An Overview of Integrtion S. F. Ellermeyer July 26, 2 The Definite Integrl of Function f Over n Intervl, Suppose tht f is continuous function defined on n intervl,. The definite integrl of f from to is denoted y fxdx. Grphiclly, the definite integrl of f from to is the signed re of the region ounded y the grph of y fx, the x xis, nd the verticl lines x nd x. Prts of the region which lie ove the x xis give positive contriution to the integrl nd prts of the region which lie elow the x xis give negtive contriution. Figure
Figure 2 In Figure 2, prt of the grph of f on the intervl, lies ove the x xis nd prt lies elow the x xis. If A nd B re the res of the regions pictured, then fxdx A B. Let us look t severl specific exmples of integrls tht we cn compute y using elementry geometry. Exmple Suppose tht f is the constnt function fx nd suppose tht we wnt to compute the definite integrl of f from x to x 6. The grph of f is shown in Figure. 5 y 2-2 5 6 7 x Figure The integrl we wnt to compute is the re of the region shown shded in Figure. 2
Figure The shded region is rectngle with re 2 so 6 fxdx 2. Another wy to write this is 6 dx 2. Exercise. Let f e the constnt function fx 8. By drwing picture nd using elementry geometry, compute the definite integrl 7 fxdx. 2 2. Let f e the constnt function fx. By drwing picture nd using elementry geometry, compute the definite integrl 8 fxdx. (Note: Rememer tht prts of the grph of y fx which lie elow the x xis give negtive contriution to the integrl.). Find generl formul for computing fxdx where f is the constnt function fx K. From Exmple nd Exercise, we see tht it is very esy to compute the integrl of constnt function. Another kind of function for which it is esy to compute definite
integrl y using geometry is liner function tht is, function of the form fx mx. Exmple 2 Suppose tht we wnt to compute the definite integrl of the function fx x 6 from x to x. The grph of f is shown in Figure 5. y 5 x - 2 5-5 - -5 Figure 5 The vlue of the integrl we wnt to compute is fxdx A B where A nd B re the res of the regions shown in Figure 6. Figure 6 Since oth of these regions re tringles, we cn use the formul for the re of tringle Are 2 seheight to find these res. Since the grph of f intersects the x xis t x /2, we hve A 2 2 6 9 2 nd
Thus, We could lso write this s B 2 5 2 25 2. fxdx 9 2 25 2 8. x 6dx 8. Exercise 2. Let f e the function fx 2x 5. Drw nice picture (using grph pper) of the grph of f over the intervl,. Then use your picture (nd geometry) to compute fxdx. 2 2. Use the picture you drew to compute fxdx.. Use the picture you drew to compute fxdx.. Suppose tht m is constnt nd suppose tht f is the liner function fx mx. Find generl formul (involving,, nd m) for computing fxdx. 5. Use the formul you found in question ove to compute 5 7xdx. Verify tht your nswer is correct y drwing picture of fx 7x nd y then computing the ove integrl y using geometry. Integrls which cn e computed y using geometry (s in the previous two exmples) re rre. In fct, liner functions nd constnt functions (which, in fct, re relly specil kind of liner function) re lmost the only exmples which we cn give. In Exmple which follows, we give one more specil exmple of n integrl which cn e computed y using geometry. Exmple Suppose we wnt to compute the integrl x 2 dx. Since the integrnd (the function whose integrl we wnt to compute) is 5
fx x 2,the grph of f is the curve y x 2. If we squre oth sides of this eqution, we otin y 2 x 2 or x 2 y 2. Thus the grph of f is the upper hlf of the unit circle (shown in Figure 7). y.5 - -.5.5 x -.5 - Figure 7 We recll tht the re of circle of rdius r is r 2. Thus the re of the unit circle is 2. The vlue of the integrl we wnt to compute is thus fxdx. 2 Of course, we cn lso write this s x 2 dx 2. Exercise. Use geometry to compute x 2 dx. (Hint: In this nd the rest of the exercises here, it will e helpful for you to drw pictures.) 2. Use geometry to compute x 2 dx.. Use geometry to compute 2 x 2 dx.. Use geometry to compute 2 x 2 6 dx. 6
5. Use geometry to compute 2 x 2 8 dx. 6. Explin why it is not so esy to determine the exct vlue of /2 x 2 dx. Why Do We Wnt to Compute Integrls? Integrls re extremely importnt ecuse they llow us to mesure ccumultion. You re fmilir with the concept of ccumultion from your everydy experience. An exmple which everyody is fmilir with is the ccumultion of rinfll. When you her on the p.m. news tht the totl rinfll in Atlnt over the pst three hours ws.5 inches, you know tht mens tht totl of.5 inches of rin hs fllen in Atlnt over the pst three hours. So wht does this hve to do with integrtion? If you think out it, there re mny different wys in which.5 inches of rin could hve fllen in Atlnt over the pst three hours. One possiility is tht there my hve een very stedy rin flling t the rte of.5 inches per hour. Another possiility is tht it might hve rined very hevily t the rte of inch per hour from 8 p.m. until 9: p.m. nd then not rined t ll from 9: p.m. until : p.m. Yet nother possiility is tht it rined somewht stedily during the three hour period with the rin eing hevy t times, light t other times, nd perhps not rining t ll for some periods of time. These three possiilities re illustrted in Figures 8, 9, nd. In ech figure, the function R is the rte of rinfll t time t where t is mesured in hours with t stnding for 8 p.m. 2.5 R(t).5-2 t Figure 8: Stedy Rinfll of.5 in/hr. 7
2.5 R(t).5-2 t Figure 9: Hrd stedy rin for.5 hrs. 2.5 R(t).5-2 t Figure : Vrile Rinfll Rte In ech of the three different rinfll scenrios (Figures 8, 9, nd ), the totl rinfll for the three hour period from 8 to p.m. is.5 inches. The reson is tht the definite integrls of ech of the different rinfll rte functions (R) over the time, is the sme. Even though the functions in Figures 8, 9, nd re oviously different functions, ech of them stisfies Rtdt.5. This simple exmple leds the wy to sic principle of Clculus: Bsic Principle: If f is function which tells us the rte t which something is hppening, then fxdx tells us the totl ccumulted effect of wht hppened s the independent vrile, x, vried from x to x. The sic principle stted ove is, dmittedly, somewht vgue. To help you to understnd it etter, we provide some more exmples. Exmple 8
. If Rt is the the rte t which rin is flling t time t where t is mesured in hours nd t stnds for noon, then 2 8 Rtdt is the totl mount of rin tht fell etween 2 nd 8 p.m. 2. If vt is the speed t which cross-country runner is running t time t where t is mesured in seconds, t t stnds for the time t which the runner strts running, nd t t stnds for the time t which the runner finishes running, then t vtdt t is the totl distnce the runner hs run.. If sphericl lloon is eing inflted nd if Vr is the volume of the lloon when the rdius of the lloon is r, then r 2 dv dr r dr is the totl increse in volume of the lloon s result of inflting the lloon from rdius r to rdius r 2.. When eker of wter is heted, the wter molecules in the eker egin to move fster. This mens tht the kinetic energy of the wter increses. If ft tells us the rte t which the kinetic energy of the wter is incresing when the temperture of the wter is T where T is mesured in degrees Celsius, then 9 ftdt tells us the totl increse in kinetic energy of the wter s the wter is heted from Cto9 C. Exercise. For the rinfll rte function shown in Figure 8, explin how you cn tell (y looking t the grph) tht Rtdt.5. 2. For the rinfll rte function shown in Figure 9, explin how you cn tell (y looking t the grph) tht Rtdt.5.. For the rinfll rte function shown in Figure, why is it more difficult to tell (just y looking t the grph) tht Rtdt.5?. Referring to the rinfll rte function shown in Figure,. At wht rte ws the rin flling t 8 p.m.?. At out wht time ws it rining the hrdest? 9
c. At out wht time did it momentrily quit rining? d. Did more rin fll etween 8 nd 9 p.m. or etween 9 nd p.m.? How cn you tell? 5. Compring the rinfll rte function in Figures 8 nd :. Which rinfll rte function indictes greter totl mount of rinfll etween 8nd9p.m.?. Which rinfll rte function indictes greter mount of totl rinfll etween nd p.m.? 6. Suppose tht ft is the rte t which the temperture in Atlnt is chnging t time t where t is mesured in hours nd t stnds for 8.m. Suppose lso tht the temperture in Atlnt t 8.m. is 75 F nd the temperture t p.m. on the sme dy is 68 F. Which of the following must e true?. The temperture etween 8.m. nd p.m. ws continuously decresing.. There must hve een t lest some time period during which ft. 7. For the function f given in the prolem ove, wht is the vlue of ftdt? 8. Referring to the sme scenrio s in prolem 6 (ove), suppose tht Tt is the temperture in Atlnt t time t. Complete the following:. T. T c. T T 9. Try to generlize wht you lerned y doing prolems 6, 7, nd 8 (ove): If Tt is the temperture t certin loction t time t nd if ft is the rte t which the temperture t tht loction is chnging t time t, nd if t nd t re two prticulr times (with t t ), then t ftdt. t. Now try to generlize the rinfll sitution: If At is the totl mount of rinfll which hs ccumulted t time t nd if Rt is the rte t which rin is flling t time t, nd if t nd t re two prticulr times (with t t ), then t Rtdt. t. Now see if you cn do this one: If KT is the kinetic energy of eker of wter when the temperture of the wter is T, nd if ft is the rte t which the kinetic energy of the wter is incresing when the temperture of the wter is T, then 9 ftdt. 2. Now generlize completely: Suppose tht F is function with independent vrile x nd suppose tht f is function which tells us the rte t which F is chnging with respect to the independent vrile x. Suppose tht nd re two numers in the domin of F (with ). Then
fxdx.. Suppose tht sphericl lloon is eing inflted nd suppose tht Vr is the volume of the lloon when the rdius of the lloon is r nd rt is the rdius of the lloon t time t. Then the volume of the lloon t time t is Vrt. Fill in the proper integrnd elow: t 2 Vrt 2 Vrt dt. t (Hint: You need to rememer the Chin Rule of differentition.) The Fundmentl Theorem of Clculus The informlly stted Bsic Principle of the previous section is ctully very centrl theme in Clculus. In fct, when stted more formlly, it is clled the Fundmentl Theorem of Clculus. The reson for this nme is tht it points out the reltionship etween the two most sic ides of Clculus - differentition nd integrtion. You will recll from your Differentil Clculus course tht the derivtive of function, F, t given point x, tells us the rte which the dependent vrile is chnging when the independent vrile is x. More specificlly, if the dependent vrile is clled y, then F x tells us the rte t which y is chnging when Fx y. In the previous section, you lerned (hopefully) from doing the exercises tht if f is function tht tells us the rte of chnge of nother function, F, nd if nd re two prticulr vlue of the independent vrile (with ), then fxdx F F. In prticulr, since F is indeed the function which tells us the rte of chnge of F, then F xdx F F. We stte the Fundmentl Theorem formlly s follows: Theorem (The Fundmentl Theorem of Clculus): If F is differentile function (with independent vrile x), nd if F x fx for ll x in the intervl,, then fxdx F F. Exmple 5 In Exmple, we sw y using simple geometry tht if f is the function fx, then 6 fxdx 2. To compute this sme integrl using the Fundmentl Theorem Clculus (FTC), we need to find n ntiderivtive of f. Tht is, we need to find function F such tht F x fx. Of course, from our differentil clculus experience, we know tht such function is Fx x. Using the FTC, we otin
6 fxdx F6 F 6 2 Exmple 6 In Exmple 2, we sw tht x 6dx 8. To do this sme prolem using the FTC, we first need to find n ntiderivtive of fx x 6. From our knowledge of differentil clculus, we know tht the function Fx 2x 2 6x is n ntiderivtive of f. Thus x 6dx F F 8 8 Exmple 7 Wht out the prolem x 2 dx which we did in Exmple y using geometry? It is not s esy to find n ntiderivtive of the function fx x 2 s it ws for the functions in the previous two exmples! Lter in this course, we will lern procedure tht will llow us to ntidifferentite the function fx x 2. For the moment, let us just give the nswer: An ntiderivtive of f is Fx 2 x x2 rcsin x. 2 Using the ntiderivtive which hs een given, we compute nd This gives us F 2 2 2 rcsin F 2 2 2 rcsin. x 2 dx F F 2 2
Plese don t worry if you don t understnd how the ove ntiderivtive ws otined. You re not supposed to t this point. Indeed, one of the min topics we will consider in this course is how to find ntiderivtives - process which itself is lso clled integrtion. Exercise 5. Use the Fundmentl Theorem Clculus to evlute the following definite integrls.. 9 5dx. 9 dx c. Kdx (where K is constnt) d. 2 2x 5dx e. 2x 5dx f. 2x 5dx g. mxdx (where m is constnt) h. mx dx (where m nd re constnts) 2. Use the ntiderivtive given in Exmple 7 ove to evlute (pproximtely) /2 x 2 dx. (Note: You should only use clcultor to compute the pproximte vlue of rcsin/2.). Use the ntiderivtive given in Exmple 7 ove to evlute (pproximtely) /2 x 2 dx.. Add the two nswers you otined in prolems 2 nd ove. Verify tht the nswer you get is the sme s the nswer you get when you compute x 2 dx. Does it mke sense to you tht these nswers should e the sme? Explin. 5. Suppose tht we hve differentile function y Fx nd suppose tht x nd x 2 re two specific points in the domin of F such tht Fx y nd Fx 2 y 2. Then x 2 x dy dx dx. 6. Explin how you cn tell, just y looking t the grph of y cosx, tht /2 cosxdx 2. (Hint: Compre the given integrl to the re of certin rectngle.)
7. Use the Fundmentl Theorem of Clculus to show tht /2 cosxdx (without using your clcultor). 8. Explin how you cn tell, just y looking t the grph of y cosx, tht cosxdx. 9. Verify tht cosxdx y using the FTC (without using your clcultor).. Bsed on your work in the pst few prolems, explin you know tht there is some numer, x, etween /2 nd, such tht x cosxdx.686. (Hint: You don t need to do ny clcultions. Just reflect on your results from the previous few prolems.). Are there re two different numers, x nd x 2, ech etween /2 nd such tht x cosxdx.686 nd x 2 cosxdx.686? Explin your nswer. 2. Drw the grph of function, f, such tht fxdx 6 fxdx. Indefinite Integrls The term indefinite integrl is often used to denote the set of ll ntiderivtives of function, f, on some intervl,. The reson for the use of this term is tht integrtion is closely relted to ntidifferentition (y the FTC). We use the nottion fxdx, to refer to the indefinite integrl of the function f on the intervl, mening the set of ll functions tht re ntiderivtives of f on the intervl,. The Men Vlue Theorem cn e used to prove tht if function, F, is n ntiderivtive of the function f on certin intervl,, then ny ntiderivtive, G, off on, must e of the form Gx Fx C where C cn e ny constnt. Thus, for exmple, since we know tht the function Fx 2 x2 is n ntiderivtive of the function fx x on the intervl,, then we know tht ny ntiderivtive, G, off must e of the form Gx 2 x2 C (where C cn e ny constnt). We descrie the indefinite integrl of f s xdx, 2 x2 C. In most cses, it is not necessry to include the intervl, in the nottion for the
indefinite integrl ecuse the ntiderivtive does not depend on the prticulr intervl in question. For exmple, it is true tht xdx, 2 x2 C no mtter wht intervl, we re considering. Thus, we cn simply write xdx 2 x2 C. However, there re some exceptions one in prticulr tht occurs frequently enough tht we should mke note of it: The function Fx lnx with domin, hs derivtive F x x nd the function Fx lnx with domin, hs the sme derivtive: F x x. Hence, the sttement, x dx lnx C is correct only if the intervl, is suset of the intervl,; wheres, if the intervl, is suset of the intervl,, then the sttement x dx lnx C, is correct. A commonly used wy of getting round this difficulty is to oserve tht the function Fx ln x with domin,, hs derivtive F x x. Thus, the sttement, x dx ln x C is correct oth when the intervl, is suset of, nd when, is suset of,. Hence, we re lwys sfe in writing x dx ln x C. Exmple 8 Suppose we wish to evlute the integrl 2 x dx. There re two wys to go out this: The first is to oserve tht since,2 is suset of,, then x dx lnx C.,2 Therefore, y the FTC, 2 x dx ln2 ln ln2 ln. 5
The other wy to pproch this prolem is to simply use the indefinite integrl x dx ln x C. Using this formul, we otin 2 x dx ln 2 ln ln2 ln. Some Bsic Integrls The following indefinite integrls (which come from sic differentition formuls from differentil clculus) should e memorized. You will need to know them s we study integrtion techniques tht will llow us to compute less sic integrls such s x 2 dx.. If K is ny constnt, then Kdx Kx C. 2. If n is ny constnt except n, then x n dx n xn C.. x dx ln x C.. If is ny positive constnt except, then x dx ln x C. 5. e x dx e x C. 6. cosxdx sinx C. 7. sinxdx cosx C. 8. sec 2 xdx tnx C. 9. csc 2 xdx cotx C.. secxtnxdx secx C.. cscxcotxdx cscx C. 2. x 2 dx rcsinx C.. x 2 dx rctnx C. 6