### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Short Calculus MAT 016B

UCD

GPA 3.88

### View Full Document

## 60

## 0

## Popular in Course

## Popular in Mathematics (M)

This 22 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 60 views. For similar materials see /class/187396/mat-016b-university-of-california-davis in Mathematics (M) at University of California - Davis.

## Reviews for Short Calculus

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/08/15

16B Midterm 2 Review Note This document has not been checked for mistakes Let me know if you nd any Midterm two will cover all aspects of integration focusing on Chapter 5 6163 and 85 You should know all of the basic integrals in Table l and know all the techniques of integration welve learned so far substitution integration by parts and partial fractions are the major techniques From the basic integrals you can gure out any others you want For example you can gure out the integral f tanz dz using what you know about the integrals of sinz and coszi How See Example 1 Other than how to integrate what do you need to know for the test Plenty of things Anything covered in the chapters or in class is fair game although it would be wiser to concentrate on what was covered in class rather than some obscure ap plication that was never talked about Some big theoremsapplicationstechniques include o The Fundamental Theorem of Calculus 0 Average Value of a Function 0 Area Bounded by Curves o The Midpoint Rule 0 Volume of Solids of Revolution llve done one example each of integration by substitution integration by parts and integration using partial fractions Examples 23 and 4 These examples are simple examples meant to show the basic steps to each method you should note that each method has tricks and twists that can come up in some problems and make them more complicated These tricks are best learned by practice You7ll need some of them to do all the practice midterm problems You can be sure you7ll also need some of them to do all of the real midterm problems What else can you do to study Go through your class notes come see me at of ce hours make an appointment to see me another time if you can t make of ce hours read the book ask a friend go to the calculus lab go to the LSC get a tutor get help online make a study group make a review sheeti But most of all practice Table 1 d In Of 1 z zin1 orna edzeEC 1 1 1 dz idzlnlzlC z coszdzsinzC sinz dz 7 cosz C 1 sec2z dz tanz c secz tanz dz secz c csc2 z dz 7 cotz c cscz cotz dz 7 cscz 0 Example 1 This example computes the integral of tanz using only basic inte grals that we have memorized Let7s walk through the process from the point of View of a master integrator one who makes mistakes along the way that you might make as well and then gures out how to x them tanz dz This isn7t one of the basic integrals l memorized So there must be some way to simplify it I can write tanz as sinz cosz so let s try that wwW mw That7s a little better Now how could 1 attack this problem Its still not one of the basic integrals that I know Perhaps a substitution Substitution is often the easiest way to do an integral so I might as well try that rst In fact l7m pretty sure it will work because I know that cosz is the derivative of sinzi If you notice that one part is the derivative of the other part that7s a good hint that substitution will work What should I substitute Let7s try u sinzi Then in substitution we have to do the following steps First we pick a u which we already did Second we take the derivative 75 and solve for dz In our case we get 7 coszi Then we solve for dz by multiplying and dividing both sides appropriately to get 1 dz The third step is to plug the new things back in cosz 7 put in u where it belongs and put in whatever dz equals mw Adzg Hm this is not quite what we wanted When we do substitution we like all the z terms to cancel lt isn7t necessary that they all cancel as you7ve seen in class and will see in the practice problems below but this is a little more complicated than we like it to be We could keep going maybe use the fact that cos2z 1 7 sin2z but instead lets try another substitution and see how it works out Let7s try u cosz instead of sinzi Then our next step is to calculate 75 For this we get 7 7sinzi Therefore solving for dz we get 3105 dz At this point let s plug everything in WW 3118 tram 41 And look this time the leftover z terms canceledl How perfect Notice that we now have a negative sign out front because we had a 7 sinz in the denominator And now we7ve reduced the problem to one of the basic integrals that we have memorized and so we can say after we plug in the u integral and substitute back in what u is that tanzdz7ldz7lnlulC7lnlcoszlC u And so we gured out the integral of tanz using only substitution and the basic integrals we know Example 2 Substitution For me substitution is broken down into four steps 1 Pick something to substitute for ie pick u 2 Calculate g7 Solve for dz 3 Plug in u and whatever dz equals 4 If all the z terms cancel proceed with the integral in terms of u and after integrating substitute back in what u is If all the z terms donlt cancel you might have to do some trickery to get rid of them or else pick something else to substitute or use another method Lets do a simple example to illustrate all of these steps zl 7z2dz 7 1 We rst need to pick u Let7s pick u l 7 z2 2 Next we have to calculate 37 If u l 7 z2 then 37 72z Solve for dz by multiplying by dz on both sides and dividing by 72z on both sides Then we get 17 dz 3 Plugging in u l 7 z2 and dz fig we get du 7l 7 2 7 7 zl z dz zu721 2 udu Notice that we were able to cancel the leftover z7s after we substituted in We also gained a constant term of 7 4 Finally we can proceed with the integral in terms of u f u du is one of the basic formulas that we have memorized so we can just say f udu 72 C The last thing is to just substitute back in what u is If the problem started with z7s we want it to end with z7s So the nal answer is 7 7 22 zl7z2dz71c Example 3 Integration by Parts Similarly we can break down integration by parts into four steps 1 Pick something for u and something for dv 2 Calculate du and 1 ie nd the derivative of u and antidifferentiate dv 3 Plug into the formula fudv uv 7 fvdu 4 Hopefully the new integral fvdu is simpler than the original Calculate fvdu either it is a basic integral you know or use substitution or lBP again etc If the second integral is not simpler you may need to pick different u and dv or try another method Tip Remember LATE Logarithm Algebraic Trigonometric Exponential When doing integration by parts the choices for u and dv will most likely be of one of these four types Pick u to be whichever type comes first For example if you have fzgehdz you have an algebraic zg and an exponential e21 func tion Since A comes before E in LATE pick u z3 the algebraic function Or if you have f5z21nzdz you have an algebraic and a logarithmic function so pick u lnz the logarithmic function Lets do a simple examp e z2 ln5z dz 7 1 First we need to pick u and dv By using LATE we pick u ln5z and dv z2 2 Next we calculate du and 1 We get if chain rule and M2 3 Plug in to the lBP formula 3 3 3 2 J Li J 1 2 z ln5z dz 7 3 ln5z 3 Idz 7 3 ln5z 3 z dz 4 Indeed we get a simpler integral so we can go ahead and evaluate it since it s one of the basic integrals we have memorized z3 1 2 z3 lz3 Eln5z7 dz 7 1n5z7 c z3 z3 z2 ln5z dz E ln5z 7 3 C Example 4 Integration by Partial Fractions Partial fractions is useful when you have a proper rational function For example is a proper rational function The idea behind partial fractions is that in most proper rational functions like the example given above the denominator is complicated We like simple denominators since theylre much easier to work with So partial fractions gives us a way of rewriting the function as a sum of things that all have simple denominators The steps to partial fractions are below 1 Factor the denominator 2 Set up the equation write the proper rational function on one side and on the other side write a sum For every factor in the denominator write an unknown like A B C above that factor as a term in the sum If a factor occurs more than once then write a term in the sum for every power of that factor This is hard to write but examples make it clear what to do 3 Now we want to solve for all the unknowns A B C To do this first clear 39 t by t39 39 of the proper rational function 4 Solve for the unknowns by plugging in various values of z Pick smart values for z usually ones that make most of the terms become 0 At some 5 point you may need to plug in unknowns that you have already solved for to nd unknowns that you havenlt gotten yet 5 Compute the integral using the new form of the proper rational function And now a simple example 31 5 d 7 13 7 31 2 I We can t do this using any of the methods we currently have on hand because the denominator is too complicated So let s use partial fractions to simplify it 1 The denominator 13 7 31 2 factors as I 7 l2z 2 This might take a little bit of work to nd you have to remember some precalculus most likely One trick is to try a few values for I If you plug in z a and get 0 then you know that z 7 a is a factor That can get you started at least 2 We rewrite 31 5 7 31 5 7 A B C 1373127z712z27171 z7l2 12 3 If we clear denominators we get 3I9AI2I71Bz2CI712 4 Then plugging in z l we get 12 3B gt B 4 Plugging in z 72 we get 3 9C gt C Finally plugging in B C and z 0 we get 1 71 9 72A 8 7 A 7 3 gt 3 Therefore we have 31 5 71 4 l 7 7 7 137312 3171 z7l2 3z2 If we wanted to doublecheck our answer we could put everything over a common denominator of z 7 l2z 2 and check that they match 31 7 71 z2z7lJr 4 z2Jr 1 I712 137312 7 3z7lz2z7l 1712z2 3z2z712 712z72 4z8 1272z1 137312 137312 137312 7 31 137312 You may want to do a few more steps in between to really convince your sel 5 Finally we compute the integral remember that s why we were doing all this in the rst placel 3x 5 d 71 4 1 d I 7 7 7 z 137312 3171 I702 3z2 ln 1 7 1 4 ln 1 2 7 l M l MC 3 7 3 I didn t show any work for the integrals but you can do each of them using substitution You should check my work to make sure its correct And there you have some basic examples of each of the major integration tech niques we7ve learned I want to reiterate that there are subtleties to each one that did not come up in the examples little tricks of the trade that you also need to knowi However it s best to start by really knowing how to do basic examples of each one rst De nite Integrals De nite integrals are sometimes just given to you eg com pute 12 dz and sometimes they arise when computing area or volume eigi compute the area under the curve 12 between 3 and 7 Usually when you are computing area or volume you can use the Fundamental Theorem of Calculus but you also should remember there are other ways to compute area eg evenodd functions or geometrically 7 what is the area under the curve V 4 7 12 from 72 to 29 My advice for de nite integrals is to rst compute the antiderivative without worrying about limits of integration and once you have the antiderivative in terms of the original variable remember that you have limits of integration and evaluate We also went over how to change variables along the way as you substitute which is a perfectly good technique if you choose to use it and can save time but can perhaps lead to more mistakes i Problems Okay it7s nally time for some problems These are the types of prob lems that may appear on the midtermi My lawyerspeak caveat 7 the midterm need not be restricted to the types of problems that you see here different things could appear harder things could appear 7 anything in the book or from class is fair game In addition you shouldnlt consider just doing the problems here enough practicei Its not You should seek out more problems like these from the book where most of these are from and keep practicing especially the types of prob lems that you have the most trouble with If you need more problems but can t nd any in the book ask me About the solutions There are too many problems for me to write detailed solutions for all of themi If you want more detail on a particular solution send me an email lld be happy to provide it Again the solutions are not checked for errors I found several errors in the statements of the problems so its reasonable to expect there will be several errors in the solutions Let me know if you nd any or if anything doesn7t seem to make sense to your 1 Compute 36 37 dzi A A g V 7 Solution Let u 731 then 71 7feudu7euC7e 3xC dz so fSE Sdz Seui g Compute ft9sect9tant9dt9 Solution Might be more than one good way to do this one One way would be to use integration by parts with u t9 and d1 sec 9tan 9 Then du 1 and v sec 9 When you plug into the integration by parts formula you end up having to integrate sec 9 Maybe you want to memorize that one since computing it depends on a trick that is not very obvious its in 85 in your book if you want to see it Find the area of the region bounded by the graphs of y sec 1 z y 0 and z 2 7r 17 Solution There were a few typos in this problem which have now been xed One typo y 2 instead of z 2 didn t make sense since the region was not well de ned The other typo was instead of E This would lead to problems using the Fundamental Theorem of Calculus since it would cause you to have to compute tan7r2 which is unde ned secz forms the top of the region the other lines form the sides and the bottom The integral to calculate the region can be set up as 734 secz dz Solving the integral gives ln l sec 1 tan zl li4 lnsec 2 tan 7 lnsec E tan E lnsec2 tan 7 ln 71 Evaluate lmy dy Solution This is similar the same to one we did in class In class we did it two different ways and got two different answers but there wasn7t time to go and check Why 1711 do it two different ways here more carefully and the answers should match up I will try to remember the two ways we did it in class For both ways we rst want to nd the antiderivative and then worry about the limits of integration a Use integration by parts u lny and d1 dy Then du i and v gySQ Therefore we get f lny dy lny y32 7 f 3 y lny y327fy12 gyS21ny7y32 we would add C if we were only computing the antiderivative but we7re actually doing a de nite integral so it doesn7t matter Then plug in the limits of integration and you get 2 ln2 7 32 7 1nl 7 2 1n2 7 gaff 3 Use substitution rst Lets say we thought to try substitution rst Let7s substitute u Then 37 y lQ and so 2y12du dz Plug this in to get f lny dy fulanQudu quanu2 du 4fu21nudu Make sure you understand what happened in each step Then we can apply integration by parts to this integral We already used the variable u so welll use w and v to be our inte 2 Then dw l gration by parts pieces Let w lnu and d1 u u A 0quot V 5 6 8 2 and v Plug this into the 1B formula to get 4fu2lnudu 4ln 1117 7 4ln 1117 7 Transform back into the original variable before applying the limits of integration when we do this we get lmy6 7 4E lny 7 ySQ If you move the constants around in the right way you see that this is the same anti derivative as we got above so plugging in the limits of integration will give the same answer Compute 61 7 6 2 dz Solution Expand rst and then integrate em7eim2dz 621dz72dze 21dz 62a 6721 7 2 2 0 Let R be the region bounded by the graphs 1 3 7 y2 and y z 71 Find the area of the region bounded by the two functions Solution Another typo 7 I accidentally combined two problems here There were other equations for which you were to nd the volume of a solid of revolution For these equations all I wanted was area you could calculate the volume of the solid but it would be hard You may have to graph the equations to get the region right The rst thing to do is nd the points of intersection To do this you could say y 1 z and then set the two equations equal to each other 3 7 y2 y l Solve this equation to get 71 72 and 2 l as the points of intersection Now you can either set it up as an integral with respect to y or two integrals with respect to I With respect to y 7123 7 y2 7 y l dy With respect to z f31z71773 7 z dzf23 v3 7 177x3 7 1 dz For the y integral we did the right function minus the left function For the z integral we did top function minus the bottom function The I in tegral is more complicated because which function is on top changes and also because we have to split y ixS 7 1 into two functions You should get the same answer in each case Compute Solution Split this in two f it from there 1 dz f 1 dz7 f 77 dz You can take Evaluate prim dz 10 11 12 13 9 Solution f 1ln dz fidz x x f ngcdz The rst one is easy the second one you can do via substitu tion u lnz should work Then plug in the limits of integration once you7ve plugged back in the original variable 1 Compute the antiderivative rst Find the average value of the function 56021710 over the interval 010 Solution Recall the formula for average value look it up if you don7t remember This leads to the integral i fem 56021 10 dz Compute the antiderivative rst then plug in the limits To compute the antiderivative it may help to use the fact that 5602E 10 5603921 2 26 Then you can factor out the constant part and just integrate e 021 State the Fundamental Theorem of Calculus Tell about one application of the FTOC Solution Look in your book The FTOC is commonly used to calculate the area under a curve Using the various formulas you now know it can help to calculate volume average value and other things Compute f 65 quot 7 cos 1 dz Solution Use substitution Notice that both sinz and its derivative cosz are in there This is a good hint that you should try u sinz ou might also have said Well sinz is the derivative of cos I so why not use u cos I It would be perfectly valid to try this substitution but it wouldnlt work out as nicely Find the area of the region bounded by the graphs I 7 13 and 91 I i 1 Solution Remember to calculate area between two curves you want to use top 7 bottom For this problem we have to break it up into two integrals since the curve that is on top changes draw the graphs to see what I mean We also have to nd the intersection points The nal setup should look like 1 2 Area 17137 zildz 1707 1713dz 0 1 1711 let you evaluate the integrals A deposit of 2250 is made in a savings account at an annual interest rate of 127 compounded continuously Find the average balance in the account during the rst 5 years Solution Recall the formula for continuously compounded interest P Cequot In this case that becomes P 225063912 The average value over the rst ve years then is if 225063912 You should be able to calculate this 14 15 16 17 18 integral email me if you need help Help has been requested To compute the integral factor out the con stant 2250 Then substitute u 12t and 12 or 71 dt to get f e 612 Now we can plug in the limits of integration and remember to divide by 5 to get e53912 7 1 308295 as the average balance for the rst ve years Find the area of the region bounded by the graphs of y sin z y tan z and z 1 Solution This one might be a little hard to see but in fact sinz is less than tanz when z is between 0 and 1 I ve had a few questions as to why this is true In fact its because tanz Sim and between 0 and 1 cosz is always less than 1 If you divide something like sin z by something less than 1 you get something bigger than the original like 5 divided by 05 is bigger than 5 On the test there probably won7t be any thing this hard to graph Therefore the integral to calculate this area is f0 tanz 7sinz dz The calculation for f tanz dz is done in Example 1 and the integral fsinz dz is one you should know Compute 7 23 dz Solution Multiply it out rst and then the integral will be easy Alter natively you can do a substitution with u z 7 2 which means you can also plug in z u 2 Evaluate f z2xz 7 2 dz Solution It s usually nice to have only one thing in a square root rather than a sum of things So let u z 7 2 then we only have one thing in the square root We have 710 1 and z u 2 Plugging all this in we get fz2xz 7 2 dz 22Hdu Multiply out the u 22 and then break it into several integrals It should be easy from there 8 18 dz Compute f Solution Notice that the derivative of the bottom equals the top There fore let u e 1 so that 7 e and 71 dz Therefore f 1 dz f fidulnlulClnlex1lC The outdoor temperature Tt at t hours after midnight is given by Tt 70 15 sin 23 What is the temperature at 10 am What is the average temperature between 10 am and 2 pm Solution To nd the temperature at 10 am simply plug in t 10 T10 70 15 sini772r 70 156 775 degrees To nd average 11 temperature use the formula for average value of a function Notice that 2 pm would correspond to t 14 in this case 1 14 78 1 1512 t78 Avg temp 110 70 15 sin dt 1WD W7 cos The symbol on the end is supposed to be the long vertical line meaning we plug in 14 and 10 as the limits of integration 1 can7t gure out how to typeset that right now and I want to post this as soon as possible If you plug in the limits and evaluate everything you should get 824 degrees as the average temperature assuming I didn t mess up anywhere while plug ging things in 19 Evaluate fil zxl 7 12 dz Solution Another typo this one was missing an 1 without the I one could still solve it but you guys havenlt learned how Its been xed now A substitution of u l 7 12 should do the trick Then 27 721 Substi tute this information back in to get Ixl 7 12 dz f 77127 du 7711 7 I232 Plug in the limits of integration and you get 0 A trick to avoid all this notice that the function is odd 20 Evaluate 4gigf dz Solution This problem was readymade for partial fractions We want to solve 4312275755 3 EELE Clear denominators and get 4127175 Azz 5 Bz5 C12 Plug in z 0 and z 75 to solve for B and C and then choose any other value of z to solve for A The answers you should get are A 0 B 71 and C 4 Therefore we can write 4127175dzi 1 4 dz 12z5 7 12 15 This integral you can solve I get i 4ln lz 5 C 21 Compute fidz x71 Solution A substitution of u z 7 1 would work nicely here 22 Use the Midpoint Rule with n 3 to approximate the area of the region between and the zaxis from I 0 to z 3 Solution The width of each rectangle would be 1 and the midpoints would be I z g and z The heights of the rectangles would be 12 432 and 452 so the total area is approximately 412 432 452 m 351 if my calculations are correct On a test the square root form is acceptable 23 Evaluate fog sinz 5 dz Solution Split it into two Final answer will be 7cos7r2 5277 7 7cos00 57WL 24 Compute fcch gdz Solution Substitute u g and then use an integral you have memo rize 25 Compute fzcoszdz Solution 1 think integration by parts would work nicely here Use LATE to gure out what u should be and what d1 should be 26 Evalulate fol tanz 7 1 dz Solution First a substitution like u z 7 1 would be in order Then the integral ftanudu is computed in Example 1 and turns out to be f tanudu 7 ln l cos ul C Plug back in the original variable and limits of integration to get 1 tanz 71dz 7lnlcos1 7 ll 7 7lnlcos07 ll 71 lncos7l 0 27 Find the volume of the solid generated by revolving the region bounded by the following functions around the zaxis y ln 1 y 1 z 1 Solution In my rst solution posted earlier I found the area of the re gion not the volume of the solid of revolution Here is the solution for the volume of revolution solution for area is below The integral you have to calculate for the volume is pretty tough I wouldn7t worry too much about knowing it for the test To nd the volume The intersection points are the same as for the area see below The integral we should set up is 7r127lnz2dz7rldz77rlnz2dz 1 1 1 The rst de nite integral is easy nal answer should be e 7 1 but the second is tough too tough for the midterm The trick is to use integration by parts with u lnz and d1 ln 1 Then we get du l and v 1 ln 171 see your book for how to integrate lnz using lBP Therefore plugging into the integration by parts formula we get lnz2 dz lnzzlnz7z7zlnz7z dz lnzzlnz7z7lnz71 dz 28 29 30 Again check your book for how to integrate flnzdz The integral above equa s lnzzlnz 7 z 7 zlnz 7 z z zlnz2 7 2zlnz2z Finally we can plug in the limits of integration and get 8 lnz2dz 672626 7 2 672 1 Don7t forget the 671 from the rst integral Therefore the volume is 2673 To nd the area y 1 forms the top of this region and lnz the bottom The left side is z 1 Therefore we need to integrate f1 7 lnz dz But what are the limits of integration We already saw that the left side is z 1 so the lower limit is 1 The upper limit is the point where y 1 and y lnz intersect lnz 1 means that z 6 So the area of the region is ff 1 7 lnz dz This can be split into two integrals We7ve seen in class and in the book how to integrate lnz using integration by parts so 1711 let you nish the problem Find the volume of the solid generated by revolving the region bounded by the following functions around the zaxis y 1 sin z y 1 cos z z I z 7r Solution This is the trouble with trying to make up problems Some times they end up being too tough When I wrote it I didn t realize the solution to this problem would require some trig identities that aren7t really part of this class But we can at least set up the integral 1sinz 1cosz when z E 1 sinz is the top function for g S z S 7r So the volume of the solid of revolution is given by 7r 1sinz27 1cosz2dz 2 You end up needing to integrate sin z and cos2 z which are beyond the material in this class A soup bowl can be modeled as the solid formed by revolving the graph of 1 for 0 S z S 3 around the zaxis How much soup can t into the bowl Solution Set it up 7r 03 12 dz Then just multiply it out and it should break into three easy to do integrals Evaluate z 71e dz Solution Find the antiderivative rst Probably easiest to multiply out rst re dz can be solved using integration by parts The other part is straightforward Then plug in the limits of integration 31 Compute f212 21 dz Solution Integration by parts might work well here 4172 T71 dz 32 Compute f Solution Slight typo has been xed Try u z 7 1 33 What is an odd function Compute the de nite integral 55 sinzz2 dz Solution An odd function is a function such that The integral is 0 since it is in fact an odd function How do we know it s odd sin7z 712 sin7z12 7 sinzz2 34 Compute ftan2z dz Solution One could use trig identities to rst simplify this sin2z 7 17cos2z 7 1 cos2 tan2 sec2z 7 1 cos2 7 cos2z cos2z 7 cos2z some of you may just remember the identity tan2z 1 sec2 and skip the middle steps Then the resulting integral is easy to do since f sec2z dz is one you have memorized There might be a good way to do this without trig identities but 1 havenlt tried to nd one 2 137412731i3 35 Evaluate 1 2731 dz Solution You need to divide the numerator by the denominator using 3 2 long division rst What 1 got was W73 z 7 1 g Then do partial fractions You should end up with g 71 Therefore we can rewrite the original integral as 2 3 7 2 7 2 7 z 41 313dz I7171 12 7 3x 1 1 PH leave it to you to calculate it from here there should be many similar examples in the boo Sgdz I 7 36 Compute 57dz Solution You could multiply out I 57 but 1 don7t recommend it unless you re very bored A better way might be to use the substitution u z 5 and solve from there 16B Midterm 2 Review Note This document has not been checked for mistakes Let me know if you nd any Midterm two will cover all aspects of integration focusing on Chapter 5 6163 and 85 You should know all of the basic integrals in Table l and know all the techniques of integration welve learned so far substitution integration by parts and partial fractions are the major techniques From the basic integrals you can gure out any others you want For example you can gure out the integral f tanz dz using what you know about the integrals of sinz and coszi How See Example 1 Other than how to integrate what do you need to know for the test Plenty of things Anything covered in the chapters or in class is fair game although it would be wiser to concentrate on what was covered in class rather than some obscure ap plication that was never talked about Some big theoremsapplicationstechniques include o The Fundamental Theorem of Calculus 0 Average Value of a Function 0 Area Bounded by Curves o The Midpoint Rule 0 Volume of Solids of Revolution llve done one example each of integration by substitution integration by parts and integration using partial fractions Examples 23 and 4 These examples are simple examples meant to show the basic steps to each method you should note that each method has tricks and twists that can come up in some problems and make them more complicated These tricks are best learned by practice You7ll need some of them to do all the practice midterm problems You can be sure you7ll also need some of them to do all of the real midterm problems What else can you do to study Go through your class notes come see me at of ce hours make an appointment to see me another time if you can t make of ce hours read the book ask a friend go to the calculus lab go to the LSC get a tutor get help online make a study group make a review sheeti But most of all practice Table 1 d In Of 1 z zin1 orna edzeEC 1 1 1 dz idzlnlzlC z coszdzsinzC sinz dz 7 cosz C 1 sec2z dz tanz c secz tanz dz secz c csc2 z dz 7 cotz c cscz cotz dz 7 cscz 0 Example 1 This example computes the integral of tanz using only basic inte grals that we have memorized Let7s walk through the process from the point of View of a master integrator one who makes mistakes along the way that you might make as well and then gures out how to x them tanz dz This isn7t one of the basic integrals l memorized So there must be some way to simplify it I can write tanz as sinz cosz so let s try that wwW mw That7s a little better Now how could 1 attack this problem Its still not one of the basic integrals that I know Perhaps a substitution Substitution is often the easiest way to do an integral so I might as well try that rst In fact l7m pretty sure it will work because I know that cosz is the derivative of sinzi If you notice that one part is the derivative of the other part that7s a good hint that substitution will work What should I substitute Let7s try u sinzi Then in substitution we have to do the following steps First we pick a u which we already did Second we take the derivative 75 and solve for dz In our case we get 7 coszi Then we solve for dz by multiplying and dividing both sides appropriately to get 1 dz The third step is to plug the new things back in cosz 7 put in u where it belongs and put in whatever dz equals mw Adzg Hm this is not quite what we wanted When we do substitution we like all the z terms to cancel lt isn7t necessary that they all cancel as you7ve seen in class and will see in the practice problems below but this is a little more complicated than we like it to be We could keep going maybe use the fact that cos2z 1 7 sin2z but instead lets try another substitution and see how it works out Let7s try u cosz instead of sinzi Then our next step is to calculate 75 For this we get 7 7sinzi Therefore solving for dz we get 3105 dz At this point let s plug everything in WW 3118 tram 41 And look this time the leftover z terms canceledl How perfect Notice that we now have a negative sign out front because we had a 7 sinz in the denominator And now we7ve reduced the problem to one of the basic integrals that we have memorized and so we can say after we plug in the u integral and substitute back in what u is that tanzdz7ldz7lnlulC7lnlcoszlC u And so we gured out the integral of tanz using only substitution and the basic integrals we know Example 2 Substitution For me substitution is broken down into four steps 1 Pick something to substitute for ie pick u 2 Calculate g7 Solve for dz 3 Plug in u and whatever dz equals 4 If all the z terms cancel proceed with the integral in terms of u and after integrating substitute back in what u is If all the z terms donlt cancel you might have to do some trickery to get rid of them or else pick something else to substitute or use another method Lets do a simple example to illustrate all of these steps zl 7z2dz 7 1 We rst need to pick u Let7s pick u l 7 z2 2 Next we have to calculate 37 If u l 7 z2 then 37 72z Solve for dz by multiplying by dz on both sides and dividing by 72z on both sides Then we get 17 dz 3 Plugging in u l 7 z2 and dz fig we get du 7l 7 2 7 7 zl z dz zu721 2 udu Notice that we were able to cancel the leftover z7s after we substituted in We also gained a constant term of 7 4 Finally we can proceed with the integral in terms of u f u du is one of the basic formulas that we have memorized so we can just say f udu 72 C The last thing is to just substitute back in what u is If the problem started with z7s we want it to end with z7s So the nal answer is 7 7 22 zl7z2dz71c Example 3 Integration by Parts Similarly we can break down integration by parts into four steps 1 Pick something for u and something for dv 2 Calculate du and 1 ie nd the derivative of u and antidifferentiate dv 3 Plug into the formula fudv uv 7 fvdu 4 Hopefully the new integral fvdu is simpler than the original Calculate fvdu either it is a basic integral you know or use substitution or lBP again etc If the second integral is not simpler you may need to pick different u and dv or try another method Tip Remember LATE Logarithm Algebraic Trigonometric Exponential When doing integration by parts the choices for u and dv will most likely be of one of these four types Pick u to be whichever type comes first For example if you have fzgehdz you have an algebraic zg and an exponential e21 func tion Since A comes before E in LATE pick u z3 the algebraic function Or if you have f5z21nzdz you have an algebraic and a logarithmic function so pick u lnz the logarithmic function Lets do a simple examp e z2 ln5z dz 7 1 First we need to pick u and dv By using LATE we pick u ln5z and dv z2 2 Next we calculate du and 1 We get if chain rule and M2 3 Plug in to the lBP formula 3 3 3 2 J Li J 1 2 z ln5z dz 7 3 ln5z 3 Idz 7 3 ln5z 3 z dz 4 Indeed we get a simpler integral so we can go ahead and evaluate it since it s one of the basic integrals we have memorized z3 1 2 z3 lz3 Eln5z7 dz 7 1n5z7 c z3 z3 z2 ln5z dz E ln5z 7 3 C Example 4 Integration by Partial Fractions Partial fractions is useful when you have a proper rational function For example is a proper rational function The idea behind partial fractions is that in most proper rational functions like the example given above the denominator is complicated We like simple denominators since theylre much easier to work with So partial fractions gives us a way of rewriting the function as a sum of things that all have simple denominators The steps to partial fractions are below 1 Factor the denominator 2 Set up the equation write the proper rational function on one side and on the other side write a sum For every factor in the denominator write an unknown like A B C above that factor as a term in the sum If a factor occurs more than once then write a term in the sum for every power of that factor This is hard to write but examples make it clear what to do 3 Now we want to solve for all the unknowns A B C To do this first clear 39 t by t39 39 of the proper rational function 4 Solve for the unknowns by plugging in various values of z Pick smart values for z usually ones that make most of the terms become 0 At some 5 point you may need to plug in unknowns that you have already solved for to nd unknowns that you havenlt gotten yet 5 Compute the integral using the new form of the proper rational function And now a simple example 31 5 d 7 13 7 31 2 I We can t do this using any of the methods we currently have on hand because the denominator is too complicated So let s use partial fractions to simplify it 1 The denominator 13 7 31 2 factors as I 7 l2z 2 This might take a little bit of work to nd you have to remember some precalculus most likely One trick is to try a few values for I If you plug in z a and get 0 then you know that z 7 a is a factor That can get you started at least 2 We rewrite 31 5 7 31 5 7 A B C 1373127z712z27171 z7l2 12 3 If we clear denominators we get 3I9AI2I71Bz2CI712 4 Then plugging in z l we get 12 3B gt B 4 Plugging in z 72 we get 3 9C gt C Finally plugging in B C and z 0 we get 1 71 9 72A 8 7 A 7 3 gt 3 Therefore we have 31 5 71 4 l 7 7 7 137312 3171 z7l2 3z2 If we wanted to doublecheck our answer we could put everything over a common denominator of z 7 l2z 2 and check that they match 31 7 71 z2z7lJr 4 z2Jr 1 I712 137312 7 3z7lz2z7l 1712z2 3z2z712 712z72 4z8 1272z1 137312 137312 137312 7 31 137312 You may want to do a few more steps in between to really convince your sel 5 Finally we compute the integral remember that s why we were doing all this in the rst placel 3x 5 d 71 4 1 d I 7 7 7 z 137312 3171 I702 3z2 ln 1 7 1 4 ln 1 2 7 l M l MC 3 7 3 I didn t show any work for the integrals but you can do each of them using substitution You should check my work to make sure its correct And there you have some basic examples of each of the major integration tech niques we7ve learned I want to reiterate that there are subtleties to each one that did not come up in the examples little tricks of the trade that you also need to knowi However it s best to start by really knowing how to do basic examples of each one rst De nite Integrals De nite integrals are sometimes just given to you eg com pute 12 dz and sometimes they arise when computing area or volume eigi compute the area under the curve 12 between 3 and 7 Usually when you are computing area or volume you can use the Fundamental Theorem of Calculus but you also should remember there are other ways to compute area eg evenodd functions or geometrically 7 what is the area under the curve V 4 7 12 from 72 to 29 My advice for de nite integrals is to rst compute the antiderivative without worrying about limits of integration and once you have the antiderivative in terms of the original variable remember that you have limits of integration and evaluate We also went over how to change variables along the way as you substitute which is a perfectly good technique if you choose to use it and can save time but can perhaps lead to more mistakes i Problems Okay it7s nally time for some problems These are the types of prob lems that may appear on the midtermi My lawyerspeak caveat 7 the midterm need not be restricted to the types of problems that you see here different things could appear harder things could appear 7 anything in the book or from class is fair game In addition you shouldnlt consider just doing the problems here enough practicei Its not You should seek out more problems like these from the book where most of these are from and keep practicing especially the types of prob lems that you have the most trouble with If you need more problems but can t nd any in the book ask me About the solutions There are too many problems for me to write detailed solutions for all of themi If you want more detail on a particular solution send me an email lld be happy to provide it Again the solutions are not checked for errors I found several errors in the statements of the problems so its reasonable to expect there will be several errors in the solutions Let me know if you nd any or if anything doesn7t seem to make sense to your 1 Compute 36 37 dzi 2 Compute ft9sect9tant9dt l 3 Find the area of the region bounded by the graphs of y sec 1 z E y 0 and z 2i 4 Evaluate 1ny dy 5 Compute fe 7 eix2 d1 6 Let R be the region bounded by the graphs 1 3 7 y2 and y z 71 Find the area of the region bounded by the two functions 7 Compute f 1 31 dzi 2 8 Evaluate 1451quot dzi ac 9 Find the average value of the function 56021 10 over the interval 010 10 State the Fundamental Theorem of Calculusi Tell about one application of the FTOCi 11 Compute fem coszdzi 12 Find the area of the region bounded by the graphs z 7 13 and g z z 7 1 13 A deposit of 2250 is made in a savings account at an annual interest rate of 127 compounded continuouslyi Find the average balance in the account during the rst 5 years 14 Find the area of the region bounded by the graphs of y sin I y tan 1 and z 1 15 Compute 7 23 dzi 16 Evaluate 23 z2xz 7 2 dzi 17 Compute f 1 dzi 18 The outdoor temperature Tt at t hours after midnight is given by Tt 70 15 sin Whig What is the temperature at 10 am What is the average temperature between 10 am and 2 pmi 19 Evaluate filzx17 12 dzi 20 Evaluate 43121 Tgfdzi 21 Compute fidzi 71 22 Use the Midpoint Rule with n 3 to approximate the area of the region between 5 and the zaxis7 from I 0 to z 3 23 Evaluate fog sinz 5 dzi 24 Compute fcch gdzi 25 Compute fzcoszdzi 26 Evalulate fol tanz 7 1 dzi 27 Find the volume of the solid generated by revolving the region bounded by the following functions around the zaxis y ln 1 y 17 z 1 28 Find the volume of the solid generated by revolving the region bounded by the following functions around the zaxis y 1 sin I y 1 cos I z I z TL 29 A soup bowl can be modeled as the solid formed by revolving the graph of 54r17 for 0 S I S 37 around the zaxis How much soup can fit into the bowl 30 Evaluate z 7 1e dzi 31 Compute 212621dzi 32 Compute f 4912 dzi 33 What is an odd function Compute the definite integral 55 sinzz2 dzi 34 Compute ftan2z dzi 35 Evaluate deL 36 Compute 57dzi

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the material...plus I made $280 on my first study guide!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.