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08A-Improper Integral Practice Solutions

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08A-Improper Integral Practice Solutions Improper Integral Practice Problems These problems are taken from old quizzes I have given on improper integrals. Solutions will be posted on the course webpage later, so you can use these to gauge your preparedness for the quiz. 1) Evaluate each improper integral below using antiderivatives. Make sure your work is correct! e ∞ ⌠ dx ⌠ dx a) ⎮ Improper at x = 1 b) ⎮ Improper at infinity x ln(x) x ln(x) ⌡1 ⌡e So this integral converges to 2/. which diverges. ∞ ⌠ dx π 2 c) ⎮ Improper at 0 and infinity d) sec (x)dx Improper at π/2 x ln(x) ∫0 ⌡1 We must split this integral into two We must split this one up into two pieces, parts, for it is improper at x = 0 and at also. So: infinity. We could split it into the integral from parts a) and b), and since the integral from part b) diverges, this whole thing diverges. Now as a gets close to π/2, tan(a) blows up, so the first piece, hence the entire integral diverges. 1) (Evaluate using antiderivatives—continued) 2 ∞ ⌠ dx ⌠ dx e) Improper at x = 1 f) Improper at infinity ⎮ x ln2 (x) ⎮ x ln2 (x) ⌡1 ⌡2 But as c goes to one, ln(c) goes to zero so this integral diverges. 2 ⌠ dz g) Improper at both endpoints ⎮ 2 ⌡−2 4 − z The antiderivative can be found by the trig substitution z = 2sin(θ); the antiderivative turns out to be sin–1(z/2). So breaking the integral into two pieces (zero seems convenient) and using appropriate limits on each part gives: 2) Test each integral for convergence or divergence. 3 ⌠ dt a) ⎮ 3 ⌡0 t t + 1 3 ⌠ dt Improper at x = 0, where the t is much larger than the t , so this “looks like” the p-type ⎮ ⌡0 t 1 1 which converges since p < 1. In fact, < (bigger denominator = smaller fraction), and 3 t t + t 1 ⌠ dt the p-type integral ⎮ converges, so by the comparison test, this integral also converges. ⌡0 t 2) (Test for convergence or divergence—continued) ∞ ⌠ dt b) ⎮ 3 ⌡3 t − t This integral is improper at infinity only, and for large t we know that t3 is the dominant part. 1/ t3 t3 − t t3 − t Let’s do limit comparison to 1/ t3 : lim = lim = lim . Now this last t→∞ 3 t→∞ 3 t→∞ 3 1/ t − t t t limit is clearly one (divide top and bottom by t3, or use continuity of the square root to move the limit inside the radical). So by the limit comparison test, our integral behaves the same as ∞ ∞ ⌠ dt ⌠ dt = , which converges because it is p-type with p > 1. ⎮ 3 ⎮ 3/2 ⌡3 t ⌡3 t ∞ ⌠ x2 c) dx ⎮ 4 4 x −1 ⌡ 2 4 2 This time, the integrand “looks like” x / x = 1/ x = 1/ x . We know the integral of 1/x out to infinity diverges. (It is p-type with p = 1, or just note its antiderivative is ln(x).) In fact, by x2 x2 1 1 direct comparison (limit comparison would also work), 4 > 4 = 2 = . Since our x −1 x x x function is larger than one whose integral diverges, our function’s integral also diverges. ∞ ⌠ dx d) ⎮ x ⌡0 e − x This integral is improper only at infinity. We expect it to behave like the integral of 1/ex, which x 1/ (e − x) ex ex ex converges. In fact, by limit comparison, lim x = lim x = lim x = lim x = 1. x→∞ 1/ e x→∞ e − x x→∞ e −1 x→∞ e (The second- and third-to-last equalities are by L’Hopital’s rule.) Since the limit of the ratio is a ∞ ⌠ dx positive constant our integral behaves like ⎮ x , which converges. ⌡0 e 2) (Test for convergence or divergence—continued) ∞ ⌠ dx e) ⎮ 3 3 2 ⌡0 x + x + x +1 This integral is only improper at infinity. For large x, the x3 term is larger than all the others, so 1 1 this looks like = . In fact, 3 3 x x 1/ 3 x3 + x2 + x +1 x 1 lim = lim = lim (the last step by x→∞ 1/ x x→∞ 3 3 2 x→∞ 3 2 3 x + x + x +1 1+1/ x +1/ x +1/ x dividing top and bottom by x), and this clearly equals one. So by the limit comparison test, this ∞ ⌠ dx integral behaves like ⎮ which is p-type and diverges. ⌡0 x 1 ⌠ dy f) ⎮ y y2 ⌡0 + This is pretty much exactly the same problem as 2)a), and converges with the same explanation. π /2 ⌠ sin(q) g) dq ⎮ q2 ⌡0 sin(q) This integral is improper only at q = 0. Now lim = 1 so near zero this integrand looks like q→0 q sin(q) / q2 sin(q) 1/q. In fact, by limit comparison, lim = lim = 1 which is a positive constant. So q→0 1/ q q→0 q π /2 ⌠ dq our integral behaves like the p-type integral that diverges. ⎮ q ⌡0 2) (Test for convergence or divergence—continued) 5 ⌠ dx h) (Extra credit) ⎮ x 2 ⌡4 2 − x This integral is improper at x = 4. Not being able to antidifferentiate, or having any other idea what to try, let’s try the trick of limit comparing with 1/x, or (because the impropriety happens at 1/ (x − 4) 2x − x2 x = 4) in this case 1/(x – 4). So: lim = lim . This is a 0/0 type of x→4 x 2 x→4 1/ (2 − x ) x − 4 indeterminate form, so using L’Hopital’s rule shows us this equals 2x ln(2) − 2x lim = 16ln(2) − 8. This is a positive constant, so by limit comparison test, out x→4 1 5 ⌠ dx integral behaves the same as ⎮ which diverges (antidifferentiate, or substitute u = x – 4 ⌡4 x − 4 to turn this into a nice p-type. 3) Approximate each integral to within 0.001 ∞ 2 a) e− x dx ∫0 a 2 ∞ 2 Rewrite as e− x dx + e− x dx and determine how large a has to be to make the second term ∫0 ∫a 2 ∞ 2 ∞ − x − x − x − x ignorable. Since e < e we know that e dx < e dx . We can compute that this last ∫a ∫a integral is simply e–a so we choose a so that e–a is less than 0.001 (a = 10 will do). Now using a 10 2 calculator or other tool we find e− x dx ≈ 0.886 so this is our approximation to within 0.001. ∫0 (The exact value of this integral is / 2 which is 0.886227 to six places. π ∞ ⌠ dx b) ⎮ x4 2tan−1(x) ⌡1 + We apply the same technique as above, only this time we will compare to 1/x4. By using limits ∞ dx 1 1000 ⌠ 3 and antidifferentiation, ⎮ 4 = 3 . To make this less than 0.001 we need a > ≈ 6.934 . ⌡a x 3a 3 Well use 7 because it’s nicer. Using our favorite electronic calculator, we find 7 ⌠ dx ≈ 0.207 to three decimal places. ⎮ x4 2tan−1(x) ⌡1 + .
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