Calculus II Review Sheet 1 Solutions

Review Sheet 1 Solutions 1. Prove by induction: n X i2 = i=1 n(n + 1)(2n + 1) 6 SOLUTION: • Prove the first case: If n = 1, then does 12 = k X 1·2·3 ? 6 Yes. k(k + 1)(2k + 1) . 6 i=1 Use the assumption to prove that the statement is true if n = k + 1. In that case, we want to show that • Assume true if n = k. That is: k+1 X i2 = i=1 i2 = (k + 1)(k + 2)(2k + 3) 6 Here now is the proof: k+1 X i2 = k X i2 + (k + 1)2 = i=1 i=1 k(k + 1)(2k + 1) + (k + 1)2 6 Get a common denominator and factor out k + 1: = (k + 1)(k(2k + 1) + 6(k + 1)) k(k + 1)(2k + 1) + 6(k + 1)2 = = 6 6 (k + 1)(k + 2)(2k + 3) (k + 1)(2k 2 + 7k + 6) = 6 6 2. State the definition of Z b f (x) dx. a The definition uses the limit of the Riemann sum. You might also state the definition using right endpoints, as we did below- It may help on other problems. Z b a n X n X ! b−a b−a f (x∗i ) · ∆xi = n→∞ lim f a+i· f (x) dx = n→∞ lim · n n i=1 i=1 KEY IDEA: If the definition is asked for, you must use a Riemann sum, and NOT the Fundamental Theorem of Calculus. 3. 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. 1 (c) Consider P an . If n→∞ lim an = 0, then the sum is said to converge. ∞ X 1 is divergent. n=1 n Although, if the terms do NOT go to zero, then the sum diverges by the Test for Divergence. FALSE. For example, 1/n → 0, but 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. 4. 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. √ SOLUTION: 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 ALTERNATE SOLUTION: If we use washers, we would have: Z 2 π  (y 2 + 1)2 − 52 dy 0 5. Write the area under y = endpoints) √ 3 1 + x, 1 ≤ x ≤ 4 as the limit of a Riemann sum (use right 3i . n SOLUTION: The ith right endpoint is: 1 + 3i f 1+ = n   s 3 The height at this endpoint is given by: 3i 1+1+ = n s 3 2+ 3i n Therefore, the integral is: lim n→∞ n X s 3 2+ i=1 3i 3 · n n Secondly, for the arc length, we need to differentiate y: 1 y 0 = (1 + x)−2/3 3 ⇒ 1 (y 0 )2 = (1 + x)−4/3 9 so the integral is: Z 4s 1+ 1 6. Find the Taylor series for f (x) = √ 1 dx 9(1 + x)4/3 x centered at a = 9. SOLUTION: The first few derivatives are: n 0 1 2 3 f (n) (x) √ x (1/2)x−1/2 −(1/4)x−3/2 (3/8)x−5/2 2 f (n) (4) 2 (1/4) −1/32 3/256 (I’ll try to make the numbers work out nicely on the exam) Therefore, √ 1 1 3 x = 2 + (x − 4) − (x − 4)2 + (x − 4)3 + · · · 4 2 · 32 6 · 256 Main Point: Recall the formula for the Taylor series, and be able to compute the first few terms of a Taylor series. 7. Find dy , dx if y = R 5x cos(x) cos(t2 ) dt SOLUTION: The general formula we got using the Fundamental Theorem of Calculus, Part I was: d Z h(x) f (t) dt = f (h(x))h0 (x) − f (g(x))g 0 (x) dx g(x) Therefore, in this particular case,   cos(25x2 ) · 5 + cos cos2 (x) · sin(x) 8. 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: 1 Zb favg = f (c) = f (x) dx b−a a So, we first compute the average, then we’ll find the c: e2 − 1 1Z 2 x e dx = 2 0 2 e2 − 1 so that c = ln 2 ! ≈ 1.16 9. Use a template series to find the series for R 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)! 10. Does the series converge (absolute or conditional), or diverge? ∞ X cos(n/2) Note that cos(n/2) < 1 for all n. Therefore, by direct comparison, 2 n=1 n + 4n cos(n/2) 1 < 2 Thus, the series converges absolutely by the direct comparison 2 n + 4n n test. ∞ X n2 + 1 (n + 1)2 + 1 5n (n + 1)2 + 1 1 (b) Using the ratio test, lim · = lim · = n→∞ 5n 5n+1 n2 + 1 n→∞ n2 + 1 5 n=1 1 The series converges absolutely by the ratio test. 5 (a) 3 (c) ∞ X 3n n2 n=1 3n+1 (n + 1)2 n! n+1 By the ratio test, lim · n 2 = lim n→∞ n→∞ n! (n + 1)! 3 n n  2 · 3 =0 n+1 11. Find the interval of convergence: ∞ X n2 xn (n + 1)2 |x|n+1 10n n + 1 2 |x| |x| (a) = The radius (Ratio) n→∞ lim · 2 n = n→∞ lim · n n+1 10 n |x| n 10 10 n=1 10 of convergence 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)   ∞ X 2 |3x − 2|n+1 n |3x − 2| (3x − 2)n n3n = x− < (b) (Ratio) n→∞ lim · = n→∞ lim n n+1 n n3 (n + 1)3 |3x − 2| n+1 3 3 n=1 1 From this, the radius of convergence is 1. The interval so far is: (−1/3, 5/3), so now check the endpoints. P (−1)n If x = − 13 , the sum becomes: , which converges (conditionally). If x = 53 , n the sum becomes the Harmonic Series, which diverges.   1 5 The interval of convergence is − , 3 3 ∞ n 2n−1 X (−1) x (c) n=1 (2n − 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: lim n→∞ x2 |x|2n+1 (2n − 1)! · = lim =0 n→∞ (2n)(2n + 1) (2n + 1)! |x|2n−1 for all x. Therefore, the interval of convergence is the set of all real numbers. 12. Evaluate the integral. You should be able to do these without the table of integrals. (a) 1 dy y 2 − 4y − 12 SOLUTION: Use partial fractions, where Z y2 1 1 A B = = + − 4y − 12 (y + 2)(y − 6) y+2 y−6 Then: 1 = A(y − 6) + B(y + 2) Now integrate: Z y2 ⇒ A = −1/8, B = −1/8 1 1 1 dy = − ln |y + 2| + ln |y − 6| − 4y − 12 8 8 4 (b) 2x + 3 2 + 2 dx 3x + 1 x + 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 Z 3 2x dx + dx x2 + 9 x2 + 9 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 hy√ 2 potenuse is x2 + 9, sec(θ) = x3 −9 , and dx = 3 sec2 (θ). The integral becomes: Z 3 9 sec2 (θ) dx = dθ = θ = tan−1 (x/3) 2 2 x +9 9 sec (θ) Z Altogether, we get: Z (c) Z 2x + 3 2 2 + 2 dx = ln |3x + 1| + ln(x2 + 9) + tan−1 (x/3) + C 3x + 1 x + 9 3 x2 cos(3x) dx Use integration by parts using a table: + x2 cos(3x) Z 1 2x 2 − 2x 13 sin(3x) ⇒ x2 cos(3x) dx = x2 sin(3x)+ cos(3x)− sin(3x)+C 1 + 2 − 9 cos(3x) 3 9 27 1 − 0 − 27 sin(3x) (d) Z 2 |x − 1| dx Here, we want to split the integral at x = 1: −2 Z 2 −2 (e) Z (f) Z 1 −2 −x + 1 dx + Z 2 x − 1 dx = 5 1 dx Use a u, du substitution with u = ln(x), du = x ln(x) Z Z |x − 1| dx = 1 x dx. This gives: Z dx 1 = du = ln |u| = ln(ln(x)) + C x ln(x) u √ x x − 1 dx It’s handy to get rid of square roots where possible. Here, try u = √ x − 1 so that 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 5 13. The velocity function is v(t) = 3t − 5, 0 ≤ t ≤ 3 (a) Find the displacement. (b) Find the distance traveled. SOLUTION: The displacement is Z 3 3t − 5 dt = − 0 3 2 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 |3t − 5| dt = 0 Z 5/3 5 − 3t dt + 0 Z 3 3t − 5 dt = 5/3 14. Use a template series to find the Maclaurin series for: 25 16 41 + = 6 6 6 R ex −1 x dx SOLUTION: Starting with the series for ex , we have: ∞ X 1 1 1 xn ex = 1 + x + x 2 + x 3 + x 4 + · · · = 2 3! 4! n=0 n! Therefore, subtract one and divide by x to get: ∞ X ex − 1 1 1 1 xn−1 = 1 + x + x2 + x3 + · · · = x 2 3! 4! n=1 n! Finally, integrate both sides: Z ∞ X ex − 1 xn dx = C + x n=1 n! · n 15. Same idea as the previous problem. Start with 1/(1 + t3 ), which looks like the sum for a geometric series (with r = (−t3 )), and: ∞ X 1 = (−1)n t3n 1 + t3 n=0 Now we can multiply both sides by t to get the integrand: ∞ X t = (−1)n t3n+1 1 + t3 n=0 Finally, integrate both sides: Z ∞ 3n+2 X t n t dt = C + (−1) 1 + t3 3n + 2 n=0 6 16. Given y below, find the arc length: y= Z x q√ t − 1 dt ⇒ y0 = q√ x−1 ⇒ 1 + (y 0 )2 = √ √ x−1+1= x 1 Therefore, L= Z 16 x1/4 dx = 1 4 5/4 x 5 16 1 4 124 = (25 − 1) = 5 5 17. Note that we’re just rotating the curve y = 1 − x2 , and not the entire region between the x−axis and the curve (like we did for a volume of revolution). Therefore, there is only one surface. r = 2 − y = 2 − (1 − x2 ) = 1 + x2 q √ ds = 1 + (f 0 (x))2 dx = 1 + 4x2 dx The integral is therefore Z 1 √ 2π(1 + x2 ) 1 + 4x2 dx 0 18. A leaky 10 kg bucket is lifted from the ground to a height of 12 m at a constant speed with a rope that weighs 0.8 kg/m. Initially, the bucket contains 36 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12 m level. Set up the computations you would make to determine how much work is done. You do not have to actually compute the work, just set up the computations and/or integrals needed. You may keep gravity as g m/s2 , and use the constant σ if you need the density of water (in kg/m3 ). SOLUTION: We compute the work separately for the bucket, the rope, then the water: • For the bucket, the mass is 10 kg, so the weight (force) is 10g. The distance is 12 m, so the work for the bucket is 120g. • The mass of the rope is 0.8 kg/m, so it actually “weighs” 0.8g N/m. So if we’re x meters from the top, the a segment ∆x meters has a force 0.8g∆x N, and it travels x meters. Add all the segments together: W = Z 12 (0.8g)x dx 0 • For the water, we lose 36 kg over a 12 meter distance, so the amount of water x meters from the top would be: 36 − 3x kg. This is mass, so we need weight- That would be (36 − 3x)g N (in this case, we don’t need σ- If the water had been given in cubic meters, we would have needed it). This force is constant over the ∆x units up, so that the overall work is: W = Z 12 0 (Also see #17 in Section 6.4) 7 (36 − 3x)g dx