4,¸I]^_`abcdefghijklmnopqrstuvwxyz{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ ¡¢£¤¥B™é ¬kBhc;4Basic Integration Formulas and the Substitution Rule kBhc:3The second fundamental theorem of integral calculus?kBhcXQRecall from the last lecture the second fundamental theorem of integral calculus.kBhcTheoremkBhc  Let kBhc f(x)kBhc1* be a continuous function on the interval kBhc [a,b]kBhc . Let kBhc F(x)kBhc.' be any function with the property thatkBhcFopxEkBhcákBhc(x) = fGrapkBhcxkBhc ThenetnI\kBhcfGrapkBhcxkBhcxkBhcbkBhcakBhc = F(b) - F(a)kBhcAntiderivativeskBhc]VLet us now see how the second fundamental theorem gets applied to solve area problems.kBhcExamplekBhc ComputeetnIzcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc 20kBhc 10kBhc<5According to the theorem, we need to find a function kBhc F(x)kBhc with the propertykBhcFopxEkBhcákBhc (x) = 1/xopxEkBhc2kBhc70Such a function is called an anti-derivative of kBhc 1/xopxEkBhc2kBhc  = xopxEkBhc -2kBhc1*. We can see that one possible solution iskBhc F(x) = -xopxEkBhc -1kBhcFrom the theoremetnIzcarF<kBhc1kBhcxopxEkBhc2kBhcxkBhc 20kBhc 10kBhc = F(20) - F(10) = GrapFkBhc-GrapkBhc 20opxEkBhc -1kBhc  - GrapFkBhc-GrapkBhc 10opxEkBhc -1kBhc  = carF%kBhc1kBhc 20kBhcQJUnfortunately, this anti-differentiation process is not always easy to do.!etnIptrqS4kBhc 1 - xopxEkBhc2kBhcxkBhc1kBhc0"kBhc0)To solve this, we need an anti-derivative#kBhcFopxEkBhcákBhc (x) = trqS4kBhc 1 - xopxEkBhc2$kBhcæßThere is no obvious anti-derivative for this function. Given the fact that many anti-derivative problems are hard to do, we will be spending a good chunk of this course working out techniques for computing anti-derivatives.,kBhcAntiderivative notation-kBhcôíSince we are going to be computing antiderivatives, one of the first things we are going to need is a convenient notation for antidifferentiation. The notation that is commonly used for the operation of antidifferentiation is called the kBhcindefinite integralkBhc..etnI/kBhc f(x)kBhcxkBhc = F(x) › FopxEkBhcákBhc (x) = f(x)%kBhc%Antiderivatives we can compute)kBhcÏÈOne way to get a start on the problem of computing antiderivatives is to write down the derivatives of some standard functions and then adjust those formulas to make them antidifferentiation formulas.(lbaTQkBhc f(x)kBhcxopxEkBhcncarF$kBhc1kBhcxkBhceopxEkBhcxkBhc cos xkBhc sin xcarF@kBhc1kBhc 1 + xopxEkBhc2kBhcF(x) = etnI/kBhc f(x)kBhcxcarFDkBhcxopxEkBhc n + 1kBhc n + 1kBhc ln VsbAkBhcxkBhceopxEkBhcxkBhc sin xkBhc -cos xkBhc tanopxEkBhc -1kBhc  x*kBhcàÙ(There is a more extensive list of anti-differentiation formulas on page 406 of the text.) Another useful tool for computing antiderivatives is related to the fact that the derivative of a sum is a sum of derivatives:+etnI6kBhc f(x) + g(x)kBhcxkBhc  = etnI/kBhc f(x)kBhcxkBhc  + etnI/kBhc g(x)kBhcx/kBhc)"Likewise, since the derivative of kBhc k f(x)kBhc  is kBhc k f opxEkBhcákBhc (x)kBhc we have0etnI1kBhc k f(x)kBhcxkBhc  = k etnI/kBhc f(x)kBhcx1kBhc Some Examples2etnI§carFƒkBhc5 - 4 xopxEkBhc3kBhc  + 2 xopxEkBhc6kBhcxopxEkBhc6kBhcxkBhc  = etnI€kBhc 5 xopxEkBhc -6kBhc  - 4 xopxEkBhc -3kBhc  + 2kBhcx3kBhc = 5 etnIEkBhcxopxEkBhc -6kBhcxkBhc  - 4 etnIEkBhcxopxEkBhc -3kBhcxkBhc  + 2 etnI,kBhc1kBhcx4kBhc = 5 GrapJcarF>kBhcxopxEkBhc -5kBhc -5kBhc  - 4 GrapJcarF>kBhcxopxEkBhc -2kBhc -2kBhc  + 2 GrapkBhcx5kBhc = -xopxEkBhc -5kBhc  + 2 xopxEkBhc -2kBhc  + 2 x:etnI‡carFckBhcxopxEkBhc2kBhcxopxEkBhc2kBhc  + 1kBhcxkBhc  = etnIšcarFvkBhcxopxEkBhc2kBhc + 1 - 1kBhcxopxEkBhc2kBhc  + 1kBhcxkBhc  = etnIskBhc 1 - carF@kBhc1kBhc 1 + xopxEkBhc2kBhcx;kBhc = x - tanopxEkBhc -1kBhc  x<etnINcarF*kBhc1kBhc2 x + 3kBhcxkBhc  = carF$kBhc1kBhc2kBhc ln(2 x + 3)=kBhcžIf you want to check whether or not the answer to an antidifferentiation problem is correct, you can always differentiate the answer to see whether or not that returns the original function.> DtotkBhcxGrap\carF$kBhc1kBhc2kBhc  lnGrapkBhc2 x + 3kBhc  = carF$kBhc1kBhc2kBhc DtotkBhcxGrap3kBhc lnGrapkBhc2 x + 3kBhc  = carF$kBhc1kBhc2kBhc carF*kBhc2kBhc2 x + 3kBhc  = carF*kBhc1kBhc2 x + 3kBhc' A method based on the chain rulekBhc¦ŸSince integration is the inverse of differentiation, many differentiation rules lead to corresponding integration rules. Consider, for example, the chain rule. DtotkBhcxkBhc kBhcfGrap,kBhcgGrapkBhcxkBhc  = fopxEkBhcáGrap,kBhcgGrapkBhcxkBhc  gopxEkBhcáGrapkBhcxkBhcÿThe chain rule says that when we take the derivative of one function composed with another the result is the derivative of the outer function times the derivative of the inner function. How does this lead to an integration formula? Well, suppose you can fkBhcigure out the original kBhc f(x)kBhc  from kBhcfopxEkBhcákBhc (x)kBhc by integrating:DtotkBhcxkBhc  fGrapkBhcxkBhc  = fopxEkBhcáGrapkBhcxkBhc2+What happens when you try to differentiate kBhcfGrap,kBhcgGrapkBhcxkBhc? GrapDkBhcfGrap,kBhcgGrapkBhcxopxEkBhcákBhc kBhc=kBhc kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc  gopxEkBhcáGrapkBhcx kBhc92What does that tell us? Integrating both sides of feR (1)kBhc  gives etnI¼kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhcxkBhc kBhc=kBhc kBhcfGrap,kBhcgGrapkBhcx kBhc^WThe formula forms the basis for a method of integration called the substitution method. kBhcSome simple examples'kBhcHAHere are some simple examples where you can apply this technique.&kBhcConsider the integral etnIVGrapkBhc2 x + 1opxEkBhc3kBhcxkBhcUNTo view the thing we are integrating as a composition, introduce the functionsAkBhcfopxEkBhcáGrapkBhcxkBhc kBhc=kBhc kBhcxopxEkBhc3kBhc  , kBhcgGrapkBhcxkBhc kBhc=kBhc  2 kBhcxkBhc  + 1BkBhc*#This does the job in the sense thatCkBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhc=kBhc Grap0kBhc 2 kBhcxkBhc  + 1opxEkBhc3DkBhcOHbut recall that the pattern we have to try and force the problem into isEetnI¼kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhcxFkBhc.'What we need to provide is the missing kBhcgopxEkBhcáGrapkBhcxkBhc-& term. From what we said above we haveGkBhcgopxEkBhcáGrapkBhcxkBhc kBhc=kBhc  2HkBhczsIntroducing a factor of 2 into the problem is easy - we just have to balance it with a corresponding factor of 1/2:IcarF$kBhc1kBhc2kBhc etnIyGrap0kBhc 2 kBhcxkBhc  + 1opxEkBhc3kBhc  2kBhcxJkBhcb[This now matches the desired form. The next thing we have to do is to recover the original kBhc f(x)kBhc  from kBhcfopxEkBhcákBhc (x)kBhc& by finding an anti-derivative.KDtotkBhcxkBhcfGrapkBhcxkBhc kBhc=kBhc kBhcfopxEkBhcáGrapkBhcxkBhc kBhc=kBhc kBhcxopxEkBhc3L kBhcfGrapkBhcxkBhc kBhc=kBhc carF$kBhc1kBhc4kBhc kBhcxopxEkBhc4MkBhcB;Putting all of this together with the formula derived aboveNetnI¼kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhcxkBhc kBhc=kBhc kBhcfGrap,kBhcgGrapkBhcxOkBhc givesPcarF$kBhc1kBhc2etnIyGrap0kBhc 2 kBhcxkBhc  + 1opxEkBhc3kBhc  2kBhcxkBhc kBhc=kBhc carF$kBhc1kBhc2kBhc Grap|carF$kBhc1kBhc4Grap0kBhc 2 kBhcxkBhc  + 1opxEkBhc4QkBhc$For the next example considerRetnIWcarF3kBhc1kBhc 1 + kBhcxkBhcxSkBhcd]Although it is a little more difficult to come up with at first, there is a composition here:TkBhcfopxEkBhcáGrapkBhcxkBhc kBhc=kBhc  1/kBhcxkBhc  , kBhcgGrapkBhcxkBhc kBhc=kBhc  1 + kBhcxU kBhc Since kBhcgopxEkBhcáGrapkBhcxkBhc kBhc=kBhc4- 1, we don't have to manufacture the missing kBhcgopxEkBhcáGrapkBhcxkBhc  termV kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhc kBhc=kBhc carF3kBhc1kBhc 1 + kBhcxkBhc  1WkBhc+$We also need the anti-derivative of kBhcfopxEkBhcákBhc (x)XetnI\kBhcfopxEkBhcáGrapkBhcxkBhcxkBhc kBhc=kBhc etnI9kBhc 1/kBhcxkBhcxkBhc kBhc=kBhc  lnVsbAkBhcxkBhc kBhc=kBhc kBhcfGrapkBhcxYetnI¼kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhcxkBhc kBhc=kBhc kBhcfGrap,kBhcgGrapkBhcxZetnIWcarF3kBhc1kBhc 1 + kBhcxkBhcxkBhc kBhc=kBhc  lnVsbA#kBhcxkBhc  + 1[kBhc`YSometimes finding the right composition takes a little extra work. Consider this problem.\etnIocarFKkBhcxkBhc 1 + kBhcxopxEkBhc4kBhcx]kBhc=6Recall that we have to match the integrand to the form^kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcx_kBhcF?It is a little hard to see, but the right thing to do is to set`kBhcgGrapkBhcxkBhc kBhc=kBhc kBhcxopxEkBhc2akBhcgopxEkBhcáGrapkBhcxkBhc kBhc=kBhc  2 kBhcxbkBhc$This shows us how to get the kBhcgopxEkBhcáGrapkBhcxkBhc term that we need.ccarF$kBhc1kBhc2kBhc etnI carFKkBhc1kBhc 1 + kBhcxopxEkBhc4kBhc Grap!kBhc 2 kBhcxkBhcxdkBhcRKWhat remains is to figure out how to write the other term as a composition kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc:ecarFKkBhc1kBhc 1 + kBhcxopxEkBhc4kBhc kBhc=kBhc carFokBhc1kBhc 1 + Grap,kBhcxopxEkBhc2opxEkBhc2fkBhcSo we see thatgkBhcfopxEkBhcáGrapkBhcxkBhc kBhc=kBhc carFKkBhc1kBhc 1 + kBhcxopxEkBhc2h kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhc=kBhc carFokBhc1kBhc 1 + Grap,kBhcxopxEkBhc2opxEkBhc2kBhc kBhc=kBhc carFKkBhc1kBhc 1 + kBhcxopxEkBhc4ikBhc2+We will need to get the anti-derivative of kBhcfopxEkBhcáGrapkBhcxkBhc:jetnIocarFKkBhc1kBhc 1 + kBhcxopxEkBhc2kBhcxkBhc kBhc=kBhc arctan kBhcxkBhc  + ckkBhcfGrapkBhcxkBhc kBhc=kBhc arctan kBhcxlkBhc)"Putting this all together gives usmetnI¼kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhcxkBhc kBhc=kBhc kBhcfGrap,kBhcgGrapkBhcxn carF$kBhc1kBhc2kBhc etnI carFKkBhc1kBhc 1 + kBhcxopxEkBhc4kBhc Grap!kBhc 2 kBhcxkBhcxkBhc kBhc=kBhc carF$kBhc1kBhc2kBhc kBhc arctanGrap,kBhcxopxEkBhc2okBhcSummary of the methodpkBhc]VLook for something in the integrand that looks like the composition of two functions, kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc.q kBhc]VFind something in the integrand that looks like the derivative of the inner function, kBhcgopxEkBhcáGrapkBhcxkBhcF?. If necessary, throw in constant factors needed to create the kBhcgopxEkBhcáGrapkBhcxkBhc  term.rkBhc"Find the antiderivative of kBhcfopxEkBhcáGrapkBhcxkBhc.s kBhcThe antiderivative of kBhcfopxEkBhcáGrap,kBhcgGrapkBhcxkBhc kBhcgopxEkBhcáGrapkBhcxkBhc  is kBhcfGrap,kBhcgGrapkBhcxkBhc.tkBhc.'Commonly used notation for substitutionukBhcI used kBhcfkBhc  and kBhcgkBhcü above to make clear the origins of this newest method in the chain rule. In practice, people tend to use slightly different notation which suggests a slightly different way to understand what is going on here. Here is an example of that method in use.vkBhcConsider the integralwetnIQkBhceopxE!kBhc 3 kBhcxkBhcxxkBhcaZThe form of the integrand suggest that we try to transform the problem into something likeykBhc etnIDkBhceopxEkBhcukBhcuzkBhc3,The substitution that gets this to happen is{kBhcukBhc kBhc=kBhc  3 kBhcx|kBhcHAWe need a derivative in order to apply this method, so we compute}Dtot(kBhcukBhcxkBhc kBhc=kBhc  3~kBhcA:With these substitutions, the original integral looks like etnIQkBhceopxEkBhcukBhc  1kBhcxkBhc kBhc=kBhc carF$kBhc1kBhc3kBhc etnIQkBhceopxEkBhcukBhc  3kBhcxkBhc kBhc=kBhc carF$kBhc1kBhc3kBhc etnI|kBhceopxEkBhcukBhc Dtot(kBhcukBhcxkBhcx€kBhc$Can we turn this integral in kBhcxkBhc into an integral in kBhcukBhcD=? The answer is yes: the key requirement is that we have the Dtot(kBhcukBhcxkBhc  dkBhcxkBhcXQ combination. By the chain rule, those two pieces can be combined into a single dkBhcukBhc5., completing the transition to an integral in kBhcukBhc instead of kBhcxkBhc.etnI|kBhcfGrapkBhcukBhc Dtot(kBhcukBhcxkBhcxkBhc kBhc=kBhc etnIDkBhcfGrapkBhcukBhcu‚kBhc:3At this point, we complete the change of variables:ƒetnIQkBhceopxE!kBhc 3 kBhcxkBhcxkBhc kBhc=kBhc carF$kBhc1kBhc3kBhc etnIQkBhceopxEkBhcukBhc  3kBhcxkBhc kBhc=kBhc carF$kBhc1kBhc3kBhc etnIDkBhceopxEkBhcukBhcukBhc kBhc=kBhc carF$kBhc1kBhc3kBhc  eopxEkBhcukBhc  + kBhcCkBhc kBhc=kBhc carF$kBhc1kBhc3kBhc  eopxE!kBhc 3 kBhcx·kBhc*#Substitution and Definite Integrals¸kBhcåÞWe have seen that an appropriately chosen substitution can make an anti-differentiation problem doable. Something to watch for is the interaction between substitution and definite integrals. Consider the following example.·etnI~kBhc x trqS4kBhc 1 - xopxEkBhc2kBhcxkBhc1kBhc -1¸kBhcD=There are two approaches we can take in solving this problem:·kBhcvoUse substitution to compute the antiderivative and then use the anti-derivative to solve the definite integral.·kBhc u = 1 - xopxEkBhc2¸Dtot(kBhcukBhcxkBhc = -2 x·etnIekBhc x trqS4kBhc 1 - xopxEkBhc2kBhcxkBhc  = - carF$kBhc1kBhc2kBhc etnIjtrqS4kBhc 1 - xopxEkBhc2kBhc (-2 x)kBhcxkBhc ¸kBhc = - carF$kBhc1kBhc2kBhc etnI<trqSkBhcukBhcukBhc  = - carF$kBhc1kBhc2kBhc etnIFkBhcuopxEkBhc 1/2kBhcukBhc  = - carF$kBhc1kBhc2kBhc  (carF$kBhc2kBhc3kBhc  uopxEkBhc 3/2kBhc ) + C¸kBhc = - carF$kBhc1kBhc3kBhc Grap0kBhc 1 - xopxEkBhc2opxEkBhc 3/2kBhc  + C¸ etnI~kBhc x trqS4kBhc 1 - xopxEkBhc2kBhcxkBhc1kBhc -1kBhc  = - carF$kBhc1kBhc3kBhc GrapGkBhc 1 - GrapkBhc1opxEkBhc2opxEkBhc 3/2kBhc  - Grap¯kBhc - carF$kBhc1kBhc3kBhc GrapHkBhc 1 - GrapkBhc -1opxEkBhc2opxEkBhc 3/2kBhc  = 0¸kBhcb[Use substitution on both the expression being integrated and on the limits of the integral.· etnI~kBhc x trqS4kBhc 1 - xopxEkBhc2kBhcxkBhc1kBhc -1kBhc  = - carF$kBhc1kBhc2kBhc etnIƒtrqS4kBhc 1 - xopxEkBhc2kBhc (-2 x)kBhcxkBhc1kBhc -1kBhc  = - carF$kBhc1kBhc2kBhc etnI¦trqSkBhcukBhcukBhc 1 - GrapkBhc1opxEkBhc2kBhc1 - (-1)opxEkBhc2kBhc  = - carF$kBhc1kBhc2kBhc etnITtrqSkBhcukBhcukBhc0kBhc0µkBhcohIn the second approach we can see that the integral will be 0 even without computing an anti-derivative.„kBhc/(Antiderivatives may differ by a constantkBhcº³When we compute an antiderivative, the answer we get can be modified by adding a constant term and still be a valid antiderivative. Here is a simple example. The antiderivative of¶kBhc f(x) = 2 x³kBhc is´kBhcF(x) = xopxEkBhc2µkBhc)"An equally valid antiderivative is²kBhcG(x) = xopxEkBhc2kBhc  + 1±kBhc’‹One way to capture this flexibility is to say that all of our antiderivatives are valid up to some constant factor by saying something like°etnI.kBhc 2 xkBhcxkBhc  = xopxEkBhc2kBhc  + C¯kBhcÿA common consequence of this effect is that doing the same problem in two different ways can produce two answers that appear to be very different. As long as we can show that the two answers differ by a constant, both answers are legitimate. This phenomenkBhc>7on is most pronounced when working with trig functions.®kBhcKDHere is an example to demonstrate this effect. Consider the integralŒetnISkBhc sinGrap!kBhc 2 kBhcxkBhcx‹kBhc[TThe form of the integrand suggests that we transform the problem into something likeŠ etnI;kBhc sin kBhcukBhcukBhc kBhc=kBhc kBhc-kBhc cos kBhcukBhc  + kBhcC‰kBhc-&The substitution we need to do this isˆkBhcukBhc kBhc=kBhc  2 kBhcx‡Dtot(kBhcukBhcxkBhc kBhc=kBhc  2†kBhce^To get the derivative term we need we will have to introduce a factor of 1/2 to balance the 2.…carF$kBhc1kBhc2etnI`kBhc sinGrap!kBhc 2 kBhcxkBhc  2kBhcxkBhc kBhc=kBhc carF$kBhc1kBhc2etnIskBhc sin kBhcukBhc Dtot(kBhcukBhcxkBhcxkBhc kBhc=kBhc carF$kBhc1kBhc2etnI;kBhc sin kBhcukBhcukBhc kBhc=kBhc kBhc-carF$kBhc1kBhc2kBhc  cosGrap!kBhc 2 kBhcxkBhc  + kBhcC·kBhcibHere is a different approach. We begin by using a trig identity to change the form of the problem.­etnIIkBhc sin GrapkBhc 2 xkBhcxkBhc  = etnI8kBhc 2 sin x cos xkBhcx¬kBhc=6The latter form can be handled by composition by using«kBhc g(x) = sin xªkBhcgopxEkBhcákBhc (x) = cos x©kBhcfopxEkBhcákBhc (x) = 2 x¨kBhcf(x) = xopxEkBhc2§kBhc This leads to¦etnI8kBhc 2 sin x cos xkBhcxkBhc  = etnI}kBhcfopxEkBhcákBhc(g(x)) gopxEkBhcákBhc (x)kBhcxkBhc = f(g(x)) + C = GrapkBhc sin xopxEkBhc2kBhc  + C¸kBhc›”This appears to be very different from the earlier answer. However, some manipulation with trig identities will bring the two forms closer together.kBhc-carF$kBhc1kBhc2kBhc  cosGrap!kBhc 2 kBhcxkBhc  = -carF$kBhc1kBhc2kBhc GrapfkBhc cosopxEkBhc2kBhc x - sinopxEkBhc2kBhc  xkBhc  = -carF$kBhc1kBhc2kBhc GrapGrap?kBhc1 - sinopxEkBhc2kBhc  xkBhc  - sinopxEkBhc2kBhc  xkBhc  = -carF$kBhc1kBhc2kBhc  + sinopxEkBhc2kBhc  xkBhc<5The difference between the two solutions is now just kBhc -1/2kBhc, which is acceptable.6kBhcHomework Problems7kBhc$Section 5.4: 6, 8, 10, 26, 369kBhc&Section 5.5: 12, 32, 44, 58, 64