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# Short Calculus MAT 016B

UCD

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This 23 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 016B at University of California - Davis taught by Staff in Fall. Since its upload, it has received 54 views. For similar materials see /class/187396/mat-016b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

Lecture Notes 7 April 167 2008 Page 1 Notes I can7t7 under any circumstances7 make special allowances if you miss or plan to miss a quiz This is the reason that the lowest quiz score will be dropped Starting with Quiz 27 I will post grading details on the quiz solutions7 so you can see exactly how points are assigned7 since I dont usually write much on your quizzes as I grade them l7ve posted a practice midterm The solutions will be posted next Monday Do not wait until then to try to do the problems 1 Exponential integrals Since fz cm is its own derivative7 it is also its own antiderivative In other words7 emdz em C We can use this fact and the idea of substitution to nd many integrals 2 Problems I 1 Find the integral ezwdz Solution Let u 2x so du 2dz and 21 u u 21 d 7 d 7 0777 6 26 u 26 26 2 Find the integral 5ze 2dz Solution Let u zz so du 72zdz and 2 5 5 5 2 5 md 77 d 77 077 m C ze x 26 u 26 26 Lecture Notes 7 April 167 2008 Page 2 3 Find the integral 672m 2 dz Solution Let u 72 so du dx and 672m 2 dxe due 0e 2m0 s 3 Logarithmic integrals Remember from before that d 1 7 lnx 7 dx z This is only true if z gt 07 since you cant take the natural logarithm of zero or negative numbers However7 if z lt 07 d d 1 d 71 1 111W g 1111 l 3 95 3 E Therefore d 1 7 1 7 ch lt um i for any x 31 0 So now we know that nd i lnlzlC ifn717 z 7 iglJrC otherwise Lecture Notes 7 April 167 2008 Page 3 4 Problems II 1 Find the integral 2 2 z dx x 1 Solution Let u x2 17 so do 2zd and 2 1 21dz7dulnlulClnx210 n u 2 Find the integral cosx dx s1n Solution Let u sin x so do cos ndx and 1 Cosxdz7dzlnlul0lnlsinxl0 u sinx 3 Find the integral 32 271 2 dz z Solution 32271 3x2 2x 1 2 1 Td9 t g m 2 1 1 3dz idzi 72dx3z2lnlli0 n n n 4 Find the integral 1 7d 16 m z Solution 1 dt6m 1 dt 51 dtidz 1 6 611 6 1 6 61 61 1 Let u em 17 so do emdx and em 1 m mdxEdulnlul0lne 1C Make sure that you can do Exercises 531 27 odd7 29 37 odd by hand7 not with a cornputer7 39 55 odd At this point you should be ready to do Homework 3 Lecture Notes 7 April 147 2008 Page 1 Quiz 1 One of the most common mistakes on Quiz 1 was to write things like 1 Sit1112 10 7 5mm 2 7 which is incorrect Note that if you get the correct answer to a problem but simplify77 it to something incorrect7 you may lose points 1 also noticed that many people who got low scores were just trying to repeat the steps 1 did in the example in class without really understanding the purpose of each step Overall7 1 was pleased with the results of this quiz 1 Differentials review The last topic you should have covered in 16A was differentials see page 244 As an example7 since d sin x cos x we can write7 in differential form7 d sin s cos m dx 2 The General Power Rule Let u x2 1 Then du 2xdx We can use this to rewrite the integral x2 13 2zd as 4 2 1 4 u3duuiCMC 4 4 This is an example of the General Power Rule7 du u 1 7d d O u cm x u u 71 1 7 for n 74 71 Lecture Notes 7 April 147 2008 Page 2 3 Problems I 1 Find the integral 2x 1 2 m dx Solution Let u x2 x so do 2x 1 dz Then 2 2 2 2x1x2dzuduCWC 2 Find the integral 3x2 3 7 2dm Solution Let u x3 7 27 so do 32dz Then 32 3 7 232 2 3 2 3213726 12d L O 07 372 C z z u u 32 32 3x 3 Find the integral 74 2d 17 2x2 Solution Let u 1 7 2x27 so do 74zdx Then 74x 7 do i 2 i 1 7 2 1 i 1 17222dz77u dui7u Ci7172x 0777172z20 4 Substitution The General Power Rule is a simple example of a process called substitution7 or a change of variables When you decide what the new variable of integration u should be7 you know what do should be If do is not already in the integral7 sometimes you can rewrite the integral so that it is Lecture Notes 7 April 14 2008 Page 3 For example to nd z 3 7 4x22 dz you might let u 3 7 1amp2 Then du 78zdz so you can write the integral in a form you can work with by multiplying and dividing by 78 1 z 3 7 4x22 dx 3 7 4x22 78 mdz 1 1 77 3 7 4x22 78x dx 77 u2du 8 8 Another way to see this is to solve du 78zdx for dz which gives dm 7871mdu so z 3 74x22dz zu2 7idu 7 uzdu 7 u2du 5 Problems II 1 Find 3 3x4 12dz Solution Let u 3x4 1 so du 12x3dx Then 3 4 2 1 4 2 3 x 3m 1 dx 53x 112xdz 1 2 13 1 4 3 7 d7 073 1 O 12uu 36 36951 2 Find 7x2mdz Solution Let u x3 1 so du 32dm Then 7xzmdz 3x2 dx 7 71132 14 32 7 d 77 07 3 1 o 3 u 332 905 1 Lecture Notes 7 April 21 2008 Page 1 New policy on grading questions After any quiz or exam is returned you have one week to ask questions in of ce hours or email about your score I will keep an electronic copy of each quiz or exam for one week after which it will be deleted I can only take very limited questions about the grading of homework assignments since I dont do this myself 1 Quiz 2 The problem was to nd elzz 32 dz z Most students saw that it would be best to split this into two integrals ell2 3x2 ell2 3x2 dz 3dz73dz z z z Most of those who did this found the second integral in some way The main problem was the rst integral At this point in the course the only things we can try to do with an integral like this is to either nd it directly and its too complicated for that or use a substitution to turn it into an integral we can nd directly It looks like there are only a few possible choices for the substitution 2 1 lw uz3 or 117 2 u 6 These are the main substitutions people tried on the quiz If we let u 61 2 then d 2 1 d M2712 d 72 wid u 6 Ch z z 6 3 z and the integral becomes 1 1 1 iidu 7511 C fig2 C This substitution works Mature mes 7 Am 21 zoos Page 2 u we m u a 011m du aw thmghke L11 annals m we 1nle so ms substitution theblv woulan do any sud Kwem d a and ms ms Mm J m ecq C Th1 suhsmuum also wads end n I pmhehlv the mml nature choice The ysas dismhuunn m nus qmz has Lher mam swaps a smasms The st mu m Pan39s Mm th mm myust Jump m and dams magnum sway The second yam spin ms leBel up end am we amply d we Wm new msgst curredW The mm youp dud ms Problem mmw mm wl he at 129st me women m Mmsm 1 xmgdy es mm es ms quiz and mm d m m 0 ms amuse ccuses an ndmg mleyels and um ewluslm m reelrwmld thlms we sm ups in ms muss based an we lsues sew an em 2 E E Advme a 1nle Do my rush m sum mg smmsm m dedde whsl we best ms Step wmld he m to unk ahead ys whet you re sham m an smngm mmPMy ms smash u mmhmg asssm walks uy mezhmg else D ry us has your rst 1m m walk Be mm m m confuse dm mmueum end musng Lecture Notes 7 April 21 2008 Page 3 2 Review of integration Exercises 5214 22 28 536 20 26 30 50 5436 44 50 3 Practice midterm 14 1 Solve each equation for x a 21nx 31n2 Inf 1n23 1n8 x28 i but you cant take the natural log of a negative number so z 2 i 39 1 5570252 7 2 1 5670255 5670255 1 670255 l 1 7025x 1n 7 5 1 z filing or 41n5 C 2 3 1n 2mg 1n3 ln2wlnew 1n3 xln21 1n3 1n3 7 1n21 Lecture Notes 7 April 21 2008 Page 4 2 Some scientists start observing a frog pond After two years they count 27 frogs After another year they count 81 frogs If the frog population is growing exponentially how many frogs were there initially 052k 27 053k 81 053k 81 052k 27 3k 57k 3 i 63k72k 3 6k 3 k ln3 0521113 27 27 7621113 or 3 Lecture Notes 7 April 217 2008 Page 5 3 Find the derivative of each function a f z2x we WW f e h wztw lm 2 2x1 2x z2x s 121nx2z xE b g cos Mk ehizlncosm d gz emmmmsw lnz 1n cos sinx 7 7 1 7 1 cos s cm z n cos s n s COS z Lecture Notes 7 April 97 2008 Page 1 Quiz 1 The rst quiz will be similar to Homework 27 which is based on this lecture 1 Exponential growth and decay In many real life situations7 a quantity changes at a rate proportional to its size In other wordls7 dyi 7k dt 9 for some constant of proportionality k which may be positive or negative If k gt 07 the quantity y is growing If y lt 07 y is shrinking7 or flecaying7 If we start with an initial value y0 C then y 06 This situation is called exponential growth or decay 2 Example Suppose the rate of change of a quantity P is four times the size of P7 so dP 7 dt 7 and P0 1 We know that 4P Pt 16 We can show that Pt satis es the two equations above7 since dP t d d 70 CT 154 54 4t 454 4Pt and Lecture Notes 7 April 9 2008 Page 2 How long does it take P to double To answer this we need to nd the time t at which Since we know Pt we can rewrite this as 154 2 1 Taking the natural logarithm ln 64 j ln 2 so 4t ln2 andt ln2 1 I Just for fun we can rewrite Pt as an exponential function with base 2 by using the properties of exponents and the natural logarithm Pa 64 614t 6111211124t 6m24th12 24tln239 Another way to do this is to notice that we found earlier that P doubles when t iln 2 so 2tiln2 because then P 31112 21 2 4 3 Example time preference lf I offered you either 100 right now or 150 a year from now you would probably choose to have 100 now However if 1 increased the amount I offered to give in a year eventually you might choose to wait for a bigger payment For example suppose I offered you either 100 right now or 1000 a year from now This might be the point at which you would have trouble deciding In other words 1000 in a year is only worth about 100 to you right now The fact that people prefer a small payment in the present over a slightly larger payment in the future is known in economics as time preference77 The reason that banks can get people to borrow money from them with interest is that they have much lower time preference than the people who borrow the money In other words to a bank money in the future is worth almost as much as money right now Lecture Notes 7 April 97 2008 Page 3 We can use a function to write the statement 100 right now is equal to 1000 in a year77 as an equation 1D0 10D1 This equation is satis ed by a simple exponential function7 Dt Exponential functions are often used by economists to think about time preference Expo nential functions have the special property that they always take the same time to grow or shrink by the same factor For example7 Dt takes 1 year to shrink by 10 times7 so mamm 1 mai Dm and so on In populations that grow exponentially7 it always takes the same amount of time to double If the doubling time of humans is exactly 61 years7 then there will be twice as many in April 2069 as right now7 and twice as many in December 2055 as there were in December of 1994 In economics7 if time preference is an exponential function7 then a person who thinks 100 right now is worth 1000 in a year will also think that 1000 in a year is worth 10000 in two years 4 Example radioactive decay Radium is radioactive see pages 259 and 3007 so it decays exponentially7 and it has a half life of 1599 years If we start with 1 gram7 the amount of radium Rt left after t years is t1599 Ra lt61ngtt1599 67m2t1599 64115 Remember that any equation of exponential decay with base 6 has a negative exponent Lecture Notes 7 April 97 2008 Page 4 5 Example population growth Suppose a population of fruit ies is growing exponentially After 2 days7 there are 100 After 4 days7 there are 300 Let Ft 06 be the number of ies after t days Then F2 052k 100 and F4 054k 300 If we solve the rst equation for C 100 Using this in the second equation7 100 4k We 3007 so 64k E 37so e4k 2k 37so 2k ln37so 1 k fl 3 2 n This means that 100 100 100 6251113 51113 3 7 Ft 06kt ge th g Make sure that you can do Exercises 461 9 odd7 23 In the next lecture7 we will start doing something completely different Lecture Notes 7 62 2008 Page 1 1 The Midpoint and Trapezoidal Rules We want to approximate an integral abfx dx One way to look at this is approximating the area the integral represents Last time we approximated integrals by dividing the area into n rectangles each with width b 7 a n and height equal to the function evaluated in the center midpoint of the rectangle b 7a fltzgtdzb f1fx2fxnl n with 1 being the center of the rst rectangle and so on A more accurate way to approximate an area is to divide it into trapezoids This process is called the Trapezoidal Rule It is based on the fact that the trapezoid has area 1 The Trapezoidal Rule To approximate the de nite integral f dz 1 Divide the interval 11 into 71 smaller intervals with width b 7 a n 2 Call the endpoints of these intervals 1 0x12z3 sn b 3 The integral is approximately equal to b 7a 0 1 1 2 nil n fdxbi f f f f f f n 2 2 2 Lecture Notes 7 62 2008 Page 2 Notice that you can rewrite the formula as 1 bia 1 1 imam n f96of961f962f71 f For example to approximate 2 2 67m dx 0 on 4 intervals 1 Divide the interval 02 into 0 12 12 1 1 32 and 32 2 each with width 2 1 4 239 2 We have x0 0 1 12 2 1 x3 32 x4 2 3 The integral is approximately equal to 1 540 1 6412 1 12 1 6432 1 642 As with the Midpoint Rule the accuracy of the estimate increases with higher 71 2 Simpson s Rule With the Trapezoidal Rule f is approximated by a straight line a rst degree polynomial on each interval meal 12 etc An even more accurate method would be to use a second degree polynomial Lecture Notes 7 62 2008 Page 3 Given three points such as 07f0717f17 and 27f27 we can nd a second degree polynomial p012 sz Bs 0 Similarly we can nd a second degree polynomial p234 that passes through 27f2737f37 and 47f4 Then if the points a zox123x4 b divide the interval 1 b into 4 intervals we have the approximation abf 96 dz 10012 x dz Arlyn 95 d This is called Simpson7s Rule The details of nding the polynomials and integrating them are complicated but the nal result can be written in the fairly simple form 17 bia afdxm 3n with a zo x1z2x3 man b chosen just like in the Trapezoidal Rule Notice that n has to be even Wm 4f961 2f962 2f72 4f71 f96nl For example to approximate 2 2 67m dx 0 on 4 intervals 1 Divide the interval using x0 0 1 12 2 1 x3 32 x4 2 as before 2 The integral is approximately equal to for 454122 25W 457632 22 Lecture Notes 7 627 2008 Page 4 The Final Insert ominous music here The practice nal gives a pretty accurate idea of what may be covered on the Final Exam problems will be based directly on this list of problems Simpson7s Rule will not be included Note that some of these problems involve some complicated arithmetic for example7 5787 5647 and 652 I will do my best to avoid situations like this7 but some kinds of problems particularly those on continuous probability and approximate integration tend to involve lots of fractions As usual7 feel free to leave your nal answer unsimpli ed unless otherwise instructed in the problem7 for example 424 I have posted some handwritten solutions to these problems The solutions dont show much work 7 only about as much as l7d expect you to show on the exam 7 so you should mainly use them to check your own answers Remember that the Final is this Saturday7 June 77 from 1030 to 1230 in the usual room Make sure you bring an ID with a visible photo Lecture Notes 7 May 5 2008 Page 1 Note Starting immediately you will be penalized for the following notation mistakes 0 Using an integral sign instead of brackets For example writing 1 1 3 1 x2dx 770 0 0 3 3 o Failing to convert the limits of integration in a substitution For example 1 2x 11 1 AmdzO adulnlul lnlx210 1 Antiderivatives of trigonometric functions Remember that d i 7 sinz cos x so dx cosz dx sinz 0 Also d icosz isinz so dx sinzdz7cos0 We can use these facts and techniques of integration like substitution to nd many inde nite and de nite integrals Lecture Notes 7 May 57 2008 Page 2 2 Problems 1 Find z2 sin z3dz Solution Let u z37 so du 3z2dz and z2dz du3 Then 1 1 1 z2sinz3dz gsinudu igcosu0 igcosz3CC 2 Find sin2 4z cos 4z dz Solution Let u sin 4z7 so du 4cos 4z dz and 1 13 1 Zuzdu1u30 sin34z0 3 Find tanz dz Solution Remember that tanz sin z cos z7 so sinz tanz dz dz cosz Let u cos z7 so du isinz dz and 1 sm dz77du71n1u1071n1cosz10 cosz u Lecture Notes 7 May 57 2008 Page 3 4 Find sec tanx dx Solution Since secx 1cosz and tanx sin x cos 7 1 sin z sin z secx tanx dx dx 2 dz cos x cos c cos c Let u cos x so du isinx dx and 1 1 1 Sm2di72dui0 OSSCO cos c u u cosx 5 Find sinx dx 0 Solution sinxdxicosxg7cos7rcos0112 0 6 Find 7r2 cot dx 7r3 Solution Since cot z cos x sin x 7r2 7r2 cot dx Cf dm Wg Wg Slni Let u sin x so du cos dz u7r3 27 and u7r2 1 Then 7r2 1 1 3 CSdz idu 1n1uH1 2 7 1n Wg sinx Z u 2 Lecture Notes 7 May 5 2008 Page 4 3 Other trigonometric integrals You might remember that d d 7 i i 7 1 7 1 2 itanx77sinxcos s icosxcos 7SIHCOS 7S1n dx dx sin2 z cos2 x sin2 z 1 2 7 sec s cos2 x cos x cos x cos2 x and similarly 1 d d 1 71 2 Ecotxcosxsin s icsc m Therefore sec2dtanCandcsczzdz7cotx0 Also secx tan seczz secztanz secx dx secx dx dz secztanz secztanz and if we choose u secx tanx then du secxtanx sec2 s dx and the integral becomes 1 7du ln C ln lsecx tanl C u By a similar process cscxdx lnlcscsicotzl 0 The four integrals above and the process we used to nd them are interesting but I wouldnt expect you to do something like this on an exam nor to memorize these integrals 4 What do you need to know in this course What you need to know You need to know meaning memorize all of the basic properties of trigonometric functions These include the de nitions77 listed on page 558 the derivative and integral of sinx and cos s and the fact that sin2 coszx 1 for any x You must also be able to apply the techniques of integration that we have learned and those we will learn in future lectures to integrals containing trigonometric functions

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