Calculus II Review Sheet 1 Solutions 2015

Review Sheet 1 Solutions Z 1. State the definition of b f (x) dx assuming equal subintervals and right endpoints. a SOLUTION: Z b f (x) dx = lim n→∞ a n X f (x∗i ) i=1   n X b−a b−a · ∆xi = lim f a+i· · n→∞ n n i=1 KEY IDEA: If the definition is asked for, you must use the Riemann sum, and NOT the Fundamental Theorem of Calculus. 2. True or False, and give a short reason: (a) The Alternating Series Test is sufficient to show that a series is conditionally convergent. FALSE, because if you go directly to the Alternating Series Test, you do not know if the series converges absolutely. (b) You can use the Integral Test to show that a series is absolutely convergent. TRUE. The Direct Comparison, Limit Comparison, Ratio Test and Integral tests were all used on positive series. P (c) Consider an . If lim an = 0, then the sum is said to converge. n→∞ FALSE. For example, 1/n → 0, but ∞ X 1 is divergent. n n=1 Although, if the terms do NOT go to zero, then the sum diverges by the Test for Divergence. 1 (d) The sequence an = 0.1n converges to 1−0.1 FALSE. This is a sequence, not a series. The sequence converges to zero. (e) All continuous functions have derivatives. FALSE: For example, f (x) = |x| is continuous, but is not differentiable at x = 0. Side remark: This problem was given to you to compare it against the next problem. (f) All continuous functions have antiderivatives. TRUE. That is the Fundamental Theorem of Calculus, Part I. √ 3. Set up an integral for the volume of the solid obtained by rotating the region defined by y = x − 1, y = 0 and x = 5 about the y−axis. Find the work involved if this was a tank filled with water and we wanted to pump it all out of the top (You may assume the measurements are in meters, and g = 9.8. The density of water is 1000 kg/m3 ). √ • For the volume, we might use shells. In that case, the height of a shell is x − 1 and the radius is x. The volume is then: Z 5 √ 2πx x − 1 dx 1 • For the work, take a disk and lift it. A disk at height y has to be lifted 2 − y units, and it has radius x = y 2 + 1. The work is therefore Z 2 (9.8)(1000)π(y 2 + 1)2 (2 − y) dy 0 1 4. Write the area under y = th SOLUTION: The i √ 3 1 + x, 1 ≤ x ≤ 4 as the limit of a Riemann sum (use right endpoints) right endpoint is: 1 +  f 3i 1+ n 3i n. The height at this endpoint is given by: r  = 3 3i 1+1+ = n r 3 2+ 3i n Therefore, the integral is: n X lim n→∞ r 3 2+ i=1 3i 3 · n n Secondly, for the arc length, we need to differentiate y: y0 = 1 (1 + x)−2/3 3 ⇒ (y 0 )2 = 1 (1 + x)−4/3 9 so the integral is: 4 Z s 1+ 1 5. Find the Taylor series for f (x) = √ 1 dx 9(1 + x)4/3 x centered at a = 9. SOLUTION: Look at the table to see if a pattern is showing up. This pattern is a little tricky, so at the least you should write down the first few terms of the sum (to get a big chunk of the points). n f (n) (x) f (n) (9) 0 x1/2 1 1 −1/2 2x 2 − 212 x−3/2 3 1·3 −5/2 23 x 4 −7/2 − 1·3·5 24 x − 1·3·5 24 37 5 .. . 1·3·5·7 −9/2 25 x 1·3·5·7 25 39 3 1 2·3 − 22133 1·3 23 35 .. . (−1)n+1 1·3·5····(2n−3) 2n 32n−1 n with the pattern starting at n = 2. If you found the pattern, write the general formula. At a minimum, write down the first few terms as shown. √ x=3+ In general: √ 6. Find dy dx , if y = R 5x cos(x) 1 1 1 (x − 9) − 2 3 (x − 9)2 + 3 3 (x − 9)3 + · · · 2·3 2 3 ·2 2 3 · 3! ∞ X 1 1 · 3 · 5 · · · · (2n − 3) x = 3 + (x − 9) + (−1)n+1 (x − 9)n n 32n−1 n! 6 2 n=2 cos(t2 ) dt 2 SOLUTION: The general formula we got using the Fundamental Theorem of Calculus, Part I was: d dx Z h(x) f (t) dt = f (h(x))h0 (x) − f (g(x))g 0 (x) g(x) Therefore, in this particular case,
cos(25x2 ) · 5 + cos cos2 (x) · sin(x) 7. Let f (x) = ex on the interval [0, 2]. (a) Find the average value of f . (b) Find c such that favg = f (c). SOLUTION: Remember the theorem: If f is continuous on [a, b], then there is a c in [a, b] so that: favg 1 = f (c) = b−a Z b f (x) dx a So, we first compute the average, then we’ll find the c: Z 1 2 x e2 − 1 e dx = 2 0 2  2  e −1 so that c = ln ≈ 1.16 2 R 8. Use a template series to find the series for cos(x2 ) dx. SOLUTION: The template series is the series for cos(x). In that formula, substitute x2 for x, then integrate: cos(x2 ) = ∞ X (−1)n n=0 (x2 )2n (2n)! Z ⇒ cos(x2 ) dx = ∞ X (−1)n n=0 x4n+1 +C (4n + 1)(2n)! 9. Does the series converge (absolute or conditional), or diverge? ∞ X cos(n/2) cos(n/2) 1 Note that cos(n/2) < 1 for all n. Therefore, by direct comparison, < 2 2 2 n + 4n n + 4n n n=1 Thus, the series converges absolutely by the direct comparison test. ∞ X (n + 1)2 + 1 5n (n + 1)2 + 1 1 1 n2 + 1 Using the ratio test, lim · = lim · = The series (b) n n+1 2 2 n→∞ n→∞ 5 5 n +1 n +1 5 5 n=1 converges absolutely by the ratio test.  2 ∞ X 3 n n2 n! 3 3n+1 (n + 1)2 n+1 (c) By the ratio test, lim · n 2 = lim · =0 n→∞ n→∞ n! (n + 1)! 3 n n n+1 n=1 (a) 10. Find the interval of convergence:  2 ∞ X (n + 1)2 |x|n+1 10n n+1 |x| |x| n2 xn (a) (Ratio) lim · 2 n = lim · = The radius of conn n+1 n→∞ n→∞ 10 10 n |x| n 10 10 n=1 vergence is 10. Check the endpoints: P P If x = 10, the sum is n2 , which diverges. If x = −10, the sum is (−1)n n2 , which still diverges. The interval of convergence is therefore: (−10, 10) 3 ∞ X |3x − 2|n+1 n3n n |3x − 2| (3x − 2)n |3x − 2| (Ratio) lim · = lim (b) = From this, n n n→∞ (n + 1)3n+1 n→∞ n + 1 n3 |3x − 2| 3 3 n=1 the radius of convergence is 3. The interval so far is: (−1/3, 5/3), so now check the endpoints. P (−1)n 5 If x = − 13 , the sum becomes: n , which converges (conditionally). If x = 3 , the sum becomes the Harmonic Series, which diverges.  1 5 The interval of convergence is − , 3 3 ∞ n 2n−1 X (−1) x (c) (2n − 1)! n=1 Be careful with the indices on this one. If the nth term has index 2n − 1, then the n + 1st index is 2(n + 1) − 1 = 2n + 1. Furthermore, 1 · 2 · 3 · · · (2n − 2)(2n − 1) 1 (2n − 1)! = = (2n + 1)! 1 · 2 · 3 · · · (2n − 2)(2n − 1)(2n)(2n + 1) (2n)(2n + 1) Now apply the ratio test: |x|2n+1 (2n − 1)! x2 · = lim =0 2n−1 n→∞ (2n + 1)! n→∞ (2n)(2n + 1) |x| lim for all x. Therefore, the interval of convergence is the set of all real numbers. 11. Evaluate the integral. Z 1 (a) dy 2 y − 4y − 12 SOLUTION: Use partial fractions, where 1 A B 1 = = + y 2 − 4y − 12 (y + 2)(y − 6) y+2 y−6 Then: 1 = A(y − 6) + B(y + 2) ⇒ A = −1/8, B = −1/8 1 1 1 dy = − ln |y + 2| + ln |y − 6| Now integrate: 2 y − 4y − 12 8 8 Z 2 2x + 3 (b) + dx 3x + 1 x2 + 9 The first integral is a straight u, du substitution with u = 3x + 1. The second integral is similar to the one on the quiz except that we don’t have to complete the square- split the integral into two: Z Z 2x 3 dx + dx x2 + 9 x2 + 9 Z On the first integral, use u, du substitution with u = x2 + 9, and on the second integral, use either trig substitution (with triangles), or recall the formula for the inverse tangent. √ To use trig substitution, we’ll use the triangle where tan(θ) = x/3, so the hypotenuse is x2 + 9, √ 2 sec(θ) = x3 −9 , and dx = 3 sec2 (θ). The integral becomes: Z Z 3 9 sec2 (θ) dx = dθ = θ = tan−1 (x/3) x2 + 9 9 sec2 (θ) Altogether, we get: Z 2 2x + 3 2 + 2 dx = ln |3x + 1| + ln(x2 + 9) + tan−1 (x/3) + C 3x + 1 x + 9 3 4 Z (c) Z x2 cos(3x) dx Use integration by parts using a table: + x2 − 2x + 2 − 19 cos(3x) − 0 1 − 27 sin(3x) cos(3x) 1 3 sin(3x) Z ⇒ x2 cos(3x) dx = 1 2 2x 2 x sin(3x) + cos(3x) − sin(3x) + C 3 9 27 2 |x − 1| dx Here, we want to split the integral at x = 1: (d) −2 Z 2 Z 1 −2 −2 2 x − 1 dx = 5 −x + 1 dx + |x − 1| dx = 1 Z dx Use a u, du substitution with u = ln(x), du = x1 dx. This gives: x ln(x) Z Z dx 1 = du = ln |u| = ln(ln(x)) + C x ln(x) u Z √ √ x x − 1 dx It’s handy to get rid of square roots where possible. Here, try u = x − 1 so that (e) (f) Z u2 = x − 1, or u2 + 1 = x. This also gives 2u du = dx. Now, Z Z Z √ 2 2 x x − 1 dx = (u2 + 1) · u · 2u · du = 2u4 + 2u2 du = (x − 1)5/2 + (x − 1)3/2 + C 5 3 Z ∞ x2 √ (g) dx 1 + x3 0 SOLUTION: Before you integrate, consider √ what the integrand does as x gets large. In this case, the fraction will behave like x2 /x3/2 = x, and that will diverge. Unfortunately, we can’t do a direct comparison, but we can integrate and expect divergence. Let u = 1 + x3 so that du = 3x2 dx, and Z Z x2 1 2 √ dx = u−1/2 du = u1/2 3 3 1 + x3 And taking the limit as u → ∞ will be infinity, so the integral diverges. Z (h) sin2 (x) cos3 (x) dx SOLUTION: Reserve one of the cosines for u = sin(x), du = cos(x) dx, and re-write the remaining cos2 (x) = 1 − sin2 (x): Z Z 1 1 2 2 u (1 − u ) du = u2 − u4 du = sin3 (x) − sin5 (x) + C 3 5 Z (i) sin−1 (x) dx SOLUTION: Use integration by parts with sin−1 (x) in the center column. sign u + −1 − sin (x) √ 1/ 1 − x2 dv ⇒ 1 x 5 x sin−1 (x) − Z √ x dx 1 − x2 For this integral, use u = 1 − x2 , du = −2x dx, and Z p sin−1 (x) dx = x sin−1 (x) + 1 − x2 + C Z (j) dx x2 − 9 SOLUTION: This is a trig substitution problem, with x = 3 sec(θ) (so we can also get the right triangle with hypotenuse x and length of the side adjacent θ is 3). Then dx = 3 sec(θ) tan(θ)dθ, and Z Z Z 3 sec(θ) tan(θ)dθ 3 sec(θ) tan(θ)dθ p = = sec(θ) dθ = ln | sec(θ) + tan(θ)| + C 3 tan(θ) 9(sec2 (θ) − 1) √ Using the right triangle described above, we back substitute to get x again: Z √ x dx = ln + 3 x2 − 9 √ x2 − 9 +C 3 12. The velocity function is v(t) = 3t − 5, 0 ≤ t ≤ 3 (a) Find the displacement. (b) Find the distance traveled. SOLUTION: The displacement function is the antiderivative of velocity: Z 3 s(t) = 3t − 5 dt = t2 − 5t + C 2 where C is the initial displacement. The distance traveled is the integral of the absolute value of velocity: Z 3 |3t − 5| dt 0 Checking 3t − 5, we see that it is negative if t < 5/3, so we break up the integral: Z 3 Z |3t − 5| dt = 0 5/3 Z 3 5 − 3t dt + 0 3t − 5 dt = 5/3 13. Set up the integral that will give the length of the graph of y = √ 41 25 16 + = 6 6 6 √ x − x2 + sin−1 ( x), with 0 ≤ x ≤ 1. We want to set up the integral for the arc length: Z L= b p 1 + (f 0 (x))2 dx a Let’s work on getting and simplifying the integrand first. 2  1 1 + (f 0 (x))2 = 1 +  (x − x2 )−1/2 (1 − 2x) + q 2 1 1 −1/2  x √ 2 2 1−( x ) This simplifies a bit 1+  2 1 1 − 2x 1 4 (1 − x)2 1 1 √ +√ =1+ =1+ −1= 4 4 x(1 − x) x x x − x2 x − x2 6 And finishing, we were not asked to evaluate the integral: Z 1 1 √ dx L= x 0 Side Note: Often on exams/quizzes, I will ask you to set up the integral only. This is because if an error is made in the setup, it can mess everything up past that point. Plus, this saves you some time on the exam. 14. (You can use a calculator at home for the following problem, which would be suitably changed for an ∞ X (−1)n+1 correct to three decimal places. exam with no calculators) Find the sum of the series n3 n=1 SOLUTION: With a calculator, we can simply keep summing until Rn ≤ bn+1 = 1 ≤ 5 × 10−4 (n + 1)3 Using a calculator, we see that 1 ≈ 4.5 × 10−4 133 so we can sum using the first 12 terms. Doing that gives us an estimate: ∞ X 1 ≈ 1.1989 n3 n=1 15. Evaluate each sum. Hint on part (c): Partial Fractions (a) ∞ X (−1)n π 2n 62n (2n)! n=0 SOLUTION: Hopefully the sum looks familiar since we have even powers and even factorials∞ ∞  π  √3 X X (−1)n x2n (−1)n π 2n cos(x) = ⇒ = cos = (2n)! 62n (2n)! 6 2 n=0 n=0 (b) ∞ X (−1)n−1 n=1 3n+2 22n SOLUTION: This is a geometric series in disguise.  n ∞ ∞ n+2 X X 3 27/4 27 n−1 3 (−1) = (−9) − = = 2n 2 4 1 + (3/4) 7 n=1 n=1 (c) ∞ X 1 n(n + 3) n=1 SOLUTION: Using the hint, we can expand the sum using partial fractions. A B 1 = + n(n + 3) n n+3 ⇒ 1 = A(n + 3) + Bn Solving for A, B, we get: 1 1 = n(n + 3) 3 7  1 1 − n n+3  Writing out the sum (factoring out the 1/3), we can write down Sn : n Sn = 1X1 1 − 3 k k+3 k=1 Focusing on this sum, terms are canceling out:         1 1 1 1 1 1 1 + − + − + − + ···+ 1− 4 2 5 3 6 4 6         1 1 1 1 1 1 1 1 ··· + − + − + − + − n−3 n n−2 n+1 n−1 n+2 n n+3 It looks like we’re left with: 1 Sn = 3  1 1 1 1 1 1+ + − − − 2 3 n+1 n+2 n+3  11 As n → ∞, Sn → 3·6 = 11 18 NOTE: This is probably more involved than anything on the exam, but it was Section 11.2, Exercise 45. 16. Determine whether the integral is convergent or divergent: Typo... Sorry about the partial question. 8