ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019


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1 ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. 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 xxis 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. 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
2 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 2. Figure 2. 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 f(x)dx = A. In this cse the re f(x)dx hs specil nme. Definition 7.. (Definite Integrl). If function f is integrble on the intervl [,b], then the number The function f is clled the integrnd. f(x)dx is clled the definite integrl of f from to b. In Figures nd 2, we hve given n exmple of function tht lies bove the xxis between the points nd b but the re is signed re. Tht is if prt of the grph of f lies below the xxis then this re is counted s negtive. For exmple in Figure 3, the integrl f(x)dx represents the re in red minus the re in 2
3 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 xxis nd which prts lie below. In the cse of Figure 3, the ctul re tht lies between the grph of f nd the xxis 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 c f(x)dx gives minus the green re. c b c f(x)dx 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 The Fundmentl Theorem of Clculus. It is ll very well defining n integrl s we did in Section 7. 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 7.2. (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 f(x)dx = [F(x)] b = F(b) F(). Remrk Although this result is tught quite erly on in your mthemticl creer, it is most remrkble nd very deep result. It connects two seemingly 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. 3
4 The function F tht ppers in Theorem 7.2. hs specil nme. Definition (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 (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. 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 > 0 x e x ex +c sin(x) cos(x)+c Note the chnge of sign cos(x) sin(x)+c Tble. Some common integrls 4
5 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. (2) 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 2. 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. 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 2 [ x 2 dx = (2) Clculte the definite integrl π 0 ] 2 3 x3 π 0 2 = = 7 3. sin(2x) dx. x 2 dx. [ sin(2x)dx = ] π 2 cos(2x) 0 = ( 2 cos(2π) ) 2 cos(0) = ( 2 ) 2 = 0. Note tht in this cse the integrl is zero since the re bove the xxis cncels out the re below the xxis. (3) Clculte the definite integrl 2 2 e 4x dx. e 4x dx = [ 4 ] e 4x = 4 e4 ( 4 ) e8 = e8 e As expected this integrl is positive since e x > 0 for ll vlues of x (i.e., the grph of f(x) = e x lies bove the xxis). 5
6 f(x) f(x) dx Comments 0 c 2 2x+c 4 4x+c x+c is just number e ex+c e is just number cos() cos()x + c cos() is just number x 2 x2 +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( 2x) cos( 2x)+c Here we tke = 2 2 sin(x) π cos(x)+c Here we tke = 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 2. Some exmples of indefinite integrls 7.4. 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. 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. 6
7 Theorem 7.4. (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 (f +g)(x)dx = f(x)dx+ provided the integrls of f nd g exist. 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. x 4 +e x dx = = [ 5 x5 x 4 dx+ ] e x dx + [ e x] = ( )5 +( e ) ( e ) = 2 5 +e e. (2) Find the indefinite integrl x +cos( 3x)dx. Provided x > 0 (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 provided the integrl of f exists. (kf)(x)dx = k 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 2 2 2x dx = 2 x dx 2 2x dx. = 2 [ln(x)]2 = 2 (ln(2) ln()) = ln(2) 2. Note tht since the grphof f(x) = lies below the xxis onthe intervl 2x 2 [, 2], the integrl dx must be negtive. 2x (2) 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 2sin(3x) 4e x dx = () Evlute the definite integrl π Provided x > 0 (so tht π 2sin(3x)dx+ = 4 ( e e π). (2) Find the indefinite integrl x π π π 4e x dx 2sin(3x) 4e x dx. = 2 sin(3x) 4 e x dx [ = 2 ] π 3 cos(3x) 4[e x ] π = 2 [ 3 ( cos(3π) 3 )] cos( 3π) 4 [ e π e ] [ = 2 3 ] 4 [ e π e ] 3 6x +5x5 dx = 6x +5x5 dx. dx = ln(x)+c), 6x dx+ = 6 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. 9