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MIT 18 01 - Exam 4 Review

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MIT OpenCourseWare http ocw mit edu 18 01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use visit http ocw mit edu terms Lecture 32 Exam 4 Review 18 01 Fall 2006 Exam 4 Review 1 Trig substitution and trig integrals 2 Partial fractions 3 Integration by parts 4 Arc length and surface area of revolution 5 Polar coordinates 6 Area in polar coordinates Questions from the Students Q What do we need to know about parametric equations A Just keep this formula in mind ds dx dt 2 dy dt 2 Example You re given x t t4 and y t 1 t Find s length ds 4t3 2 1 2 dt Then integrate with respect to t Q Can you quickly review how to do partial fractions A When nding partial fractions rst check whether the degree of the numerator is greater than or equal to the degree of the denominator If so you rst need to do algebraic longdivision If not then you can split into partial fractions Example x2 x 1 x 1 2 x 2 We already know the form of the solution x2 x 1 A B C x 1 2 x 2 x 1 x 1 2 x 2 There are two coe cients that are easy to nd B and C We can nd these by the cover up method 12 1 1 3 B x 1 1 2 3 1 Lecture 32 Exam 4 Review 18 01 Fall 2006 To nd C C 2 2 2 1 1 2 1 2 3 x 2 To nd A one method is to plug in the easiest value of x other than the ones we already used x 1 2 Usually we use x 0 1 A 1 1 3 1 2 2 1 1 2 2 and then solve to nd A The Review Sheet handed out during lecture follows on the next page 2 Lecture 32 Exam 4 Review 18 01 Fall 2006 Exam 4 Review Handout 1 Integrate by trigonometric substitution evaluate the trigonometric integral and work backwards to the original variable by evaluating trig trig 1 using a right triangle a a2 x2 use x a sin u dx a cos u du b a2 x2 use x a tan u dx a sec2 u du c x2 a2 use x a sec u dx a sec u tan u du 2 Integrate rational functions P Q ratio of polynomials by the method of partial fractions If the degree of P is less than the degree of Q then factor Q completely into linear and quadratic factors and write P Q as a sum of simpler terms For example 3x2 1 A B1 B2 Cx D 2 2 2 2 x 1 x 2 x 9 x 1 x 2 x 2 x 9 Terms such as D x2 9 can be integrated using the trigonometric substitution x 3 tan u This method can be used to evaluate the integral of any rational function In practice the hard part turns out to be factoring the denominator In recitation you encountered two other steps required to cover every case systematically namely completing the square1 and long division 2 3 Integration by parts a b b b uv dx uv u vdx a a This is used when u v is simpler than uv This is often the case if u is simpler than u 4 Arclength ds dx2 dy 2 Depending on whether you want to integrate with respect to x t or y this is written ds 1 dy dx 2 dx ds dx dt 2 dy dt 2 dt ds dx dy 2 1 dy 5 Surface area for a surface of revolution a around the x axis 2 yds 2 y 1 dy dx 2 dx requires a formula for y y x b around the y axis 2 xds 2 x dx dy 2 1 dy requires a formula for x x y 6 Polar coordinates x r cos y r sin or more rarely r x2 y 2 tan 1 y x a Find the polar equation for a curve from its equation in x y variables by substitution b Sketch curves given in polar coordinates and understand the range of the variable often in preparation for integration 7 Area in polar coordinates 2 1 1 2 r d 2 Pay attention to the range of to be sure that you are not double counting regions or missing them 1 For example we rewrite the denominator x2 4x 13 x 2 2 9 u2 a2 with u x 2 and a 3 division is used when the degree of P is greater than or equal to the degree of Q It expresses P x Q x P1 x R x Q x with P1 a quotient polynomial easy to integrate and R a remainder The key point is that the remainder R has degree less than Q so R Q can be split into partial fractions 2 Long 3 Lecture 32 Exam 4 Review 18 01 Fall 2006 The following formulas will be printed with Exam 4 sin2 x cos2 x 1 sin2 x sec2 x tan2 x 1 1 1 cos 2x 2 2 cos2 x cos 2x cos2 x sin2 x d d tan x sec2 x sec x sec x tan x dx dx tan x dx ln cos x c 1 1 cos 2x 2 2 sin 2x 2 sin x cos x d 1 d 1 tan 1 x sin 1 x dx 1 x2 dx 1 x2 sec x dx ln sec x tan x c See the next page for a review on integration of rational functions 4 Lecture 32 Exam 4 Review 18 01 Fall 2006 Postscript Systematic integration of rational functions For a general rational function P Q the rst step is to express P Q as the sum of a polynomial and a ratio in which the numerator has smaller degree than the denominator For example x3 3x 2 x 2 2 x2 2x 1 x 2x 1 To carry out this long division do not factor the denominator Q x x2 2x 1 just leave it alone The quotient x 2 is a polynomial and is easy to integrate The remainder term 3x 2 x 1 2 has a numerator 3x 2 of degree 1 which is less than the degree 2 of the denominator x 1 2 Therefore there is a partial fraction decomposition In fact 3x 2 3x 3 1 3 1 x 1 2 x 1 2 x 1 x 1 2 In general if P has degree n and Q has degree m then long division gives P x R x P1 x Q x Q x in which P1 the quotient in the long division has degree n m and R the remainder in the long division has degree at most m 1 Evaluation of the simple pieces The integral dx 1 x a 1 n c x a n n 1 if n 1 and ln x a c if n …


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MIT 18 01 - Exam 4 Review

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