Math 1B: Discussion ExercisesGSI: Theo Johnson-Freydhttp://math.berkeley.edu/~theojf/09Summer1B/Find two or three classmates and a few feet of chalkboard. Introduce yourself to your newfriends, and write all of your names at the top of the chalkboard. As a group, try your hand atthe following exercises. Be sure to discuss how to solve the exercises — how you get the solutionis much more important than whether you get the solution. If as a group you agree that you allunderstand a certain type of exercise, move on to later problems. You are not expected to solve allthe exercises: some are very hard.Many of the exercises are from Single Variable Calculus: Early Transcendentals for UC Berkeleyby James Stewart; these are marked with an §. Others are my own, are from the mathematicalfolklore, or are independently marked.Here’s a hint: drawing pictures — e.g. sketching graphs of functions — will always make theproblem easier.Partial Fraction DecompositionWhen we add fractions, we go through all sorts of steps to find a common denominator, becausefor some purposes it’s best to have an expressing written as a single fraction rather than as multiplefractions:12x + 1+xx2+ 1=x2+ x + 12x3+ x2+ 2x + 1But for integration, complicated fractions like (x2+ x + 1)/(2x3+ x2+ 2x + 1) are less than useful,whereas the simpler fractions are perfectly tractable:Zx2+ x + 12x3+ x2+ 2x + 1dx =Z12x + 1+xx2+ 1dx =12ln(2x + 1) +12ln(x2+ 1) + C(We compute the second integral by substituting u = x2+ 1.)In fact, every rational function (the ratio of two polynomials) can be decomposed uniquely intoa sum of (a polynomial plus) simple fractions. A “simple fraction” is one where the denominator isa power of a linear or irreducible-quadratic polynomial, and the numerator is of lower degree thanthe root of the denominator.The zeroth step is to perform long division so that your fraction is written as a “mixed number”:a polynomial plus a fraction in which the numerator has lower degree than the denominator. Thefirst step is to factor the denominator completely: 2x3+ x2+ 2x + 1 = (x2+ 1)(2x + 1). Overthe real numbers, every polynomial has a unique factorization into linear and irreducible-quadraticparts, but finding such a factorization in general can be very hard.Then each factor in the denominator corresponds to a potential term in the decomposition, withrepeated factors counting multiply. Write the most general decomposition given the factorization,and then solve for the unknown coefficients.11. Decompose the following fractions into sums of simple fractions:(a)x(x + 2)(x − 1)(b)x − 1x3+ x(c)2x + 1(x + 1)3(x2+ 4)21For the more curious of you, a few years ago (back when I kept a semi-regular blog) I wrote a very short proof thatpartial fraction decompositions always exist: http://theojf.blogspot.com/2007/09/partial-fractions.html.12. § Compute the following integrals. Remember thatRdx/(x2+ 1) = arctan x and thatR2x dx/(x2+ 1) = ln(x2+ 1). If necessary, don’t forget to complete the square in thedenominator:(a)Zx − 1x2+ 3x + 2dx (b)Zx2(x − 3)(x + 2)2dx (c)Zx2+ x + 1(x2+ 1)2dx(d)Z3x2+ x + 4x4+ 3x2+ 1dx (e)Z10xx2+ 4x + 13dx (f)Zx3x3+ 1dx3. SolveRx/(x2−1) dx (a) with a u-substitution, (b) with a trig substitution, (c) with a partial-fractions decomposition.4. EvaluateRsec θ dθ as follows: write the integrand in terms of sines and cosines; multiplythe numerator and denominator by cos θ; use trig identities to rewrite the denominator interms of sin θ; make a u-substitution; decompose the resulting integrand into partial fractions;integrate and substitute back for θ.5. (a) ComputeRtan2θ sec θ dθ by first performing a u-substitution and then decomposinginto partial fractions.(b) ComputeR(x2+ 4)3/2dx by first using a trig sub, then a u-sub, then partial fractions.6. § EvaluateZdx√x −3√x. Hint: substitute u =6√x.7. § Factor x4+ 1 as a difference of squares by first adding and subtracting the same quantity.Use this factorization to evaluateR1/(x4+ 1) dx.8. Any rational expression in sines and cosines can be made into a rational function via theWeierstrass substitution: u = tan(θ/2).(a) Using trig identities, find general formulae for sin θ, cos θ, and dθ in terms of u.(b) § Compute (i)Zdθ3 cos θ − 4 sin θand (ii)Zπ/2π/3dθ1 + sin θ − cos θ.9. § If f is a quadratic function such that f(0) = 1 andZf(x)x2(x + 1)3dx is a rational function(no ln and arctan terms), find the value of f0(0).10. This problem is only for those who know complex numbers. Normally we integrateRdx/(x2+ 1) = arctan(x) with a trig substitution. Factor the denominator with complexnumbers, decompose it as partial fractions, and integrate: what does this say about therelationship between arctan and
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