ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 2 MATH00040 SEMESTER /2018

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1 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH4 SEMESTER 7/8 DR. ANTHONY BROWN 4.. Introduction to Integrtion. 4. Integrl Clculus As ws the cse with the chpter on differentil clculus, for most of this chpter we will concentrte on the mechnics of how to integrte functions. However we will first give n indiction s to wht we re ctully doing when we integrte functions. This cn be mde rigorous mthemticlly but in this course we just wnt to get n intuitive ide of wht is going on. Suppose we wnt to find the re lying between the grph of function nd the x-xis between two given points nd b. Then one wy of doing this would be to pproximte this re by the re of rectngles which lie under the grph, s shown in Figure. The reson we use rectngles is becuse it is esy to clculte their re, it is simply their height times their width. Figure. An underestimtion of the re under the grph of the function f.

2 Of course the problem with this pproch is tht we will usully underestimte the re under the curve since we re not including the re bove the rectngles nd under the grph. One possible solution would beto mke the width of the rectngles smller nd smller. In this wy we would hopefully get better pproximtion to the re under the curve. However we cn not be sure tht this would be the cse if we re deling with relly strnge function. Another pproch is to overestimte the re by putting the rectngles bove the curve s Shown in Figure. Figure. An overestimtion of the re under the grph of the function f. You might point out tht this doesn t get us ny further nd you would be correct. Clerly it is no better to hve n overestimtion of the re. However the clever bit is tht we cn try nd reduce the overestimtion by chnging the widths of the rectngles nd we cn try nd reduce the underestimtion the sme wy (using different rectngles). If we cn get both the overestimtion nd the underestimtion of the re to be close to given number A then we sy tht the function f is integrble on the intervl [,b] nd we write under the curve is A. The number b b f(x)dx = A. In this cse the re f(x)dx hs specil nme. Definition 4.. (Definite Integrl). If function f is integrble on the intervl [,b], then the number b The function f is clled the integrnd. f(x)dx is clled the definite integrl of f from to b. In Figures nd, we hve given n exmple of function tht lies bove the x-xis between the points nd b but the re is signed re. Tht is if prt of the grph of f lies below the x-xis then this re is counted s negtive. For exmple

3 in Figure 3, the integrl b f(x)dx represents the re in red minus the re in green. This mens tht if we re going to use integrls to clculte res rther thn signed res, we hve to first find which prts of the grph lie bove the x-xis nd which prts lie below. In the cse of Figure 3, the ctul re tht lies between the grph of f nd the x-xis between the points nd b (i.e., the re of the red portion plus the re of the green portion) is we hve to put minus sign before the integrl b c f(x)dx gives minus the green re. c b c f(x)dx b c f(x)dx. Note tht f(x)dx to llow for the fct tht Figure 3. Signed re under the grph of the function f. 4.. The Fundmentl Theorem of Clculus. It is ll very well defining n integrl s we did in Section 4. but in prctice it is lmost impossible to use this definition to ctully clculte res. Luckily, the Fundmentl Theorem of Clculus comes to our rescue. There re severl slightly different forms of this theorem tht you my meet in your studies but the one we re going to use is the following. Theorem 4.. (The Fundmentl Theorem of Clculus). Let F nd f be functions defined on n intervl [,b] such tht f is continuous nd such tht the derivtive of F is f. Then b f(x)dx = [F(x)] b = F(b) F(). Remrk 4... Although this result is tught quite erly on in your mthemticl creer, it is most remrkble nd very deep result. It connects two seemingly 3

4 completely unrelted concepts. Firstly there is the derivtive of function, which gives the slope of tngent to curve nd then there is the integrl of function, which clcultes the re under the curve. The function F tht ppers in Theorem 4.. hs specil nme. Definition 4..3 (Antiderivtive). Let F be ny function such tht the derivtive of F is equl to nother function f. Then F is sid to be n ntiderivtive of f. Note tht the ntiderivtive of function is not unique. If F is ny ntiderivtive of f nd if c is constnt, then it follows from the sum rule nd the fct tht the derivtive of constnt is zero, tht F +c is lso n ntiderivtive of f. However, when using The Fundmentl Theorem of Clculus, it doesn t mtter if we use F or F +c since (F +c)(b) (F +c)() = F(b)+c (F()+c) = F(b) F(). Tht is the constnt will lwys cncel out. The function F +c, where c is rbitrry constnt, lso hs specil nme. Definition 4..4 (Indefinite integrl). Let F be ny function such tht the derivtive of F is equl to nother function f nd let c be n rbitrry constnt. Then F +c is sid to be n indefinite integrl of f nd the c is sid to be constnt of integrtion. This is written s f(x)dx = F(x)+c. Tht is, there is no or b on the integrl sign. Although we hve lot of progress theoreticlly, we hve still not ctully clculted ny integrls nd tht is wht we will turn our ttention to next Some Common Integrls. As with differentition, we strt with some bsic integrls nd then use these to integrte wide rnge of functions using vrious rules nd techniques. The bsic integrls tht you will need in this course re collected together in Tble. The min thing is to lern how to use them rther thn lerning them off by hert, since this tble will be included in the exm pper. Note tht in the tble, c will stnd for n rbitrry constnt. Wrning () As ws the cse with derivtives, the integrls of sin(x) nd cos(x) re only vlid if x is in rdins. If x is in degrees then extr constnts re needed. () Note tht the minus sign occurs with the integrl of sin(x), rther thn the integrl of cos(x), where it ppered when we were differentiting. As lwys, some exmples will mke things clerer. First of ll we will give some indefinite integrls in Tble. Remrk If you wnt to check your nswer when you hve foundn indefinite integrl then ll you need to do is to differentite your nswer. You should lwys get bck to the function you strted with. 4

5 f(x) f(x) dx Comments k kx+c Here k is ny rel number x n n+ xn+ +c Here we must hve n ln(x)+c Here we must hve x > x e x ex +c sin(x) cos(x)+c Note the chnge of sign cos(x) sin(x)+c Tble. Some common integrls In Exmple I hve given few exmples of definite integrls but relly finding the indefinite integrl is the hrd prt. Once you hve this, finding the definite integrl is just mtter of substituting numbers into the formul. Plese do remember however tht the vlue of the ntiderivtive t the lower limit hs to be subtrcted from the vlue of the ntiderivtive t the upper limit. Also note tht when clculting definite integrls, we ignore the constnt of integrtion c since it lwys cncels out. Exmple () Clculte the definite integrl [ x dx = () Clculte the definite integrl π ] 3 x3 π = = 7 3. sin(x) dx. x dx. [ sin(x)dx = ] π cos(x) = ( cos(π) ) cos() = ( ) =. Note tht in this cse the integrl is zero since the re bove the x-xis cncels out the re below the x-xis. (3) Clculte the definite integrl e 4x dx. e 4x dx = [ 4 ] e 4x = 4 e4 ( 4 ) e8 = e8 e

6 f(x) f(x) dx Comments c x+c 4 4x+c π πx+c π is just number e ex+c e is just number cos() cos()x + c cos() is just number x x +c Since x = x, n = x 3 4 x4 +c Here we tke n = 3 x 4 3 x 3 +c = +c Here we tke n = 4 3x3 x π π + xπ+ +c π is just number x e e+ x e+ +c e is just number e x e x +c Here we tke = e 5x 5 e5x +c Here we tke = 5 e 7x 7 e 7x +c Here we tke = 7 e ex e eex +c = e ex +c Here we tke = e sin(x) cos(x)+c Here we tke = sin(3x) cos(3x)+c Here we tke = 3 3 sin( x) cos( x)+c Here we tke = sin( πx) π cos( πx)+c Here we tke = π cos(x) sin(x)+c Here we tke = cos(4x) sin(4x)+c Here we tke = 4 4 cos( 5x) sin( 5x)+c Here we tke = 5 5 cos(πx) π sin(πx)+c Here we tke = π Tble. Some exmples of indefinite integrls As expected this integrl is positive since e x > for ll vlues of x (i.e., the grph of f(x) = e x lies bove the x-xis) The Sum nd Multiple Rules. As ws the cse with differentition, lthough the integrls in Tble re very useful, we would not get very fr if these were the only functions we could integrte. 6

7 Luckily there re rules tht llow us to integrte more complicted functions. The first two of these re lmost identicl to the equivlent ones for differentition. Theorem 4.4. (The Sum Rule for Integrtion). Let f: (,b) R nd g: (,b) R, then the definite integrl of f +g on the intervl [,b] is given by b (f +g)(x)dx = provided the integrls of f nd g exist. b f(x)dx+ b g(x) dx, All this sys is tht if we wnt to integrte sum of two functions then ll we hve to do is integrte them seprtely nd dd the integrls. Remrk As you might expect there is n equivlent rule for indefinite integrls: (f +g)(x)dx = f(x)dx+ g(x) dx. Note tht when you hve sum like this you only need to include one constnt of integrtion. This is since if you dd n rbitrry constnt to n rbitrry constnt you just get n rbitrry constnt. Here re couple of exmples of the use of the Sum Rule. Exmple () Evlute the definite integrl x 4 +e x dx = = [ 5 x5 x 4 dx+ ] e x dx + [ e x] x 4 +e x dx. = ( )5 +( e ) ( e ) = 5 +e e. () Find the indefinite integrl x +cos( 3x)dx. Provided x > (so tht dx = ln(x)+c), x x +cos( 3x)dx = x dx+ cos( 3x) dx = ln(x) 3 sin( 3x)+c. As ws the cse with differentition, the second rule tht will enble us to integrte lrger rnge of functions is the Multiple Rule. 7

8 Theorem (The Multiple Rule for Integrtion). Let f: (,b) R nd let k R (here I will use k insted of c to void confusion with the constnt of integrtion c). Then the definite integrl of kf over the intervl [,b] is given by b provided the integrl of f exists. (kf)(x)dx = k b f(x)dx, All this sys is tht if we wnt to integrte constnt multiple of function, then ll we hve to do is first integrte the function nd then multiply by the constnt. Remrk Of course, there is corresponding Multiple Rule for indefinite integrls: (kf)(x)dx = k f(x)dx. Here re couple of exmples of how the Multiple Rule works. Exmple () Evlute the definite integrl x dx = x dx x dx. = [ln(x)] = (ln() ln()) = ln(). Note tht since the grphof f(x) = lies below the x-xis onthe intervl x [, ], the integrl dx must be negtive. x () Find the indefinite integrl 3e 4x dx. 3e 4x dx = 3 e 4x dx = 3 4 e4x +c = 3e4x 4 +c. Here we just write c rther thn 3c since three times n rbitrry constnt is still just n rbitrry constnt. As you would expect, both the sum nd multiple rules cn be used t the sme time. Here re couple of exmples of this. 8

9 Exmple π π sin(3x) 4e x dx = () Evlute the definite integrl π π π sin(3x)dx+ = 4 ( e π e π). () Find the indefinite integrl Provided x > (so tht π π π π π 4e x dx sin(3x) 4e x dx. = sin(3x) 4 e x dx π π [ = ] π 3 cos(3x) 4[e x ] π π π = [ 3 ( cos(3π) 3 )] cos( 3π) 4 [ e π e π] [ = 3 ] 4 [ e π e π] 3 6x +5x5 dx = 6x +5x5 dx. dx = ln(x)+c), x = 6 6x dx+ x dx+5 5x 5 dx x 5 dx = 6 ln(x)+5 ( 6 x6 )+c = 5x6 ln(x) +c. 6 Agin note we only hve the one rbitrry constnt Integrtion by Substitution. Unfortuntely the sum nd multiple rules re the only rules tht crry over directly from differentition to integrtion. While there re rules for integrtion, they re not quite s direct s the rules for differentition nd it cn often be hrder to decide which rule to use. Becuse of this it is even more importnt to do lots of prctice problems for integrtion, since it is only through experience tht you will lern which rule is likely to be the best one to use. It is lso the cse tht it is esy to write down functions tht cn t be integrted lgebriclly (lthough there re numericl methods tht cn be used, we won t be looking t these in this course). For exmple the function f(x) = e x cn t be integrted. The first technique we will look t is integrtion by substitution. This cn be written down s theorem but I feel tht pproching it in this wy cn mke it look more 9

10 difficult thn it ctully is. It is fr better in my opinion to lern this technique through looking t some exmples nd this is wht we will do now. Exmple Evlute the definite integrl (x+3) dx. One wy of pproching this problem would be to expnd (x + 3) nd then integrte the resulting expression. However this would tke lot of work nd there would be lots of scope for errors. Insted we will use integrtion by substitution. The key to the technique is to note tht f(x) = (x + 3) cn lso be written s f(x) = u, if we let u = x+3. The three things we hve to do now re: (i) Express the function f(x) = (x+3) in terms of u. Note tht in generl in this step, there will be some function of x left over. With luck this left over bit will cncel with the expression we obtin in (ii). (ii) Express dx in terms of du, u nd x. (iii) Chnge the limts of integrtion to be in terms of u rther thn x. If this method is going to work we will now hve n integrl with respect to u nd with no x s ppering nywhere. If there re x s still left then we will either hve to try different substitution or we will hve to try different method ltogether. In this cse we do ech of these s follows: (i) Since u = x+3, the function (x+3) becomes u. (ii) For this we use the formul dx = dx du du du =. In this cse =, so du du/dx dx dx = du = du. Note tht dx nd du re relly not vribles but in this prticulr sitution, you cn tret them s if they were. (iii) When x =, u = +3 = 3 nd when x =, u = +3 = 4. Thus the lower nd upper limits of integrtion become 3 nd 4 respectively. Putting ll this together we obtin (x+3) dx = = 4 3 u du Note there re no x s here. [ u ] 4 3 = 4 3 = 4 3.

11 Exmple Find the indefinite integrl 6x +4x+ x 3 +x +x+ dx. First note tht when we re using integrtion by substitution to find n indefinite integrl, then clerly we only need to perform the first two steps bove. However, the min question is wht substitution will we mke? Looking t the integrnd, we see tht the numertor is multiple of the derivtive of the denomintor. In cses like this, when one prt of the integrnd is multiple of the derivtive of nother prt, then good strtegy is to let u equl the undifferentited bit. So we let u = x 3 +x +x+. Then the steps re: (i) We hve 6x +4x+ x 3 +x +x+ = 6x +4x+. Note tht in this cse we still hve u function of x in the numertor. (ii) Since du dx = 3x +x+, we hve dx = du du/dx = Putting this together we hve 6x +4x+ 6x x 3 +x +x+ dx = +4x+ u = u du = u du = ln(u)+c du 3x +x+. du 3x +x+ = ln ( x 3 +x +x+ ) +c. Remrk () The step where we hve written 6x +4x+ du isbit dubious nottionlly, since theintegrl u 3x +x+ is with respect to u but there re lso x s in the expression. However I recommend tht you include this step, t lest in this course. () Once we hve definite integrl then it is good ide to differentite it nd mke sure we get the originl function. I will leve it to you to differentite f(x) = ln(x 3 +x +x+)+c nd mke sure you get f (x) = 6x +4x+ x 3 +x +x+. Exmple Evlute the definite integrl π xcos ( x ) dx. Agin in this exmple we see tht x is multiple of the derivtive of x, so let us try the substitution u = x. The three steps re: (i) Since u = x, the function xcos(x ) becomes xcos(u). We hve n x left over here. In this cse we could write x = u but this would not be good ide since we re hoping tht the x will cncel with prt of the expression we get from (ii).

12 (ii) Since du du = x, dx = dx du/dx = du x. (iii) When x =, u = = nd when x = π, u = ( π lower nd upper limits of integrtion become nd π 4 respectively. Putting this together we obtin ) = π. Thus the 4 π xcos ( x ) dx = = u= π 4 u= π 4 xcos(u) du x cos(u) du = [sin(u)]π 4 = [ ( π ] sin sin() 4) = = 4. Exmple Find the indefinite integrl cos(x)e sin(x) dx. Here we note tht cos(x) is the derivtive of sin(x), so we try the substitution u = sin(x). The steps re s follows. (i) We hve cos(x)e sin(x) = cos(x)e u. Note tht gin we hve function of x left over. It is possible to express cos(x) s function of u but gin this would not be good ide. (ii) Since du du = cos(x), we hve dx = dx du/dx = du cos(x). Putting this together we obtin cos(x)e sin(x) dx = cos(x)e u du cos(x) = e u du = e u du = e u +c = e sin(x) +c.

13 Agin you should check tht on differentiting e sin(x) +c you get cos(x)e sin(x) Integrtion by Prts. The next technique is used to trnsform integrls of the form f(x)g (x)dx into something esier to integrte. In this course f(x) will usully be power of x (or polynomil in x) but we will lso look t one other interesting cse. Provided ll the integrls exist then the integrtion by prts formul for indefinite integrls sys tht f(x)g (x)dx = f(x)g(x) f (x)g(x)dx nd the integrtion by prts formul for definite integrls sys tht b f(x)g (x)dx = [f(x)g(x)] b b f (x)g(x)dx. Note tht in the originl integrl, we hve f times the derivtive of g, while in the integrl on the right we hve g times the derivtive of f. Remrk () There is no constnt of integrtion in the integrtion by prts formul for definite integrls since we cn just include it in the integrl f (x)g(x)dx. () In order to use the method of integrtion by prts we hve to be ble to integrte the function tht we cll g (x). Sometimes this will give us clue s to which function we will cll f nd which function we will cll g. (3) It is possible when using integrtion by prts to end up with more difficult integrl thn we strted with. If this hppens then it mens we will hve to go bck to the drwing bord - either we will hve to choose different functions for f nd g or mybe integrtion by prts is not suitble method for the prticulr function we re trying to integrte. Now we will look t some exmples to see how the method works in prctice. Exmple Find the indefinite integrl xe x dx. If we look t this integrl we note tht if we could get rid of the x then we could integrte the e x, so this suggests tht we let f(x) = x nd g (x) = e x. (Note tht if we let f(x) = e x nd g (x) = x, then lthough we could perform integrtion by prts, we would end up with more complicted integrl since we would hve g(x) = x nd f (x) = e x, so we would hve f(x)g (x)dx = x e x dx). With f(x) = x nd g (x) = e x, we obtin f (x) = nd g(x) = e x, so the integrtion by prts formul yields xe x dx = xe x e x dx. 3

14 We cn now esily finish the integrtion to get xe x dx = xe x e x +c. Let us now use integrtion by prts with definite integrl. Exmple Evlute the definite integrl π x sin(3x) dx. Here we let f(x) = xndg (x) = sin(3x), so tht f (x) = ndg(x) = 3 cos(3x). Thus, using the integrtion by prts formul, we obtin [ x3 ]π cos(3x) π xsin(3x)dx = = π 3 cos ( 3π = π 3 cos(3x)dx π ) ( )+ [ ]π = 9 sin(3x) = ( ) 3π 9 sin 9 sin() = 9. 3 cos(3x)dx π 3 cos(3x)dx The next exmple is little bit different since it ppers t first sight tht we cn t use integrtion by prts since we don t hve product of functions. Exmple Find the indefinite integrl ln(x) dx. The key here is to note tht we cn write ln(x) = ln(x) nd let f(x) = ln(x) nd g (x) =, so tht f (x) = nd g(x) = x. Then, using the integrtion by prts x formul, we obtin ln(x)dx = xln(x) xdx = xln(x) dx = xln(x) x+c. x Remrk Note tht differentiting on the log term will lso enble us to integrte functions of the form y = x n ln(x). We will finish this section by looking t n exmple where we hve to use the integrtion by prts formul twice. Exmple Evlute the definite integrl π x cos(x)dx. For the first ppliction of the formul, we let f(x) = x nd g (x) = cos(x), so tht 4

15 f (x) = x nd g(x) = sin(x). Then, using the integrtion by prts formul, we obtin π x cos(x)dx = [ x sin(x) ] π π π xsin(x)dx = + x sin(x) dx. For the second ppliction of the formul, we let f(x) = x nd g (x) = sin(x), so tht f (x) = nd g(x) = cos(x). Then, using the integrtion by prts formul gin, we obtin π π xsin(x)dx = [xcos(x)] π cos(x)dx Thus π x cos(x)dx = π. = π [sin(x)] π = π ( ) = π Integrtion Using Prtil Frctions. The lst technique of integrtion we will look t is integrtion using prtil frctions. Relly this is not so much technique of integrtion, it is more technique used to express lgebric expressions in different form. It is used in situtions where we re deling with functions of the form f(x), where f(x) nd g(x) re both polynomils g(x) with the degree of f being less thn the degree of g nd where g cn be fctorised. Remrk This technique cn be extended to the cse where the degree of f is greter thn or equl to the degree of g but in this cse we first hve to divide f by g using polynomil long division. We covered this in the first semester, but I won t give you ny questions on integrtion requiring it since it would mke the questions too long. Now let us look t some exmples to see how this technique works in prctice. The esiest cse is where g is qudrtic nd cn be fctored into two different liner fctors 4 Exmple Find the indefinite integrl x x 3 dx. Here we note tht x x 3 = (x+)(x 3), so we let 4 () x x 3 = A x+ + B x 3, where A nd B re constnts we hve to find. Multiplying both sides of () by x x 3 we obtin () 4 = A(x 3)+B(x+). There re now two wys we cn proceed. There is quick wy tht will work in lots of cses nd there is slightly longer method tht will work in ll cses. 5

16 We will first look t the quick method. If we let x = 3 in (), we obtin 4 = B(3+), so B =. Then if we let x = in (), we obtin 4 = A( 3), so A =. 4 Thus we hve x x 3 = x+ + x 3. Before we go hed nd perform the integrtion, we will look t the other method (which will work in ll cses no mtter how complicted). Wht we do is to rewrite () s (3) 4 = (A+B)x+( 3A+B). Tht is we collect the terms in x together nd the constnt terms together. If we now look t the coefficient of x on either side of (3) we see tht = A + B (the term on the left is x) nd if we look t the constnt term on either side of (3), we see tht 4 = 3A + B. Thus we obtin the simultneous equtions = A + B nd 4 = 3A+B. Solving these we get A = nd B = s bove. We will now finish the job nd perform the integrtion. 4 x x 3 dx = x+ dx+ dx = ln(x+) ln(x 3)+c. x 3 I performed the integrtions by inspection, but if you cn t spot them, you cn use the substitutions u = x+ nd u = x 3, respectively. Next let us look t n exmple where p(x) hs three liner fctors. Note tht I won t expect you to fctor cubic - in cses like this I will lwys give you t lest one of the fctors. I will lso keep ll the exmples in this section s indefinite integrtion exmples, so we cn focus on the prtil frctions nd not hve to worry bout substituting in the upper nd lower limits. There is nothing difficult bout this however, once you hve done the integrtion, it is not problem. Exmple Find the indefinite integrl 3x +x+ (x+)(x+)(x+3) dx. First note tht the numertor is not multiple of the derivtive of the denomintor, so integrtion by substitution will not work nd we hve to use prtil frctions. In this cse we let (4) 3x +x+ (x+)(x+)(x+3) = A x+ + B x+ + C x+3, where A, B nd C re constnts we hve to find. Multiplying both sides of (4) by (x+)(x+)(x+3) we obtin (5) 3x +x+ = A(x+)(x+3)+B(x+)(x+3)+C(x+)(x+). The quick method will work gin in this cse, so let us do it this wy first. If we let x = in (5), we obtin 3 + = A( +)( +3)), so A =. Then if we let x = in (5), we obtin 4+ = B( +)( +3), so B =. Finlly if we let x = 3 in (5), we obtin = C( 3 + )( 3 + ), so C =. 6

17 3x +x+ Thus we hve (x+)(x+)(x+3) = x+ + x+ + x+3. The other method is s follows. If we expnd the right hnd side of (5) nd collect together terms in x, x nd the constnt terms we obtin (6) 3x +x+ = (A+B +C)x +(5A+4B +3C)x+(6A+3B +C). Then compring terms in x, x nd the constnt terms in (6), we obtin the simultneous equtions A +B + C = 3, 5A+4B + 3C = nd 6A+3B + C =. These cn be solved to obtin A = B = C = s before. After using either of these methods, we cn perform the integrtion: 3x +x+ (x+)(x+)(x+3) dx = x+ dx+ x+ dx+ = ln(x+)+ln(x+)+ln(x+3)+c. x+3 dx Next we will look t the cse where g(x) hs repeted fctor. x Exmple Find the indefinite integrl (x ) dx. Agin note tht the numertor is not multiple of the derivtive of the denomintor, so integrtion by substitution will not work nd we hve to use prtil frctions. When we hve repeted fctor in the denomintor, we let (7) x (x ) = A (x ) + B (x ). Note tht similr method is used for higher powers, we just need three terms for third power nd so on. Multiplying both sides of (4) by (x ) we obtin (8) x = A+B(x ). Unfortuntely when we hve repeted root then the simple method of finding A nd B will not work completely. We cn let x = in (8) to obtin = A, so tht A = but then we still need to use the other method of compring coefficients to find B. We could use hybrid method but perhps it is more strightforwrd to go stright for the compring coefficients method. Rerrnging (8) we obtin x = Bx+(A B). Then we obtin the simultneous equtions B = nd = A B. These hve solution A = nd B =, so x (x ) = (x ) +. We cn now perform (x ) the integrtion. x (x ) dx = (x ) dx+ (x ) dx = x +ln(x )+c. I performed the lst two integrtions by inspection, but you cn use the substitution u = x if you cn t spot them. 7

18 For our finl exmple, we will look t the cse where g(x) hs qudrtic fctor tht cn t be fctorised. 3x 4x+ Exmple Find the indefinite integrl (x x+)(x ) dx. In this cse we let (9) 3x 4x+ (x x+)(x ) = Ax+B x x+ + C x. Agin the simple method for finding A, B nd C will not work here, so we hve to use the method of compring coefficients. Multiplying both sides of (9) by (x x+)(x ) we obtin 3x 4x+ = (Ax+B)(x )+C(x x+). Multiplying this out nd collecting terms we get 3x 4x+ = (A+C)x +( A+B C)x+( B +C). This yields the simultneous equtions 3 = A + C, 4 = A + B C nd = B + C. These cn be solved to give A =, B = nd C =, so we 3x 4x+ hve (x x+)(x ) = x x x+ +. We cn now perform the integrtion to get x 3x 4x+ (x x+)(x ) dx = x x x+ dx+ x dx = ln(x x+)+ln(x )+c. Note you cn use the substitution u = x x+ for the second lst integrl if you cn t spot it Finding Ares nd Volumes. In the introduction to this chpter, we noted tht integrtion is essentilly method of clculting the re between the grph of function nd the x-xis. However we lso showed tht if we wnt to clculte res then we hve to be little creful since integrtion finds signed res. Tht is if the function lies below the x-xis then the integrl is negtive. We will now look t couple of exmples where we wnt to clculte ctul res rther thn signed res. Exmple Find the re lying between the grph of f(x) = x 3 nd the x-xis between the points x = nd x =. We first note tht if x < then x 3 < nd if x > then x 3 >. So the grph of f(x) lies below the x-xis for x [,) nd bove the x-xis for x (,]. We hve to tke ccount of the fct tht it lies below the x-xis for x [,) by tking minus the integrl for tht rnge of x (since we re looking for the ctul re rther thn the signed re). 8

19 Hence the ctul re is x 3 dx+ ] ] [ [ x x 3 4 x 4 dx = ) ( ) = ( ( ) = =. Wrning Ifweclcultingnctulre(rtherthnsignedre),thenwe must get positive nswer. If we don t then we must hve gone wrong somewhere. Here is nother exmple. Exmple Find the re lying between the grph of f(x) = sin(x) nd the x-xis between the points x = π nd x =. In this cse the grph of f lies below the x-xis throughout the region of interest. Hence we just tke minus the integrl. So the re is [ sin(x)dx = ] π cos(x) π = [ ( cos() )] cos( π) [ = ( )] =. Here is one finl exmple. Exmple Find the re lying between the grph of f(x) = x 3 +x 9x 9 nd the x-xis between the points x = nd x = given tht the grph of this function only crosses the x-xis t x = in the intervl [,]. Here we re given tht the grph of f only cuts the x-xis once in the region of interest. Thus it either lies bove or below the x-xis between nd nd the opposite between nd. To decide which, we just hve to check nother point. In this cse zero is esiest to check. Since f() = 9 <, it follows tht f lies below the x-xis in the intervl (,] nd bove the x-xis in the intervl [, ). Thus 9

20 the re is x 3 +x 9x 9dx ] x 3 +x 9x 9dx [ = 4 x4 + 3 x3 9 [ x 9x 4 x4 + 3 x3 9 x 9x [( = 4 ( )4 + 3 ( )3 9 ) ( ) 9( ) ( 4 ( )4 + 3 ( )3 9 )] ( ) 9( ) [( 4 ()4 + 3 ()3 9 ) () 9() ( 4 ( )4 + 3 ( )3 9 )] ( ) 9( ) [ 53 = 4 ] [ ] = 6. We cn lso use definite integrtion to find volumes. Wht we will look t in this section is finding the volumes creted when grphs re rotted bout the x-xis but there re mny other techniques vilble. ] Figure 4. The volume of revolution of the function f(x) = x 4 x between x = nd x =.

21 Figure 4 shows the grph of the the function f(x) = x 4 x rotted bout the x-xis between the points x = nd x =. Luckily there is simple formul tht llows us to find the volume of the body obtined when ny function is rotted bout the x-xis. Theorem (Volume of Solid of Revolution). The volume V obtined when function f is rotted bout the x-xis between the points x = nd x = b is given by b V = π f(x) dx. Remrk () Since the function is squred, we don t hve to worry bout whether the function is positive or negtive, s we did when clculting res. () Since the re of circle of rdius f(x) is πf(x), this formul is relly just obtined by summing up the volumes of series of disks perpendiculr to the x-xis, in similr fshion to the wy the re under curve ws obtined by summing up the res of series of rectngles. Let us now show how this works in prctice by clculting the volume of the solid shown in Figure 4 Exmple Find the volume of revolution of the function f(x) = x 4 x bout the x-xis between x = nd x =. In this cse we use Theorem with =, b = nd f(x) = x 4 x. Hence the volume is ( V = π x 4 x ) dx = π x 8 x 6 +x 4 dx [ = π 9 x9 7 x7 + ] 5 x5 [( = π ) ] ( +) 5 = 8π 35. Wrning As ws the cse when clculting ctul res, we must get positive nswer when clculting volumes. If we don t then we hve gone wrong somewhere. Let us do one more exmple. Exmple Find the volume of revolution of the function f(x) = e x bout the x-xis between x = nd x = 3. In this cse we use Theorem with =, b = 3 nd f(x) = e x. Hence the

22 volume is V = π 3 ( e x ) dx = π 3 4e 4x dx = π [ e 4x] 3 = π( e e 4).