Review for Exam II: Integration Tips and Techniques - Solutions, Lecture notes of Calculus

Download Review for Exam II: Integration Tips and Techniques - Solutions and more Lecture notes Calculus in PDF only on Docsity! Math 106: 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′ dx = uv − ∫ u′v dx 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, arcsinx) Algebraic (such as x, x2, x3 + 4) Trig (such as sin x, cos 2x) Exponentials (such as ex, e3x) • 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. 3x2 + 11 (x + 1)(x − 3)2(x2 + 5) = A x + 1 + B x − 3 + C (x − 3)2 + Dx + E x2 + 5 Each linear term in the denominator on the left gets a constant above it on the right; the squared linear factor (x− 3) 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 √ a2 − x2 √ a2 + x2 √ x2 − a2 Substitution x = a sin t x = a tan t x = a sec t sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x sin2 x = 1 2 − cos(2x) 2 cos2 x = 1 2 + cos(2x) 2 sin(2x) = 2 sin x cos x • Powers of Trigonometric Functions: here are some strategies for dealing with these. ∫ sinm x cosn x dx Possible Strategy Identity to Use m odd Break off one factor of sin x and substitute u = cos x. sin2 x = 1 − cos2 x n odd Break off one factor of cos x and substitute u = sin x. cos2 x = 1 − sin2 x m, n even Use sin2 x + cos2 x = 1 to reduce to only powers of sin x sin2 x = 1 2 − cos(2x) 2 or only powers of cos x, then use table of integrals #39-42 cos2 x = 1 2 + cos(2x) 2 or identities shown to right of this box. ∫ tanm x secn x dx Possible Strategy Identity to Use m odd Break off one factor of sec x tanx and substitute u = sec x. tan2 x = sec2 x − 1 n even Break off one factor of sec2 x and substitute u = tan x. sec2 x = tan2 x + 1 m even, n odd Use identity at right to reduce to powers of sec x alone. tan2 x = sec2 x − 1 Then use table of integrals #51. Useful Trigonometric Derivatives and Antiderivatives d dx tan x = sec2 x d dx sec x = sec x tanx ∫ sec x dx = ln | 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 = 0 Note: this is the same as lim x→−∞ ex. lim x→∞ 1/x = 0 Note: the answer is the same for lim x→∞ 1/x2 and similar functions. lim x→0+ 1/x = ∞ Note: the answer is the same for lim x→0+ 1/x2 and similar functions. lim x→∞ lnx = ∞ lim x→0+ lnx = −∞ lim x→∞ arctan x = π/2 1. Evaluate the following. (a) Let u = sinx, so du = cos x dx. ∫ sin6 x cos3 x dx = ∫ sin6 x(1 − sin2 x) cos x dx Use cos2 x = 1 − sin2 x. = ∫ u6(1 − u2) du = ∫ (u6 − u8) du = u7 7 − u9 9 + C = sin7 x 7 − sin9 x 9 + C (b) Let x = 10 tan t, so dx = 10 sec2 t dt.      10 x y t x2 + 102 = y2 ⇒ y = √ x2 + 100 sec t = hyp adj = √ x2 + 100 10 tan t = opp adj = x 10 2. Solve the differential equation dy/dx = 2xy + 6x if the solution passes through (0, 5). dy dx = 2xy + 6x dy dx = 2x(y + 3) dy y + 3 = 2x dx Separate the variables. ∫ dy y + 3 = ∫ 2x dx ln |y + 3| = x2 + C |y + 3| = ex2+C Exponentiate each side to remove the ln. y + 3 = ±eCex2 |w| = z means w = ±z. y = −3 + Aex2 Replace ±eC with A. Now we use the initial condition y(0) = 5 to find the value of A. We have 5 = −3 + Ae0 ⇒ A = 8, so the solution is y = −3 + 8ex2 . 3. Find the second-degree Taylor polynomial for f(x) = √ x centered at x = 100. f(x) = x1/2 f(100) = 10 f ′(x) = 1 2 x−1/2 = 1 2 √ x f ′(100) = 1 2 √ 100 = 1 20 f ′′(x) = −1 4 x−3/2 = −1 4x3/2 f ′′(100) = −1 4 · 1003/2 = −1 4000 P2(x) = f(100) + f ′(100)(x − 100) + f ′′(100) 2! (x − 100)2 = 10 + x − 100 20 − (x − 100)2 8000 4. What is the maximum possible error that can occur in your Taylor approximation from the previous problem on the interval [100, 110]? We know that |f(x) − Pn(x)| ≤ Kn+1 (n + 1)! |x− x0|n+1. In this case, n = 2, x0 = 100, and x = 110 (the farthest from x0 that we are considering). K3 = max of |f ′′′(x)| on [100, 110] = max of | 3 8x5/2 | on [100, 110] = 3 8 · 1005/2 = 3 800, 000 Putting this all together, we have |f(x) − P2(x)| ≤ 3 800,000 3! |110− 100|3 = 1 1600 . 5. 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) ∫ ∞ 1 6 + cos x x0.99 dx For all x ≥ 1, we have 6 + cos x x0.99 ≥ 6 − 1 x0.99 = 5 x0.99 because the minimum value of cos x is −1. Since ∫ ∞ 1 5 dx x0.99 diverges (compute yourself or notice that p = 0.99 < 1), we know that the integral in question must diverge too. (b) ∫ ∞ 1 4x3 − 2x2 2x4 + x5 + 1 dx For all x ≥ 1, we have 4x3 − 2x2 2x4 + x5 + 1 ≤ 4x3 x5 = 4 1 x2 . (We’ve made the denominator smaller and the numerator larger, so the new fraction is larger.) 4 ∫ ∞ 1 dx x2 = 4 lim t→∞ ∫ t 1 dx x2 = 4 lim t→∞ −1 x ∣∣∣ t 1 = 4 lim t→∞ [ −1 t − −1 1 ] = 4[0 − (−1)] = 4 Therefore, the original integral in question must converge to a value less than 4. 6. (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−0.2x where x is the number of minutes. Note that the domain is x ≥ 0 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 1 (because it must account for 100% of events.) ∫ ∞ 0 ke−0.2x dx = lim t→∞ ∫ t 0 ke−0.2x dx = lim t→∞ ke−0.2x −0.2 ∣∣∣ t 0 = lim t→∞ ke−0.2t −0.2 − ke0 −0.2 = 0 − k −0.2 = 5k So, we have 5k = 1 or k = 0.2. (b) What fraction of calls last more than 3 minutes? ∫ ∞ 3 0.2e−0.2x dx = lim t→∞ ∫ t 3 0.2e−0.2x dx = lim t→∞ 0.2e−0.2x −0.2 ∣∣∣ t 3 = lim t→∞ −e−0.2t − (−e−0.6) = 0 + e−0.6 = e−0.6 ≈ 0.5488 Note that we could instead have computed 1 − ∫ 3 0 0.2e−0.2x dx 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.