Mthemtics Lerning Centre Introduction to Integrtion Prt : The Definite Integrl Mr Brnes c 999 Universit of Sdne
Contents Introduction. Objectives...... Finding Ares 3 Ares Under Curves 4 3. Wht is the point of ll this?... 6 3. Note bout summtion nottion... 6 4 The Definition of the Definite Integrl 7 4. Notes..... 7 5 The Fundmentl Theorem of the Clculus 8 6 Properties of the Definite Integrl 7 Some Common Misunderstndings 4 7. Arbitrr constnts..... 4 7. Dumm vribles...... 4 8 Another Look t Ares 5 9 The Are Between Two Curves 9 Other Applictions of the Definite Integrl Solutions to Eercises 3
Mthemtics Lerning Centre, Universit of Sdne Introduction This unit dels with the definite integrl. Iteplins how it is defined, how it is clculted nd some of the ws in which it is used. We shll ssume tht ou re lred fmilir with the process of finding indefinite integrls or primitive functions (sometimes clled nti-differentition) nd re ble to ntidifferentite rnge of elementr functions. If ou re not, ou should work through Introduction to Integrtion Prt I: Anti-Differentition, nd mke sure ou hve mstered the ides in it before ou begin work on this unit.. Objectives B the time ou hve worked through this unit ou should: Be fmilir with the definition of the definite integrl s the limit of sum; Understnd the rule for clculting definite integrls; Know the sttement of the Fundmentl Theorem of the Clculus nd understnd wht it mens; Be ble to use definite integrls to find res such s the re between curve nd the -is nd the re between two curves; Understnd tht definite integrls cn lso be used in other situtions where the quntit required cn be epressed s the limit of sum.
Mthemtics Lerning Centre, Universit of Sdne Finding Ares Ares of plne (i.e. flt!) figures re firl es to clculte if the re bounded b stright lines. The re of rectngle is clerl the length times the bredth. The re of right-ngled tringle cn be seen to be hlf the re of rectngle (see the digrm) nd so is hlf the bse times the height. The res of other tringles cn be found b epressing them s the sum or the difference of the res of right ngled tringles, nd from this it is cler tht for n tringle this re is hlf the bse times the height. Are of rectngle length bredth A B D C Are of tringle re of rectngle length bredth A B C D Are of ABC Are of ABC re of ABD re of ABD + re of ACD re of ACD Using this, we cn find the re of n figure bounded b stright lines, b dividing it up into tringles (s shown). Ares bounded b curved lines re much more difficult problem, however. In fct, lthough we ll feel we know intuitivel wht we men b the re of curviliner figure, it is ctull quite difficult to define precisel. The re of figure is quntified b sking how mn units of re would be needed to cover it? We need to hve some unit of re in mind (e.g. one squre centimetre or one squre millimetre) nd imgine tring to cover the figure with little squre tiles. We cn lso imgine cutting these tiles in hlves, qurters etc. In this w rectngle, nd hence n figure bounded b stright lines, cn be delt with, but curviliner figure cn never be covered ectl. We re therefore forced to rel on the notion of limit in order to define res of curviliner figures. To do this, we mke some simple ssumptions which most people will ccept s intuitivel obvious. These re:-. If one figure is subset of second figure, then the re of the first will be less thn or equl to tht of the second.
Mthemtics Lerning Centre, Universit of Sdne 3. If figure is divided up into non-overlpping pieces, the re of the whole will be the sum of the res of the pieces. Using these ssumptions, we cn pproimte to curved figures b mens of polgons (figures with stright line boundries), nd hence define the re of the curved figure s the limit of the res of the polgons s the pproch the curved figure (in some sense et to be mde precise).
Mthemtics Lerning Centre, Universit of Sdne 4 3 Ares Under Curves Let us suppose tht we re given positive function f() nd we wnt to find the re enclosed between the curve f(), the -is nd the lines nd b. (The shded re in the digrm.) If the grph of f() isnot stright line we do not, t the moment, know how to clculte the re precisel. f() Are b We cn, however, pproimte to the re s follows: First we divide the re up into strips s shown, b dividing the intervl from to b into equl subintervls, nd drwing verticl lines t these points. f() subdivisions b Net we choose the lest vlue of f() inech subintervl nd construct rectngle with tht s its height (s in the digrm). The sum of the res of these rectngles is clerl less thn the re we re tring to find. This sum is clled lower sum. lower sum b Then we choose the gretest vlue of f() inech subintervl nd construct rectngle with tht s its height (s in the digrm opposite). The sum of the res of these rectngles is clerl greter thn the re we re tring to find. This sum is clled n upper sum. upper sum b Thus we hve sndwiched the re we wnt to find in between n upper sum nd lower sum. Both the upper sum nd the lower sum re esil clculted becuse the re sums of res of rectngles.
Mthemtics Lerning Centre, Universit of Sdne 5 Although we still cn t s precisel wht the re under the curve is, we know between wht limits it lies. If we now increse the number of strips the re is divided into, we will get new upper nd lower sums, which will be closer to one nother in size nd so closer to the re which we re tring to find. In fct, the lrger the number of strips we tke, the smller will be the difference between the upper nd lower sums, nd so the better pproimtion either sum will be to the re under the curve. division into nrrower strips It cn be shown tht if f() is nice function (for emple, continuous function) the difference between the upper nd lower sums pproches zero s the number of strips the re is subdivided into pproches infinit. We cn thus define the re under the curve to be: the limit of either the upper sum or the lower sum, s the number of subdivisions tends to infinit (nd the width of ech subdivision tends to zero). Thus finding the re under curve boils down to finding the limit of sum. Now let us introduce some nottion so tht we cn tlk more precisel bout these concepts. Let us suppose tht the intervl [, b] is divided into n equl subintervls ech of width. Suppose lso tht the gretest vlue of f() in the ith subintervl is f( i ) nd the lest vlue is f( i). Then the upper sum cn be written s: f() f(*) f(') f( ) + f( ) +... + f( n) or, using summtion nottion: n i f( i ). Similrl, the lower sum cn be written s: b f( ) + f( ) +... + f( n) or, using summtion nottion: n i f( i). With this nottion, nd letting A stnd for the re under the curve f() from to b, we cn epress our erlier conclusions in smbolic form. The re lies between the lower sum nd the upper sum nd cn be written s follows: n n f( i) A f( i ). i i
Mthemtics Lerning Centre, Universit of Sdne 6 The re is equl to the limit of the lower sum or the upper sum s the number of subdivisions tends to infinit nd cn be written s follows: or n A lim f( n i) i n A n lim f( i ). i 3. Wht is the point of ll this? Well, firstl it enbles us to define precisel wht up till now hs onl been n imprecise intuitive concept, nmel, the re of region with curved lines forming prt of its boundr. Secondl it indictes how we m clculte pproimtions to such n re. B tking firl lrge vlue of n nd finding upper or lower sums we get n pproimte vlue for the re. The difference between the upper nd lower sums tells us how ccurte this pproimtion is. This, unfortuntel, is not ver good or ver prcticl w of pproimting to the re under curve. If ou do course in Numericl Methods ou will lern much better ws, such s the Trpezoidl Rule nd Simpson s Rule. Thirdl it enbles us to clculte res precisel, provided we know how to find finite sums nd evlute limits. This however cn be difficult nd tedious, so we need to look for better ws of finding res. This will be done in Section 5. At this stge, mn books sk students to do eercises clculting upper nd lower sums nd using these to estimte res. Frequentl students re lso sked to find the limits of these sums s the number of subdivisions pproches infinit, nd so find ect res. We shll not sk ou to do this, s it involves gret del of computtion. 3. Note bout summtion nottion The smbol (pronounced sigm ) is the cpitl letter S in the Greek lphbet, nd stnds for sum. The epression 4 i f(i) isred the sum of f(i) from i toi 4, or sigm from i to4off(i). In other words, we substitute,, 3 nd 4 in turn for i nd dd the resulting epressions. Thus, 4i i stnds for + + 3 + 4, 5i i stnds for + +3 +4 +5, nd i f( ) stnds for f( ) + f( ).
Mthemtics Lerning Centre, Universit of Sdne 7 4 The Definition of the Definite Integrl The discussion in the previous section led to n epression of the form n A n lim f( i ) () i where the intervl [, b] hs been divided up into n equl subintervls ech of width nd where i is point in the ith subintervl. This is ver clums epression, nd mthemticins hve developed simpler nottion for such epressions. We denote them b f()d which is red s the integrl from to b of f()d. The sign is n elongted s nd stnds for sum, just s the did previousl. The difference is tht in this cse it mens the limit of sum rther thn finite sum. The d comes from the s we pss to the limit, just s hppened in the definition of d. d Thus the definite integrl is defined s the limit of prticulr tpe of sum i.e. sums like tht given in () bove, s the width of ech subintervl pproches zero nd the number of subintervls pproches infinit. 4. Notes. Although we used the re under curve s the motivtion for mking this definition, the definite integrl is not defined to be the re under curve but simpl the limit of the sum ().. Initill, when discussing res under curves, we introduced the restriction tht f() hd to be positive function. This restriction is not necessr for the definition of definite integrl. 3. The definition cn be mde more generl, b removing the requirement tht ll the subintervls hve to be of equl widths, but we shll not bother with such generlistions here. 4. Sums such s () re clled Riemnn sums fter the mthemticin Georg Riemnn who first gve rigorous definition of the definite integrl. 5. The definition of definite integrl requires tht f() should be defined everwhere in the intervl [, b] nd tht the limit of the Riemnn sums should eist. This will lws be the cse if f is continuous function.
Mthemtics Lerning Centre, Universit of Sdne 8 5 The Fundmentl Theorem of the Clculus So fr, we hve defined definite integrls but hve not given n prcticl w of clculting them. Nor hve we shown n connection between definite integrls nd differentition. Let us consider the specil cse where f(t) iscontinuous positive function, nd let us consider the re under the curve f(t) from some fied point t c up to the vrible point t. For different vlues of we will get different res. This mens tht the re is function of. Let us denote the re b A(). Clerl, A() increses s increses. Let us tr to find the rte t which it increses, tht is, the derivtive of A() with respect to. c A() f(t) t At this point, recll how we find derivtives from first principles: Given function f(), we let chnge b n mount, sotht f()chnges to f(+ ). The derivtive of f() isthe limit of P Q f( + ) f() s. We shll go through this process with A() in plce of f(). c S R + t When we increse b, A() increses b the re of the figure PQRS. Tht is, (see the digrm) A( + ) A() re PQRS. Now tht the re PQRS is bounded b curved line t the top, but it cn be seen to lie in between the res of two rectngles: T P Q U re PURS < re PQRS < re TQRS. Both of these rectngles hve width. Let the height of the lrger rectngle be f( ) nd the height of the smller rectngle f( ). (In other words, nd re the vlues of t which f() ttins its mimum nd minimum vlues in the intervl from to +.) f(') f(*) Thus re PURS f( ) nd re TQRS f( ). So, f( ) A( + ) A() f( ). S R
Mthemtics Lerning Centre, Universit of Sdne 9 Now if we divide these inequlities ll through b, we obtin f( ) A( + ) A() f( ). Finll, if we let, both f( ) nd f( ) pproch f(), nd so the epression in the middle must lso pproch f(), tht is, the derivtive of A(), da f(). d This result provides the link we need between differentition nd the definite integrl. If we recll tht the re under the curve f(t) from t to t is equl to f(t)dt, the result we hve just proved cn be stted s follows: d f(t)dt f(). () d In words This is the Fundmentl Theorem of the Clculus. If we differentite definite integrl with respect to the upper limit of integrtion, the result is the function we strted with. Youm not ctull use this result ver often, but it is importnt becuse we cn derive from it the rule for clculting definite integrls: Let us suppose tht F () isnnti-derivtive of f(). Tht is, it is function whose derivtive is f(). If we nti-differentite both sides of the eqution () we obtin f(t)dt F ()+c. Now we cn find the vlue of c b substituting in this epression. Since Thus f(t)dt is clerl equl to zero, we obtin F ()+c, nd so c F (). f(t)dt F () F (), or, letting b, f(t)dt F (b) F (). This tells us how to evlute definite integrl first, find n nti-derivtive of the function then, substitute the upper nd lower limits of integrtion into the result nd subtrct. Note A convenient short-hnd nottion for F (b) F () is[f ()] b.
Mthemtics Lerning Centre, Universit of Sdne To see how this works in prctice, let us look t few emples: i Find d. An nti-derivtive of is 3 3,sowewrite [ ] d 3 3 3 ()3 3 ()3 3. ii Find π sin tdt. π sin tdt [ cos t] π cos(π)+ cos ( )+. iii Find the re enclosed between the -is, the curve 3 +5 nd the ordintes nd. In question like this it is lws good ide to drw rough sketch of the grph of the function nd the re ou re sked to find. (See below) If the required re is A squre units, then ( A 3 +5 ) d [ ] 4 4 +5 ( ) (4 4+) 4 +5 5 3 4. 3 +5 Eercises 5.. [ 3] 4 [ ] 3 b. c. [ ] 6 9 d. [ln ] 4.. b. 9 4 π π d cos tdt
Mthemtics Lerning Centre, Universit of Sdne c. d. d (s +s +)ds 3. Find the re of the shded region in ech of the digrms below:. b. + / 3 c. d. v sint v (4 u) 4 u π t 4. Evlute. e d b. c. d. π 5 3 d sin d t 4+t dt
Mthemtics Lerning Centre, Universit of Sdne 6 Properties of the Definite Integrl Some simple properties of definite integrls cn be derived from the bsic definition, or from the Fundmentl Theorem of the Clculus. We shll not give forml proofs of these here but ou might like to think bout them, nd tr to eplin, to ourself or someone else, wh the re true.. f()d. If the upper nd lower limits of the integrl re the sme, the integrl is zero. This becomes obvious if we hve positive function nd cn interpret the integrl in terms of the re under curve. b. If b c, c f()d f()d + c b f()d. This ss tht the integrl of function over the union of two intervls is equl to the sum of the integrls over ech of the intervls. The digrm opposite helps to mke this cler if f() is positive function. b c c. d. cf()d c f()d for n constnt c. This tells us tht we cn move constnt pst the integrl sign, but bewre: wecn onl do this with constnts, never with vribles! (f()+g())d f()d + g()d. Tht is, the integrl of sum is equl to the sum of the integrls. e. If f() g() in[, b] then f()d g()d. Tht is, integrtion preserves inequlities between functions. The digrm opposite eplins this result if f() nd g() re positive functions. g() f() b
Mthemtics Lerning Centre, Universit of Sdne 3 f. cd c(b ). This tells us tht the integrl of constnt is equl to the product of the constnt nd the rnge of integrtion. It becomes obvious when we look t the digrm with c>, since the re represented b the integrl is just rectngle of height c nd width b. c c b b g. We cn combine (e) nd (f) to give the result tht, if M is n upper bound nd m n lower bound for f() inthe intervl [, b], so tht m f() M, then m(b ) f()d M(b ). This, too, becomes cler when f() is positive function nd we cn interpret the integrl s the re under the curve. M m f() b h. Finll we etend the definition of the definite integrl slightl, to remove the restriction tht the lower limit of the integrl must be smller number thn the upper limit. We do this b specifing tht f()d f()d. For emple, b f()d f()d.
Mthemtics Lerning Centre, Universit of Sdne 4 7 Some Common Misunderstndings 7. Arbitrr constnts When ou first lerned how to find indefinite integrls (nti-derivtives), ou probbl lso lerned tht it ws importnt to remember lws to dd n rbitrr constnt to the nswer. There is no rbitrr constnt in definite integrl. If we interpret definite integrl s n re, it is cler tht its vlue is fied number (the number of units of re in the region). There is no mbiguit, nd so no need to dd n rbitrr constnt - in fct, it is wrong to do so. When we ppl the Fundmentl Theorem of the Clculus to finding definite integrl, however, the possibilit of n rbitrr constnt ppers to rise. For emple, in clculting d, wehve to find n nti-derivtive for. The most nturl choice would be 3 3, but insted of tht we could choose 3 3 + c, where c is n constnt. Then, [ d 3 3 + c] ( ) ( ) 3 ()3 + c 3 ()3 + c 3. Note tht the constnts cncel one nother out, nd we get the sme nswer s we did before. Thus we might s well tke the simplest course, nd forget bout rbitrr constnts when we re clculting definite integrls. 7. Dumm vribles Wht is the difference between f()d nd Let s work them both out in specil cse. 4 4 f(t)dt? d [ln ]4 ln4 ln. t dt [ln t]4 ln4 ln. So both integrls give the sme nswer. It is cler tht the vlue of definite integrl depends on the function nd the limits of integrtion but not on the ctul vrible used. In the process of evluting the integrl, we substitute the upper nd lower limits for the vrible nd so the vrible doesn t pper in the nswer. For this reson we cll the vrible in definite integrl dumm vrible -wecn replce it with n other vrible without chnging thing. Thus, f()d f()d f(t)dt f(θ)dθ.
Mthemtics Lerning Centre, Universit of Sdne 5 8 Another Look t Ares We hve defined the definite integrl f()d s the limit of prticulr tpe of sum, without plcing n restrictions on whether the function f() ispositive or negtive. We know tht, if f() ispositive, f()d is equl to the re between the curve f(), the -is nd the ordintes nd b, (which we refer to s the re under the curve ). The nturl question to sk now is: wht does f()d equl if f() is negtive? Cn we represent it s n re in this cse too; perhps the re bove the curve? If we go bck to the definition of f()d s the limit of sum, we cn see clerl tht if f() islws negtive then ech of the terms f( i ) will lso be negtive (since is positive). f() f() n So the sum f( i ) will be sum of negtive terms nd so will be negtive too. And i when we let n pproch infinit nd pss to the limit, tht will be negtive lso. Thus, if f() isnegtive for between nd b, f()d will lso be negtive. Now res re, b definition, positive. Remember tht, in section, we eplined tht we cn mesure the re of region b counting the number of little squre tiles (ech of unit re) needed to cover it. Since we cn t cover region with negtive number of tiles (it doesn t mke sense to tlk of it) we cn t hve negtive re. On the other hnd, if we ignore the fct tht ech of the terms f() is negtive, nd consider its numericl vlue onl, we cn see tht it is numericll equl to the re of the rectngle shown. And, if we go through the usul process, dding up the res of ll the little rectngles nd tking the limit, we find tht f()d is numericll equl to the re between the curve nd the -is. So to find the re, we clculte tke its numericl (i.e. bsolute) vlue. f()d, which will turn out to be negtive, nd then
Mthemtics Lerning Centre, Universit of Sdne 6 To see this more clerl, let s look t n emple. Consider the curve, ( ). This is cubic curve, nd cuts the -is t, nd. A sketch of the curve is shown below. Let us find the shded re. First we clculte the definite integrl ( )d. ( )d ( 3 )d [ 4 4 ] ( 4 ( ) ) 4. Since ( ) is negtive when lies between nd, the definite integrl is lso negtive, s epected. We cn conclude tht the re required is squre units. 4 As check, let us find the re of the other loop of the curve, i.e. the re between the curve nd the -is from to. Since ( ) is positive for this rnge of vlues of, the re will be given b ( )d [ 4 4 ] ( ( ) 4 ) 4. This is the nswer we would epect, since glnce t the digrm shows tht the curve hs point smmetr bout the origin. If we were to rotte the whole grph through 8, the prt of the curve to the left of the origin would fit ectl on top of the prt to the right of the origin, nd the unshded loop would fit on top of the shded loop. So the res of the two loops re the sme. Now let us clculte ( )d. ( )d [ 4 4 ] ( 4 ( ) 4 ). This mkes it ver cler tht definite integrl does not lws represent the re under curve.
Mthemtics Lerning Centre, Universit of Sdne 7 We hve found tht. If f() ispositive between nd b, then the curve.. If f() isnegtive between nd b, then curve, since the vlue of f()d is negtive. f()d does represent the re under f()d represents the re bove the 3. If f() issometimes positive nd sometimes negtive between nd b, then f()d mesures the difference in re between the prt bove the -is nd the prt below the -is. (In the emple bove, the two res were equl, nd so the difference cme out to be zero.) Let s look t nother emple. Consider the function ( + )( )( ) 3 +. This is cubic function, nd the grph crosses the -is t, nd. A sketch of the grph is shown. A B The re mrked A is given b ( 3 +)d [ 4 4 3 3 +] ( 4 3 ) ( + 4 + 3 ) 4 3 + 4 3. So the re of A is squre units. 3 The re mrked B cn be found b evluting ( 3 + ) d. This works out s 5. (The detils of the clcultion re left to ou.) So the re of B is 5 squre units.
Mthemtics Lerning Centre, Universit of Sdne 8 If we clculte ( 3 +)d the nswer will be the difference between the re of A nd the re of B, tht is, squre units. (Check it out for ourself.) 4 If we wnt the totl re enclosed between the curve nd the -is we must dd the re of A nd the re of B. i.e. + 5 3 3 squre units. WARNING In working out re problems ou should lws sketch the curve first. If the function is sometimes positive nd sometimes negtive in the rnge ou re interested in, it m be necessr to divide the re into two or more prts, s shown below. c b The re between the curve nd the -is from to b is NOT equl to f()d. c Insted, it is f()d + f()d. c Before ou cn clculte this, ou must find the vlue of c, i.e. find the point where the curve f() crosses the -is. Eercises 8. Find the re enclosed b the grph of 3 ( 4) nd the -is.. i Find the vlue of π sin d. ii Find the re enclosed between the grph of sin nd the -is from to π. 3. Find the totl re enclosed between the grph of ( + )( ) nd the -is.
Mthemtics Lerning Centre, Universit of Sdne 9 9 The Are Between Two Curves Sometimes we wnt to find, not the re between curve nd the -is, but the re enclosed between two curves, s between f() nd g(). We cn pproch this problem in the sme w s before b dividing the re up into strips nd pproimting the re of ech strip b rectngle. The lower sum is found b clculting the re of the interior rectngles s shown in the digrm. f() g() The height of ech interior rectngle is equl to the difference between the lest vlue of f(), f( ), nd the gretest vlue of g(), g( ), in the rectngle. The re of the ith rectngle is (f( i) g( i )). n The lower sum (f( i) g( i )). i The upper sum cn be found in the sme w. The re enclosed between the curves is sndwiched between the lower sum nd the upper sum. f(') i g(*) i f() g() When we pss to the limit s, we get Are enclosed between the curves (f() g())d. Note tht the height is lws f() g(), even when one or both of the curves lie below the -is. For emple, if for some vlue of, f() nd g() 3, the distnce between the curves is f() g() ( 3) 5, or, if f() nd g() 3, the distnce between the curves is ( ) ( 3) (see the digrm). (,) (4, ) (, 3) (4, 3) So, to find the re enclosed between two curves, we must:. Find where the curves intersect.. Find which is the upper curve in the region we re interested in.
Mthemtics Lerning Centre, Universit of Sdne 3. Integrte the function (upper curve lower curve) between the pproprite limits. In other words, if two curves f() nd g() intersect t nd b, nd f() g() for b, then Eercises 9 Are enclosed between the curves (f() g())d. (Remember to drw digrm first, before beginning n problem.). Find the re enclosed between the prbol ( ) nd the line +.. Find the re enclosed between the two prbols 4 +nd. 3. Check tht the curves sin nd cos intersect t π nd 5π, nd find the 4 4 re enclosed b the curves between these two point. 4. i Sketch the grphs of the function 6 nd 3 7 +6. ii Find the points of intersection of the curves. iii Find the totl re enclosed between them.
Mthemtics Lerning Centre, Universit of Sdne Other Applictions of the Definite Integrl The problem with which we introduced the ide of the definite integrl ws tht of finding the re under curve. As result, most people tend to think of definite integrls lws in terms of re. But it is importnt to remember tht the definite integrl is ctull defined s the limit of sum: n lim f( i ) n i nd tht n other problem which cn be pproimted b similr sum will give rise to definite intregrl when we tke the limit. Emples. Volume of solid If we wnt to find the volume of solid, we cn imgine it being put through bred slicer, nd cut into slices of thickness. If A() is the cross sectionl re t distnce long the -is, the volume of the slice will be pproimtel A(), nd the totl volume of the solid will be pproimtel n A( i ). i When we pss to the limit s nd n, this becomes the definite integrl Are A() A()d.. Length of curve We cn pproimte to the length of curve b dividing it up into segments, s shown, nd pproimting the length of ech segment b replcing the curved line with stright line joining the end points. If the length of the ith stright line segment is l i, the totl length of the curve will be pproimtel n l i. i If we tke the limit of this sum s the length of ech segment pproches zero nd the number of segments pproches infinit, we gin get definite integrl. The detils re rther complicted nd re not given here. P i P i l i P n P
Mthemtics Lerning Centre, Universit of Sdne 3. Mss of bod of vring densit Suppose we hve br, rope or chin whose liner densit (mss per unit length) vries. Let the densit t distnce long the -is be d(). If we subdivide the object into smll sections of length, the totl mss cn be pproimted b the sum n d( i ). When we tke the limit s n,weobtin the definite integrl i d()d. 4. Work done b vrible force In mechnics, the work done b constnt force is defined to be the product of the mgnitude of the force nd the distnce moved in the direction of the force. If the force F ()isvring, we cn pproimte to the work b dividing up the distnce into smll subintervls. If these re smll enough, we cn regrd the force s effectivel constnt throughout ech intervl nd so the work done in moving through distnce is pproimtel F (). The totl work is thus pproimtel n i F ( i ) nd when we tke the limit s n,wefind tht the work done in moving the force from to b is F ()d. Mn other emples could be given, but these four should be sufficient to illustrte the wide vriet of pplictions of the definite integrl.
Mthemtics Lerning Centre, Universit of Sdne 3 Solutions to Eercises Eercises 5.. (4 3 ) ( 3 ) b. 9 8 9 c. 6 9 d. ln 4 ln ln 4 ln.. b. c. d. 9 4 π π [ ] 9 d 9 4 4 cos tdt [sin t] π π sin π ( sin π ) ( ) d [ ] ( ) [ (s +s +)ds 3 s3 + s +s] ( ) 3 + ( 8 ) 3 +4 4 3 3.. Are b. Are c. Are d. Are 3 4 π [ ( ) ( ) 8 ( +)d 3 3 + ] 3 + 3 + 3 3 d [ln]3 ln3 ln ln3 4 [ ] (4 u)du (4 u) 4 ( )du 3 (4 u) 3 5 3 sin tdt [ cos t] π cos π +cos 4 4.. b. c. d. π 5 e d e d [ e ] (e ) 3 d ( )d [ln(3 )] 3 (ln 4 ln 5) ln 5 ln 4ln5 4 sin d π sin d [ cos ] π ( cos π + cos ) t 4+t dt 5 t 4+t dt [ ln(4 + t ) ] 5 9 ln 5
Mthemtics Lerning Centre, Universit of Sdne 4 Eercises 8. First, drw grph. The re is below the -is, so we first clculte 4 3 ( 4)d. 4 3 ( 4)d 4 (3 3 )d 4 [ 3 4 4 4 3 ] 4 64. The required re is therefore 64 units.. i ii π sin d [ cos ] π cos π + cos + π π From the grph we see tht the re π π Are sin d + sin d π [ cos ] π + [ cos ]π π ( cos π + cos ) + cos π + cos π ( ( )+)+ +( ) 4. 3. The grph of the curve cuts the -is t, nd. The totl re re A + re B. Are A ( + )( )d ( 3 + +4)d [ 3 4 +4 3 + ] ( 3 4+) 5 5. B A Are B ( + )( )d [ 3 4 +4 3 + ] ( 48 + 3 + 48) 3.
Mthemtics Lerning Centre, Universit of Sdne 5 Therefore the totl re is 5 + 3 37 squre units. Eercises 9. The curves nd + intersect where +. i.e. t or. The upper curve is +. (,3) Are (( +) ( ))d ( + )d [ + 3 3 ] (,) (4+ 8 3 ) ( + + 3 ) 4.. The curves intersect where 4 + i.e. 4 i.e. or. The upper curve is (see sketch). Are (( ) ( 4 + ))d (4 )d [ + 3 3 ] (,) (8 3 8) 3. (, ) 3. When π 4, sin nd cos. When 5π 4, sin nd cos. So the curves sin nd cos intersect t π 4 nd 5π 4. π/4 5π/4 Are 5π 4 π 4 (sin cos )d [ cos sin ] 5π 4 π 4
Mthemtics Lerning Centre, Universit of Sdne 6 ( cos 5π 4 sin 5π 4 )+(cos π 4 + sin π 4 ) 4. 4. (i) nd (ii) The curves re esier to sketch if we first find the points of intersection: the meet where 3 7 + 66. Tht is, 3 + 6 or ( )( +3). So the points of intersection re (, 6); (, ); nd ( 3, ). The first curve is n upside-down prbol, nd the second cubic. Totl re re A+re B. A 3 B Are A 3 3 (( 3 7 +6) (6 ))d ( 3 + 6)d [ 4 4 + 3 3 3 ] 5 3 4. 3 Are B 5 3. ((6 ) ( 3 7 + 6))d (6 3 )d.. the totl re 5 3 4 +5 3 squre units.