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: udv uv vdu 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 ln x) 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 tan t x a sec t sin x + cos x tan x + sec x sin x cos(x) cos x + cos(x) sin(x) sin x cos x 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 cos x and substitute u sin x. cos x sin x m, n even Use sin x + cos x to reduce to only powers of sin x sin x cos(x) or only powers of cos x, then use table of integrals #9-4 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 tan x and substitute u sec x. tan x sec x n even Break off one factor of sec x and substitute u tan x. 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 table of integrals #5. d tan x sec x Useful Trigonometric Derivatives and Antiderivatives d sec x sec x tan x sec xln sec x + tan x + 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 Note: this is the same as lim x ex. lim /x x lim /x x + lim ln x x lim ln x x + lim arctan x π/ x. Evaluate the following. 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 sin x, sodu 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 tan t, so sec tdt. y x t x + y y x + sec t hyp adj x + tan t opp adj x

3 + x sec tdt + tan t sec tdt + tan t sec tdt sec t sec tdt ln sec t + tan t + C x + ln + x + C Now use + tan t sec t. Now use triangle above. (c) This is an improper integral, so we need to use a limit. x(ln x) t xt x u 99 xt x [ x(ln x) u du Substitute u lnx, so du x. 99 t 99(ln x) 99 99(ln t) 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 e x t [ e 4e ( ) ( /4) /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 ln u + 5 arctan x +ln x + D ln(x +) + 5 arctan x +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 5ln x 6 + C (g) This integral is improper at x because the integrand has a vertical asymptote there. x t + t x ln x t + t [ln ln t t + Since lim t + ( ln t ), this integral diverges (to ).

5 . Solve the differential equation / xy +6x if the solution passes through (, 5). xy +6x x(y +) y + x Separate the variables. y + x ln y + x + C y + e x +C Exponentiate each side to remove the ln. y +±e C e x w z means w ±z. y +Ae x Replace ±e C with A. Now we use the initial condition y() 5 to find the value of A. We have 5 +Ae A 8, so the solution is y +8e x.. Find the second-degree Taylor polynomial for f(x) x centered at x. f(x) x / f() f (x) x / x f () f (x) 4 x / f () 4x / 4 / 4 P (x) f() + f ()(x ) + f () (x )! + x (x ) 8 4. 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 5/ on [, 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. 6 + cos x (a) x.99 For all x, we have 6 + cos x 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.

6 (b) 4x x x 4 + x 5 + For all x, we have 4x x x 4 + x 5 + 4x x 5 the numerator larger, so the new fraction is larger.) 4. (We ve made the denominator smaller and x t 4 x 4 lim 4 lim [ x 4 lim t x t 4[ ( ) 4 Therefore, the original integral in question must converge to a value less than (Sections A and B may omit this question.) 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.) ke.x t ke.x ke.x t k. 5k. ke.t. ke. So, we have 5k ork.. (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.