Final Exam Review Sheet:Time: 1:30-4:30 PM on Monday, May 11th, 2015Place: Foellinger AuditoriumSections: 7.1-7.5,7.7- 7.8, 8.1 - 8.3,10.1-10.4 , 11.1 - 11.11• You have three hours for the exam.• I recommend going over the list of materials mentioned in this sheet, then identifyingwhich topics you are not comfortable with and working on the worksheets related tothose topics. You should also solve all the sample exam questions.• Take your time and go over the worksheets we did in class. In particular, you can workon the questions we didn’t solve in class.• Half of the exam is multiple choice so make sure you plan your time accordingly.• Test taking strategy: Go through the multiple choice questions and solve only thoseyou can solve without thinking for long. Then move on to the free response questionsand solve them as careful as possible. Then, if you still have extra time go back to themultple choice questions you skipped.• Don’t forget to ask me if you have any questions.Review (Things you should definitely know):The topics are split into three parts.Part I (Midterm 1):1. Substitution: Although not officially part of the syllabus as it is in chapter 6 yet Ithink it is the cornerstone of everything we did in part I. Make sure you are comfortablewith substitution techniques. This is very important. If you are not then you shouldmost certainly work on it. You can also come and ask me if you think you need somehelp.2. Integration by Parts: Remember the formula for the integration by parts:Zudv = uv −ZvduYou can use it very often when we have a product of a polynomial and a trig (orexponential) function. You need to practice a lot so that you can identify u and vcorrectly, however, if you are not sure you can always try to use this formula:LIATEwhich stands for logarithmic (L), inverse trigonometric (I), algebraic (A), trigonomet-ric (T), and exponential (E) and the rule is:Pick u in the order of LIATE.However, remember there are always exceptions.13. Trig Integrals: There are several standard trig integrals:•Zsinnx cosmx dxIf either n or m are odd, then you have to use substitution.If n or m are both even, then we have to use the following formulas:sin2x =12(1 − cos 2x) cos2x =12(1 + cos 2x) sin x cos x =12sin 2x•Ztanmx secnx dxIf m is odd or n is even, then use substitution.If m is even and n is odd then there is no easy way. (But you should rememberZsec x dx andZsec3x dx).There are trig integrals which are not of this form where determining the correctsubstitution needs some experience, see question 3 of worksheet 3.4. Trig Substitution: Some integrals can be solved by substituting a trig function.These are the following cases:Expression Substitution dx Identity Angle√a2− x2x = a sin θ dx = a cos θdθ a2− a2sin2θ = a2cos2θ −π2≤ θ ≤π2√a2+ x2x = a tan θ dx = a sec2θdθ a2+ a2tan2θ = a2sec2θ −π2< θ <π2√x2− a2x = a sec θ dx = a sec θ tan θdθ a2sec2θ − a2= a2tan2θ 0 ≤ θ <π2So, whenever you encounter these cases you can use trig substitution. However, youhave to be careful as there are exceptions. Also, sometimes you have to first completea square before you can use trig sub.5. Partial Fractions: We introduce techniques to solve integrals of the formPQsuch thatP and Q are polynomials. First, we have to use polynomial division to make sure thedegree of P is lower than Q. Then we split Q = Q1Q2...Qninto its irreducible factorswhich can have three forms 1) x −a or 2) (x −a)n(where n positive) or 3) ax2+ bx + c(where b2−4ac < 0). Depending on the form we now using following partical fractionequation:• If Qi= x − a then we addAx−a• If Qi= (x − a)nthen we addAn(x−a)n+ ... +A1x−a• If Qi= ax2+ bx + c where b2− 4ac < 0 then we addCx+Dax2+bx+cFor example:P (x)(x − a)(x − b)2(x2+ 1)=Ax − a+B2(x − b)2+B1(x − b)+Cx + Dx2+ 1From there on we use ln and arctan to solve the integrals.26. Improper Integrals: There are two general types:Type 1: •Z∞af(x) dx = limt→∞Ztaf(x) dx•Zb−∞f(x) dx = limt→−∞Zbtf(x) dx•Z∞−∞f(x) dx =Zc−∞f(x) +Z∞cf(x) dxType 2: •Zbaf(x) dx = limt→b−Ztaf(x) dx if f discontinuous at b•Zbaf(x) dx = limt→a+Zbtf(x) dx if f discontinuous at a•Zbaf(x) dx =Zcaf(x) +Zbcf(x) dx If f discontinuous at cThere are some improper integrals which can’t be solved so easily. For those cases wehave some comparison theorems: Let f and g be continuous functions. Suppose thatfor all x ≥ a, 0 ≤ f(x) ≤ g(x). Then:• IfZ∞ag(x)dx is convergent, then so isZ∞af(x)dx, andZ∞af(x)dx ≤Z∞ag(x)dx;• ifZ∞af(x)dx is divergent, then so isZ∞ag(x)dx.7. Strategies: This list is not exhaustive and just covers general ideas. It is very im-portant to practice a lot and gain as much experience as possible. There are manyintegrals which don’t exactly fall into any of the mentioned categories. Also, manyintegrals require the use of several methods.Part II (Midterm 2):8. Approximating Integrals: There are some integrals we cannot calculate easily andso we can use approximation formulas.The Trapezoidal Rule:Zbaf(x) dx =b − a2n(f(x0) + 2f(x1) + ... + 2f(xn−1) + f(xn))Simpson’s Rule:Zbaf(x) dx =b − a3n(f(x0) + 4f(x1) + 2f(x2) + ... + 2f(xn−1) + 4f(xn−1) + f(xn))where xi= a + i∆x and in the second case n has to be even.39. Arc Length and Surface Area: The general formula for arc length isZbads, whereds =pdx2+ dy2. Depending on the function ds has to be interpreted in the correctway. If y = f(x) then dx = 1 and dy = f0(x) and so ds =p1 + (f0(x))2dx. If x = g(y)then dy = 1 and dx = g0(y) and so ds =p1 + (g0(y))2dy.The surface area is calculated in a very similar fashion. We only calculate it for surfaceswhich come from rotations. The general formula is 2πxds (if rotated around the y-axis)or 2πyds (if rotated around the x-axis).10. Hydrostatic Force: The formula for hydrostatic force is:F = mg = ρgAdwhere F is force, m mass, g standard gravity acceleration (9.8ms2= 32fts2), ρ density ofthe water (1000kgm3), A area and d distance. As we know P =FAwhere P is pressure,which gives us:P = ρgdAs ρ and g are constant all we have to do is to find a formula for the distance and areaof the shape and take integral.11. Moments and Centroids: Every shape determined by two functions f (x) and g(x)has two moments:Mx= ρZba12([f(x)]2− [g(x)]2) dx and My= ρZbax(f(x) − g(x)) dxwhere Mxis the moment of the system around the x-axis and Myaround the y-axis.Based
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