Final Exam Review Calculus II Sheet 2 - Answer Key

Final Exam Review Calculus II Sheet 2 1. True or False, and give a short reason: 2n+3 (a) The Ratio Test will not give a conclusive result for 3n4 +2n 3 +3n+5 TRUE. The ratio test fails for p−like series (the limit will be 1). To show convergence, use a direct or limit comparison (Limit comparison with 1/n3 ) P (b) If ∞ X an converges for some large k, then so will ∞ X an . n=1 n=k TRUE. The first few terms of a sum are irrelevant when looking at whether or not the sum converges (although they will effect what the sum converges to). (c) If f is continuous on [0, ∞) and lim f (x) = 0, then x→∞ R∞ 0 f (x) dx converges. FALSE. For example, 1/(x − 1). (The idea here is that functions must go to zero fast enough). (d) If f is continuous and Z 9 f (x) dx = 4, then Z 3 xf (x2 ) dx = 4. 0 0 FALSE. Z 3 0 u (1/2) du xf (x2 ) dx ⇒ x=0 x=3 = x2 1Z 9 1 = dx ⇒ f (u) du = · 4 = 2 ⇒u=0 2 0 2 ⇒u=9 2. Short Answer: (a) Suppose the series cn 3n converges. Will cn (−2)n also converge? For what P values of x will the series cn (x − 2)n converge? SOLUTION: For the first part of the question, we can look as if it were a power P series cn xn that converged at x = 3. Therefore, the series would converge for all |x| < 3, and x = −2 is within that range. On the other hand, if we think of the P series as cn (x − 2)n , then the series converges for all x so that |x − 2| < 3, or at least within the interval (−1, 5] (the convergence at x − 2 = 3 might be conditional, that’s why we did not include x = −1). P P (b) If an , bn are series with positive terms, and an , bn both go to zero as n → ∞, an = 0? then what can we conclude if n→∞ lim bn P SOLUTION: We can conclude that the terms of an are going to zero faster than P P P P bn . Thus, if bn is convergent, so is an , and if an is divergent, so is bn . P P (c) What is the derivative of sin−1 (x)? Of tan−1 (x)? What is the antiderivative of each? 1 1 −1 SOLUTION: The derivative of sin−1 (x) is √1−x (x) is 1+x 2 2 . The derivative of tan To integrate either, use integration by parts. For sin−1 (x), −1 + sin √ (x) 1 ⇒ − 1/ 1 − x2 x Z sin−1 (x) dx = x sin−1 (x) − 1 Z √ x dx 1 − x2 For this integral, use u = 1 − x2 , du = −2x dx to get a final answer: Z sin−1 (x) dx = x sin−1 (x) + √ 1 − x2 + C −2n (d) Find the sum: ∞ n=1 e SOLUTION: The sum of a geometric series, in its general form is: P ∞ X arn = n=k In this case, r = e−2 , so the sum is: ark 1−r e−2 1−e−2 3. Suppose h(1) = −2, h0 (1) = 2, h00 (1) = 3, h(2) = 6, h0 (2) = 5, and h00 (2) = 13, and h00 R 2 00 is continuous. Evaluate 1 h (u) du. Z 2 h00 (u) du = h0 (2) − h0 (1) = 5 − 2 = 3 1 n X s 3i 3 4. Determine a definite integral representing: lim 1+ [For extra practice, try n→∞ n i=1 n writing the integral so that a = 5]. SOLUTION: We need to find f so that 3i f 5+ = n  Here is one: f (x) = √  s 1+ 3i n x − 4. Our solution is: Z 8√ x − 4 dx 5 5. Evaluate Z 5 (1 + 2x) dx by using the definition of the integral (use right endpoints). 2 SOLUTION: The ith right endpoint is 2 + 3i . Evaluating f at this endpoint gives the n following, from which we get the Riemann sum:  3i 1+2 2+ n   6i 6i =1+4+ =5+ n n ⇒ n  X i=1 6i 3 5+ n n  Now break apart the sum to evaluate: n n X 3 6X 3 6 n(n + 1) lim 5 1+ i = lim 5n + n→∞ n n→∞ n n i=1 n 2 i=1 ! ! = lim 15 + 9 · n→∞ n+1 = 24 n (Note that geometrically, the area of the trapezoid is also 24). 6. For each function, find the Taylor series for f (x) centered at the given value of a: SOLUTION: 2 (a) f (x) = 1 + x + x2 at a = 2 We need f (2), f 0 (2), f 00 (2): f (2) = 7. f 0 (x) = 1 + 2x, so f 0 (2) = 5. f 00 (x) = 2 Now, 1 + x + x2 = 7 + 5(x − 2) + (b) f (x) = 1 x n 0 1 2 3 4 .. . 2 (x − 2)2 = 7 + 5(x − 2) + (x − 2)2 2! at a = 1. We need to compute derivatives: f n (x) x−1 −x−2 2x−3 −(3 · 2)x−4 4 · 3 · 2x−5 .. . f n (1) 1 −1 ∞ X 2 f (n) (1) n (−1)n (x − 1)n −(3 · 2) ⇒ n! = (−1) ⇒ n=0 4·3·2 .. . n (−1)n n!x−(n+1) (−1)n n! Alternatively, we could use the geometric series: ∞ ∞ X X 1 1 (−1)n (x − 1)n (1 − x)n = = = x 1 − (1 − x) n=0 n=0 7. Find a so that half the area under the curve y = the area lies in the interval [a, 4]. 1 x2 lies in the interval [1, a] and half of SOLUTION: We could set this up multiple ways- here is one way to do it: Z a 1 1Z 4 1 1 3 8 1 dx = dx ⇒ − + 1 = ⇒ a = 2 2 x 2 1 x a 8 5 8. Compute the limit, by using the series for sin(x): lim x→0 sin(x) x SOLUTION: The series for the sine function is: sin(x) = ∞ X (−1)n n=0 x(2n+1) 1 1 1 = x − x3 + x5 − x7 + · · · (2n + 1)! 3! 5! 7! Therefore, the series for sin(x)/x is: sin(x) 1 1 = 1 − x2 + x4 + · · · x 3! 5! To find the limit as x → 0, we can evaluate the series at x = 0, which leaves the limit as 1. 9. Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by y = x, y = 4x − x2 , about x = 7. SOLUTION: First, find the region of interest. y = 4x − x2 is an upside down parabola with x−intercepts at x = 0, x = 4. The point of intersection is x = 4x−x2 ⇒ 0 = 3x−x2 , or x = 0 and x = 3. Now the region of interest is between x = 0, x = 3, above the 3 line y = x and below the parabola y = 4x − x2 . Rotate about x = 7, and we will use cylindrical shells (Washers would be possible, but messy!). The height of the cylinder is (4x − x2 ) − x = 3x − x2 . The radius is 7 − x. Therefore, the integral for the volume is: Z 3 2π(7 − x)(3x − x2 ) dx 0 10. Evaluate each of the following: [The purpose of this problem is to get you to see the differences in notation] d Z sin(x) 3 (a) t dt. By FTC, part I: sin3 (x) · cos(x) − (3x)3 · 3 dx 3x d Z5 3 x dx = 0 (this is the derivative of a constant) (b) dx 1 Z 5 5 d 3 (c) x dx = x3 = 53 − 1 = 124. This is FTC, part II. 1 1 dx 11. Converge (absolute or conditional) or Diverge? (a) ∞ X (−1)n n P This will behave like (−1)n n1 , which only converges conditionn=1 (n + 1)(n + 2) ally. We can use the limit comparison test (with n1 ) to show that the series does not converge absolutely: n n · =1 lim n→∞ (n + 1)(n + 2) 1 The two series will diverge together, so the given series diverges. Now we use the Alternating Series Test to show that it converges conditionally: Each term is clearly positive, for n > 0. Is it decreasing? x 2 − x2 0 f (x) = f (x) = (x + 1)(x + 2) (x + 1)2 (x + 2)2 √ so the derivative is negative for x > 2 (or the terms of the series are decreasing for n > 2). Finally, show that the terms are going to zero: lim n→ n n 1 = lim 2 = lim =0 n→∞ n→∞ (n + 1)(n + 2) n + 3n + 2 2n + 3 (the last equality by l’Hospital’s rule). √ ∞ X n2 − 1 (b) 3 2 n=1 n + 2n + 5 P 1 It looks like it should converge by comparing it to , so we’ll try the limit n2 comparison test: √ √ √ √ n2 − 1 n2 n2 − 1 n4 n6 − n4 lim 3 · = lim · = lim n→∞ n + 2n2 + 5 n→∞ n3 + 2n2 + 5 n→∞ n3 + 2n2 + 5 1 1 4 (Don’t use l’Hospital’s rule!) Divide top and bottom by n3 : q lim n→∞ 1 n2 2 + n53 n 1− 1+ =1 By the limit comparison test, the given series converges (absolutely, but that is irrelevant since the terms are all positive anyway). (c) ∞ X 4k k=1 +k Use the ratio test: k! 4 + 4kk + 41k 4k+1 + (k + 1) k! 4k+1 + k + 1 · k = = (k + 1)! 4 +k (k + 1)(4k + k) (k + 1)(1 + 4kk ) The numerator approaches 4 as k → ∞ and the denominator goes to ∞ as k → ∞, so overall, the limit is 0. Therefore, this series converges (absolutely) by the Ratio Test. 12. Find the interval of convergence. (a) ∞ X nn xn By the root test, lim (nn xn )1/n = lim nx = ∞ Therefore, the only point n=1 n→∞ n→∞ of convergence is when x = 0. (The radius of convergence is also 0). Note: The root test is not used very often, but in this situation (where everything is raised to the nth power), this will make quick work of the problem. ∞ X (x + 2)n n4n n=1 Use the Ratio Test, as usual: |x + 2|n+1 n4n n + 1 |x + 2| |x + 2| lim · = n→∞ lim = < 1 This means that the n n→∞ (n + 1)4n+1 |x| n 4 4 radius of convergence is 4, and the interval so far is (−6, 2). P1 Check the endpoints: If x = 2, then the sum is which diverges. If x = −6, n P (−1)n , which converges. The interval of convergence is therefor then the sum is n −6 ≤ x < 2. ∞ X 2n (x − 3)n √ (c) n+3 n=1 Use the Ratio Test: √ s n+3 n+3 2n+1 |x − 3|n+1 √ · n = lim lim · 2|x − 3| = 2|x − 3| < 1 n n→∞ n→∞ 2 |x − 3| n+4 n+4 Therefore, the radius of convergence is 1/2 and the interval is 5/2 < x < 7/2. Now check endpoints: P (−1)n If x = 25 , the sum becomes √ , which converges by the Alternating Series test, n+3 P 1 7 and if x = 2 , the sum becomes √n+3 which diverges (p-series). (b) 13. Evaluate: 5 (a) Z ∞ 0 Z 1 dx By partial fractions, (x + 2)(x + 3) Z x+2 1 1 1 dx = − dx = ln |x + 2| − ln |x + 3| = ln (x + 2)(x + 3) x+2 x+3 x+3 As x → ∞, ln x+2 → ln(1) = 0. Altogether we get: x+3 Z ∞ 0 (b) Z √ √ R u( u + 3 u) du Simplify algebraically first, to get u3/2 + u4/3 du = 3 7/3 u 7 (c) 1 dx = 0 − ln(2/3) = ln(3/2) (x + 2)(x + 3) x2 dx (4 − x2 )3/2 Use √ θ is x, and side adjacent is √ a triangle whose hypotenuse is 2, side opposite 4 − x2 . Then, substitute 2 sin(θ) = x, 2 cos(θ) = 4 − x2 , and we get: Z Z Z 4 sin2 (θ) · 2 cos(θ) 2 dθ = tan (θ) dθ = sec2 (θ) − 1 dθ = tan(θ) − θ 23 cos3 (θ) Convert back using triangles to get: √ Z x − sin−1 (x/2) + C 4 − x2 tan−1 (x) dx Let u = tan−1 (x), so du = 1 + x2 Z (e) + +C Z (d) 2 5/2 u 5 1 1+x2 dx. Then the integral becomes 1 1 u du = u2 + C = (tan−1 (x))2 + C 2 2 1 dx − 4x ”Complete the Square” in the denominator to get x2 − 4x = (x − 2)2 − 4. Now, use q a triangle whose hypotenuse is x − 2, side adjacent is 2, and side opposite is (x − 2)2 − 22 . Then, Z √ x2 2 tan(θ) = q (x − 2)2 − 22 , 2 sec(θ) = x − 2, 2 sec(θ) tan(θ)dθ = dx Substituting, we get: Z √ Z Z 2 sec(θ) tan(θ) 1 dx = dθ = sec(θ) dθ = ln | sec(θ) + tan(θ)| + C 2 tan(θ) x2 − 4x [NOTE: You’ll be given the formulas as on the previous exam]. Final answer: x−2 ln + 2 6 q (x − 2)2 − 4 2 +C (f) Z x4 ln(x) dx Use integration by parts 1 5 1Z 4 1 1 + ln(x) x4 x dx = x5 ln(x) − x5 + C 5 ⇒ x ln(x) − − 1/x (1/5)x 5 5 5 25 (g) Z e−x sin(2x) dx. This is the type of integral for which we perform integration by parts twice to get the same integral on both sides of the equation: + sin(2x) e−x Z Z − 2 cos(2x) −e−x ⇒ e−x sin(2x) dx = −e−x sin(2x)−2e−x cos(2x)−4 e−x sin(2x) dx + −4 sin(2x) e−x so that Z (h) Z 3 1 √ dx x 0 Note that we have a vertical asymptote at x = 0, so Z 3 0 (i) 1 2 e−x sin(2x) dx = − e−x sin(2x) − e−x cos(2x) 5 5 Z Z 3 1 √ dx = lim+ x−1/2 dx = lim+ 2x1/2 T →0 T →0 x T 3 T √ √ =2 3−0=2 3 sin2 x cos5 x dx Recall our rules for dealing with powers of sine and cosine: If both are even, use the formulas for sin2 (x) and cos2 (x). If one (or both) are odd, try substitution: Z sin2 (x) cos4 (x) · cos(x) dx which means we want to write u = sin(x). Use the Pythagorean Identity: cos4 (x) = (1 − sin2 (x))2 , so that: Z 2 4 sin (x) cos (x)·cos(x) dx = Z 2  2 2 sin (x) 1 − sin (x) ·cos(x) dx = Z u2 (1−u2 )2 du Simplify this last integral, and integrate: Z 2 1 1 u6 − 2u4 + u2 du = u7 − u5 + u3 + C 7 5 3 so our final answer is: 1 7 2 1 sin (x) − sin5 (x) + sin3 (x) + C 7 5 3 14. Find a power series for x2 ln(5 − x) by first finding a series for 1/(5 − x) and determine the radius of convergence. SOLUTION: We start with the series suggested. I want to get it in the form 1/(1 − r), so I need to factor out the from the numerator: ∞ ∞ X 1 1 1 1 X (−1)n xn (−1)n xn   = = = 5−x 5 1− x 5 n=0 5n 5n+1 n=0 5 7 with radius of convergence: |x|/5 < 1, or |x| < 5. To get ln(5 − x), we integrate: Z ∞ X 1 (−1)n xn+1 dx = − ln(5 − x) = C + n+1 (n + 1) 5−x n=0 5 We see that − ln(5) = C, so we write that, and multiply both sides by −1 to get the series for ln(5 − x): ∞ X (−1)n+1 xn+1 ln(5 − x) = ln(5) + n+1 (n + 1) n=0 5 Finally, multiply through by x2 : x2 ln(5 − x) = ln(5) · x2 + ∞ X (−1)n+1 xn+3 n+1 (n + 1) n=0 5 The radius of convergence remains 5. 15. Find the Maclaurin series for xe−x . SOLUTION: Start with the series for ex , and substitute −x for x: e −x = Multiply both sides by x: xe−x = ∞ X (−1)n n x n! n=0 ∞ X (−1)n n+1 x n! n=0 16. Prove the following by induction: 1 · 2 + 2 · 3 + 3 · 4 + · · · + n · (n + 1) = n(n + 1)(n + 2) 3 SOLUTION: • Prove it for a first case: If n = 1, the 1·2= 1·2·3 3 Which is true. • Assume the statement is true for n = k, then use that to prove it true for n = k +1: Assume true for n = k: 1 · 2 + 2 · 3 + 3 · 4 + · · · + k · (k + 1) = k(k + 1)(k + 2) 3 And, we want to show that this implies that: 1 · 2 + 2 · 3 + · · · + k · (k + 1) + (k + 1)(k + 2) = 8 (k + 1)(k + 2)(k + 3) 3 So, starting with the left side of the equation, we want to get the right side. As is our usual practice, break up the sum to use the assumption: 1 · 2 + 2 · 3 + · · · + k · (k + 1) + (k + 1)(k + 2) = [1 · 2 + 2 · 3 + · · · + k · (k + 1)] + (k + 1)(k + 2) = k(k + 1)(k + 2) + (k + 1)(k + 2) 3 Factor out (k + 1)(k + 2) ! (k + 1)(k + 2)(k + 3) k +1 = = (k + 1)(k + 2) 3 3 Therefore, the statement is true for all positive integers n. 17. Find an integral for the surface area. SOLUTION: y = e−x so that ds = q 0≤x≤1 1 + (y 0 )2 dx = √ 1 + e−2x dx The radius is y = e−x , so we have: Z 1 √ 2πe−x 1 + e−2x dx 0 18. A bucket weighing 100 lbs is filled with sand weighing 500 lbs. A crane lifts the bucket from the ground to a point 80 ft in the air at a rate of 2 ft/s, but sand is leaking out of a hole at a rate of 3 lbs/s. Neglecting friction and the weight of the cable, find an expression (integral) that will compute the amount of work being done. SOLUTION: Another leaky bucket. We can compute the work of the bucket itself and the work from the sand separately. We’ll also ignore work from the cable. • For the bucket, the force is 100 lbs, it travels 80 ft. The work is 8000 ft-lbs. • For the sand, at a rate of 2 ft/s, it takes 40 seconds to travel 80 feet. If we lose 3 lbs per second, in 40 seconds we’ll lose 120 lbs (over the distance of 80 feet). Therefore, if we are x feet in the air, we have: 500 − 120 x lbs of sand, or 80 3 500 − x lbs 2 That’s the (constant) force over ∆x feet, so then all told, we integrate: Z 80 0 9 3 500 − x dx 2