Math 1B Discussion Exercises GSI Theo Johnson Freyd http math berkeley edu theojf 09Summer1B Find two or three classmates and a few feet of chalkboard Introduce yourself to your new friends and write all of your names at the top of the chalkboard As a group try your hand at the following exercises Be sure to discuss how to solve the exercises how you get the solution is much more important than whether you get the solution If as a group you agree that you all understand a certain type of exercise move on to later problems You are not expected to solve all the exercises some are very hard Many of the exercises are from Single Variable Calculus Early Transcendentals for UC Berkeley by James Stewart these are marked with an Others are my own are from the mathematical folklore or are independently marked Here s a hint drawing pictures e g sketching graphs of functions will always make the problem easier Partial Fraction Decomposition When we add fractions we go through all sorts of steps to find a common denominator because for some purposes it s best to have an expressing written as a single fraction rather than as multiple fractions 1 x x2 x 1 2 3 2x 1 x 1 2x x2 2x 1 But for integration complicated fractions like x2 x 1 2x3 x2 2x 1 are less than useful whereas the simpler fractions are perfectly tractable Z Z x2 x 1 1 x 1 1 dx dx ln 2x 1 ln x2 1 C 2x3 x2 2x 1 2x 1 x2 1 2 2 We compute the second integral by substituting u x2 1 In fact every rational function the ratio of two polynomials can be decomposed uniquely into a sum of a polynomial plus simple fractions A simple fraction is one where the denominator is a power of a linear or irreducible quadratic polynomial and the numerator is of lower degree than the 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 The first step is to factor the denominator completely 2x3 x2 2x 1 x2 1 2x 1 Over the real numbers every polynomial has a unique factorization into linear and irreducible quadratic parts 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 with repeated factors counting multiply Write the most general decomposition given the factorization and then solve for the unknown coefficients 1 1 Decompose the following fractions into sums of simple fractions a x x 2 x 1 x 1 x3 x b 1 c 2x 1 x 1 3 x2 4 2 For the more curious of you a few years ago back when I kept a semi regular blog I wrote a very short proof that partial fraction decompositions always exist http theojf blogspot com 2007 09 partial fractions html 1 R 2 R Compute the following integrals Remember that dx x2 1 arctan x and that 2x dx x2 1 ln x2 1 If necessary don t forget to complete the square in the denominator Z 2 Z Z x x 1 x 1 x2 dx c dx a dx b 2 2 x 3x 2 x 3 x 2 x2 1 2 Z Z 1 Z 3x2 x 4 x x3 d dx e dx f dx 2 x4 3x2 1 x3 1 0 x 4x 13 R 3 Solve x x2 1 dx a with a u substitution b with a trig substitution c with a partialfractions decomposition R 4 Evaluate sec d as follows write the integrand in terms of sines and cosines multiply the numerator and denominator by cos use trig identities to rewrite the denominator in terms of sin make a u substitution decompose the resulting integrand into partial fractions integrate and substitute back for R 5 a Compute tan2 sec d by first performing a u substitution and then decomposing into partial fractions R b Compute x2 4 3 2 dx by first using a trig sub then a u sub then partial fractions Z dx 6 Evaluate Hint substitute u 6 x 3 x x 7 Factor x4 1 as a difference of squares by first adding and subtracting the same quantity R Use this factorization to evaluate 1 x4 1 dx 8 Any rational expression in sines and cosines can be made into a rational function via the Weierstrass substitution u tan 2 a Using trig identities find general formulae for sin cos and d in terms of u Z 2 Z d d and ii b Compute i 3 cos 4 sin 3 1 sin cos Z f x 9 If f is a quadratic function such that f 0 1 and dx is a rational function x2 x 1 3 no ln and arctan terms find the value of f 0 0 10 This problem is only for those who know complex numbers Normally we integrate R dx x2 1 arctan x with a trig substitution Factor the denominator with complex numbers decompose it as partial fractions and integrate what does this say about the relationship between arctan and ln 2
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