Math 106: Review for Exam II - SOLUTIONS

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1 Math 6: Review for Exam II - SOLUTIONS INTEGRATION TIPS Substitution: usually let u a function that s inside another function, especially if du (possibly off by a multiplying constant) is also present in the integrand. Parts: u dv uv v du or uv uv u v How to choose which part is u? Let u be the part that is higher up in the LIATE mnemonic below. (The mnemonics ILATE and LIPET will work equally well if you have learned one of those instead; in the latter A is replaced by P, which stands for polynomial.) Logarithms (such as lnx) Inverse trig (such as arctan x, arcsin x) Algebraic (such as x, x, x + 4) Trig (such as sin x, cos x) Exponentials (such as e x, e x ) Rational Functions (one polynomial divided by another): if the degree of the numerator is greater than or equal to the degree of the denominator, do long division then integrate the result. Partial Fractions: here s an illustrative example of the setup. x + (x + )(x ) (x + 5) A x + + B x + C (x ) + Dx + E x + 5 Each linear term in the denominator on the left gets a constant above it on the right; the squared linear factor (x ) on the left appears twice on the right, once to the second power. Each irreducible quadratic term on the left gets a linear term (Dx + E here) above it on the right. Trigonometric Substitutions: some suggested substitutions and useful formulae follow. Radical Form a x a + x x a Substitution x a sin t x a tant x a sec t sin x + cos x tan x + sec x sin x cos(x) cos x + cos(x) sin(x) sin x cosx Powers of Trigonometric Functions: here are some strategies for dealing with these. sin m x cos n x Possible Strategy Identity to Use m odd Break off one factor of sin x and substitute u cos x. sin x cos x n odd Break off one factor of cosx and substitute u sin x. cos x sin x m, n even Use sin x + cos x to reduce to only powers of sinx sin x cos(x) or only powers of cosx, then use integration by parts cos x + cos(x) or identities shown to right of this box.

2 tan m x sec n x Possible Strategy Identity to Use m odd Break off one factor of sec x tanx and substitute u sec x. tan x sec x n even Break off one factor of sec x and substitute u tanx. sec x tan x + m even, n odd Use identity at right to reduce to powers of sec x alone. tan x sec x Then use integration by parts or reduction formula (if allowed). d tan x sec x Useful Trigonometric Derivatives and Antiderivatives d sec x sec x tanx sec x ln sec x + tanx + C Improper integrals: look for as one of the limits of integration; look for functions that have a vertical asymptote in the interval of integration. It may be useful to know the following limits. lim x ex lim x e x lim /x x lim /x x + lim lnx x lim lnx x + lim arctan x π/ x. Evaluate the following. Note: this is the same as lim x ex. Note: the answer is the same for lim x /x and similar functions. Note: the answer is the same for lim x /x and similar functions. + (a) Let u sinx, so du cos x. sin 6 x cos x sin 6 x( sin x)cos x Use cos x sin x. u 6 ( u )du (u 6 u 8 )du u7 7 u9 9 + C sin7 x 7 sin9 x 9 + C (b) Let x tant, so sec t dt. y x t x + y y x + sec t hyp adj x + tant opp adj x

3 + x sec t dt + tan t sec t dt + tan t sec t dt sec t sec t dt Now use + tan t sec t. ln sec t + tant + C Now use triangle above. x + ln + x + C (c) This is an improper integral, so we need to use a limit. x(lnx) t xt x u 99 xt x [ x(lnx) u du Substitute u lnx, so du x. 99 t 99(lnx) 99 99(lnt) 99 99(ln) 99 99(ln) 99 99(ln ) 99 So, the integral converges (to this value). (d) We ll use integration by parts: u x du and dv e x v e x. xe x t [x e x xe x [x e x t t e x 4 [ t x e x 4e x [ t e t 4e t ( ) ( /4) e x t [ e 4e /4 So, the integral converges (to this value). (e) Partial Fractions: Write x + x (x )(x + ) Ax + B x + + C x. Now multiply both sides by (x )(x + ) to get

4 x + x (Ax + B)(x ) + C(x + ). Let x. Then C(), so C. Let x. Then B( ) + (), so B 5. Let x. Then 8 (A() + 5)( ) + (), so A. x [ + x x + 5 (x )(x + ) x + + x [ x x x + + x du [ u + 5 x + + x lnu ln(x + ) + 5 arctanx + ln x + D + 5 arctanx + ln x + D Let u x +, so du x. (f) Since the degree of the numerator is greater than or equal to the degree of the denominator, we do long division. 4x x x 6 x 6 4x 7x + x 7 4x 4x x x +8x x x 5 Now, we compute the integral. 4x 7x [ + x 7 4x x + 5 4x x 6 x 6 x + x 5 ln x 6 + C (g) This integral is improper at x because the integrand has a vertical asymptote there, so we split into two integrals. 5 (x ) 6 a a a (x ) 6 + a 5 (x ) 6 (x ) 6 + lim b + 5 a 5(x ) + lim 5 b [ + 5(a ) 5 5( ) 5 Since lim and lim a 5(a ) 5 b + b (x ) 6 5 5(x ) 5 b + lim b + [ 5(5 ) 5 5(b ) 5, this integral diverges (to ). 5(b ) 5 u 6 du u C

5 . Find the second-degree Taylor polynomial for f(x) x centered at x. f(x) x / f() f (x) x/ x f (x) 4 x / f () f () 4x / 4 / 4 P (x) f() + f ()(x ) + f () (x )! + x (x ) 8. What is the maximum possible error that can occur in your Taylor approximation from the previous problem on the interval [,? We know that f(x) P n (x) K n+ (n + )! x x n+. In this case, n, x, and x (the farthest from x that we are considering). K max of f (x) on [, max of 8x on [, 5/ 8 5/ 8, Putting this all together, we have f(x) P (x) 8,! Use comparisons to show whether each of the following converges or diverges. If an integral converges, also give a good upper bound for its value. (a) (b) 6 + cos x x.99 For all x, we have 6 + cosx 6 x.99 x 5 because the minimum value of cos x is..99 x.99 5 Since diverges (compute yourself or notice that p.99 < ), we know that the integral x.99 in question must diverge too. 4x x x 4 + x 5 + 4x x For all x, we have x 4 + x 5 + 4x x 4 5 x. (We ve made the denominator smaller and the numerator larger, so the new fraction is larger.) t 4 x 4 lim x 4 lim t x [ 4 lim t 4[ () 4 Therefore, the original integral in question must converge to a value less than 4.

6 5. The probability density function (pdf) of the length (in minutes) of phone calls on a certain wireless network is given by f(x) ke.x where x is the number of minutes. Note that the domain is x since we can t have a negative number of minutes. (a) What must be the value of k? We know that the total area under any pdf must be (because it must account for % of events.) So, we have 5k or k.. ke.x t ke.x ke.x t k. 5k. ke.t. ke. (b) What fraction of calls last more than minutes?.e.x t.e.x.e.x t. e.t ( e.6 ) + e.6 e Note that we could instead have computed.e.x and gotten the same answer, but the point of introducing pdf s in this text seems to be to show how improper integrals are used.