then: = F (b) F () ntiderivtive of f on [, b] For F to be n ntiderivtive of f on [, b], not only must F (x) = f(x) for ll x (, b), but F must lso be continuous on the closed intervl [, b]. In prticulr, F must behve properly t the endpoints nd b.
3 copyright Dr. Crol JV Fisher Burns some nottion used in connection with the definite integrl; F (x) b Let F be n ntiderivtive of f. The following nottion is used in connection with evluting the definite integrl: = F (x) b = F (b) F () Tht is, the nottion F (x) b is used to represent the opertion of evluting the ntiderivtive F t b, evluting it t, nd then subtrcting these two numbers, s illustrted in the next exmple. EXAMPLE Problem: Compute the definite integrl: x dx Solution: Find n ntiderivtive of x, nd use the Fundmentl Theorem. The simplest ntiderivtive of f(x) = x is F (x) = x. Using this ntiderivtive to evlute the definite integrl yields: x dx = x 2 = (2) () = = Since x is positive on [, 2], the number f on [, 2]. gives the re under the grph of fctor constnts out first Note tht the constnt ppers in both terms (shown bold bove) in the evlution process. It is usully esiest to fctor this constnt out first, nd more simply write: x dx = x 2 = (2 ) = (6 ) = 5 = EXAMPLE Problem: Compute the definite integrl: x dx Solution: Find n ntiderivtive of x, nd use the Fundmentl Theorem. Observe tht, this time, x is NOT positive over the entire intervl of integrtion. Applying the fundmentl theorem: x dx = x = ( ( ) ) = ( ) = () = Momentrily, it will be mde cler why the nswer is. Any specultion?
4 copyright Dr. Crol JV Fisher Burns EXERCISE 2 Evlute the following definite integrls. Be sure to write complete mthemticl sentences. When possible, interpret your nswer in terms of re.. x5 dx 2. ex dx. ex dx. 5. /2 x dx x dx dummy vrible of integrtion Since the definite integrl is number, the vrible of integrtion is irrelevnt. Tht is, once is evluted, the letter x is gone. Any letter my be used; for exmple, one cn write or f(t) dt or f(ω) dω ; they re ll equl. Just be sure to crry this sme letter through your computtions; for exmple: t dt = t 2 = (2 ) = The letter used in is clled the dummy vrible of integrtion. EXERCISE Suppose you KNOW. Then, do you know f(t) dt? How bout b f(s) ds? How bout d c? How bout g(x) dx? To begin to better understnd the definite integrl, some properties tht it stisfies re stted next. Properties of the Definite Integrl linerity Suppose tht f is continuous on [, b]. For ll constnts k : k = k Tht is, constnts cn be pulled out of the definite integrl. Also: f(x) ± g(x) dx = ± Tht is, the integrl of sum is the sum of the integrls. g(x) dx Together, these two properties re referred to s the linerity of the (definite) integrl.
5 2 copyright Dr. Crol JV Fisher Burns Properties of the Definite Integrl dditivity There re lso properties tht relte to the limits of integrtion. Agin suppose tht f is continuous on [, b]. For ny c (, b): = c + c Tht is, to integrte from to b, one cn choose ny number c between nd b, nd integrte insted in two pieces: from to c, nd then from c to b. This property is referred to s the dditivity of the integrl. Finlly, for ll rel numbers nd b: = nd = b The lst property sys tht if you integrte bckwrds, you must introduce minus sign. If F is n ntiderivtive of f, then the fundmentl theorem cn be used to find the definite integrl, nd we see tht: This explins the lst property bove. b = F () F (b) = (F (b) F ()) = f(x)dx the definite integrl trets re under the xxis s negtive Let s investigte the property: k = k This property shows tht if negtive function is integrted, then negtive number will be obtined. The mgnitude of this negtive number corresponds to the mgnitude of the re beneth the xxis. To see this, suppose tht g is positive on [, b]. Then, g is negtive on [, b], nd: ( g(x)) dx = g(x) dx The grphs of g nd g re symmetric bout the xxis; g(x) dx gives the re under the grph of g. Thus, the definite integrl trets re under the xxis s negtive.
6 copyright Dr. Crol JV Fisher Burns EXAMPLE x 2 dx = x 2 = (2 ) = 7 Cution!! = does not imply tht f(x) = nd x 2 dx = x 2 dx = 7 Here, since the function x 2 is negtive over the entire intervl [, 2], the result 7 is interpreted s follows: The mgnitude of the result, 7 = 7, indictes tht there is 7 units of re trpped between the grph of x 2 nd the xxis. The fct tht the nswer 7 is negtive indictes tht this re lies beneth the xxis. If one integrtes over n intervl [, b] on which the re trpped between the grph of f nd lying bove the xxis is the sme s tht re below the xxis, then the dditivity of the integrl shows tht the definite integrl will hve vlue. For exmple, it ws seen erlier tht x dx =. This is becuse, by dditivity: x dx = x dx + = ( A) + (A) =, x dx where A represents the mgnitude of the re trpped between the grph of x nd the xxis on [, ]. So, just becuse = does not necessrily men tht f(x) = on [, b]. Insted, it mens tht the re trpped between the grph of f nd lying bove the xxis, is the sme s the re trpped between the grph of f nd lying below the xxis, on the intervl [, b].
7 copyright Dr. Crol JV Fisher Burns EXERCISE The grph of function f is shown below, nd certin res re lbeled. Bsed on this informtion, evlute the following integrls, if possible. If this is not possible bsed on the given informtion, so stte f(t) dt. f(s) ds f(t) dt 2 f(y) dy This section is concluded with some exmples tht illustrte how the properties of the definite integrl cn be used to help in its evlution. EXAMPLE Problem: Evlute (x2 2x + ) dx. Solution: By linerity: (x 2 2x + ) dx = = x x 2 dx 2 2 x2 2 x dx + + x = ( ) 2 ( ) + ( ) 2 () dx = + = 2 The solution is usully written down in much more bbrevited form: (x 2 2x + ) dx = ( x 2 x2 2 + x) = ( + ) () = 2
8 copyright Dr. Crol JV Fisher Burns 5 EXAMPLE find the indefinite integrl first; then use ny ntiderivtive to find the definite integrl When the integrnd in definite integrl problem is complicted, some people prefer to first solve the compnion indefinite integrl, nd then use ny ntiderivtive to find the definite integrl. This prevents hving to crry round the limits of integrtion. Problem: Find the re under the grph of x+ on [, 2]. Solution: It is not necessry to grph x+ ; it is only necessry to recognize tht whenever x [, 2], x+ >. Thus, the grph lies entirely bove the xxis on this intervl, nd the desired re is given by the definite integrl: x + dx In future section, we will discuss how to use the technique of substitution directly with definite integrls. For now, find n ntiderivtive by first solving the compnion indefinite integrl problem: x + dx = x + dx = u du = ln u + C = ln x + + C Use the simplest ntiderivtive to evlute the desired definite integrl: x + dx = ln x + 2 = (ln 7 ln ) = ln 7.65 One more time! Since vriety is the spice of life, the previous problem is solved in different wy: 2 x + dx = (x + ) dx = x + dx = ln x + 2 = (ln 7 ln ) = 7/ (ln / ) = (ln 7).65 EXERCISE 5 Evlute dx in two wys. 5x +
9 6 copyright Dr. Crol JV Fisher Burns EXAMPLE Problem: Determine the re of the region bounded by the grph of nd the xxis on the intervl [, ]. y = (x 2) 2 + Solution: A quick sketch is esy to get nd helpful. View y s being built up s follows: The xxis intercepts occur when y is zero: y = (x 2) 2 + = (x 2) 2 = x 2 = x = or x = Here we used the fcts tht: For ll rel numbers x, x 2 = x ; nd x 2 tells us how fr x is from 2. Thus, x 2 = is true exctly when x is number whose distnce from 2 equls. Now, to find the desired re, we MUST integrte in two pieces: ( (x 2) 2 + ) (x 2) dx = + x = ( ( 2) + ) ( ( 2) + ) = 8 = The nswer is negtive, becuse the re is beneth the xxis. Also: ( (x 2) 2 + ) dx =... = The desired re is therefore: + = 8 EXERCISE 6 Wht would hve hppened if, in the previous problem, you hd tried to compute the desired re by finding ( (x 2) 2 + ) dx? Evlute this integrl to confirm your nswer. EXERCISE 7 Determine the re of the region bounded by the grph of y = (x + 2) 2 on the intervl [, ]. Mke sketch showing the re tht you re finding.
10 copyright Dr. Crol JV Fisher Burns 7 QUICK QUIZ smple questions. In few words, explin why there is such similr ppernce between the indefinite integrl nd the definite integrl. 2. Give precise sttement of the Fundmentl Theorem of Integrl Clculus.. Wht does the nottion F (x) b men, when used in the context of evluting definite integrls?. Compute: x2 dx 5. Show tht: x dx = Interpret your nswer in terms of re. KEYWORDS for this section Nottion for the definite integrl, upper nd lower limits of integrtion, compring the definite nd indefinite integrls, the Fundmentl Theorem of Integrl Clculus, the nottion F (x) b, dummy vrible, properties of the definite integrl, linerity, dditivity, integrting bckwrds introduces minus sign, the definite integrl trets re under the xxis s negtive. ENDOFSECTION EXERCISES Evlute the following integrls. Use ny pproprite methods. Be sure to write complete mthemticl sentences ln ln 2 2 x dx t / dt (2x ) dx (x + b) dx x 2 + x dx e 2t dt Find the re bounded by the grph of the given function nd the xxis on the stted intervl. Mke sketch showing the re tht you re finding. You my hve to evlute more thn one integrl to obtin your finl nswer. 7. f(x) = + e x ; [, 2] 8. f(x) = (x )(x + ); [ 2, 2] 9. f(x) = 2x 2 + 5x ; [, 2]
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