Calculus II Solutions to the Review Questions Exam 3

Solutions to the Review Questions, Exam 3 1. Write the partial fraction decomposition for each of the following (do not actually solve for the coefficients): 3 − 4x2 B A C + = + (2x + 1)3 2x + 1 (2x + 1)2 (2x + 1)3 A B 7x − 41 C = + (b) + 2 2 (x − 1) (2 − x) x − 1 (x − 1) 2−x Fx + G x+1 A B C Dx + E + 2 (c) 3 2 = + 2+ 3+ 2 2 x (x − x + 10) x x x x − x + 10 (x − x + 10)2 We note that x2 − x + 10 is irreducible, since b2 − 4ac = 1 − 4 · 10 < 0. (a) 2. Integrate the following: Z 2x3 − x2 − 4x − 13 dx x2 − x − 2 SOLUTION: Do long division first: 2x3 − x2 − 4x − 13 x − 11 x − 11 = 2x + 1 + 2 = 2x + 1 + 2 x −x−2 x −x−2 (x + 1)(x − 2) Now expand the last term: x − 11 A B = + (x + 1)(x − 2) x+1 x−2 Solve for A, B: x−11 = A(x−2)+B(x+1). If we substitute x = −1, we get −12 = −3A, or A = 4. If we substitute x = 2, we get −9 = 3B, or B = −3. Therefore, 1 1 2x3 − x2 − 4x − 13 = 2x + 1 + 4 −3 2 x −x−2 x+1 x−2 and the integral is x2 + x + 4 ln |x + 1| − 3 ln |x − 2| + C 3. Suppose I made the substitution x = tan(θ), and after integration I got the expression θ + sin(2θ). Convert it back to x. SOLUTION: From the substitution, we note two thingsθ = tan−1 (x) and the substitution gives the relationship on a triangle, where if θ is an angle, then x is the length of √ the side opposite and 1 is the length of the side adjacent, so the hypotenuse has length 1 + x2 . We can’t get sin(2θ) directly, but we can use the identity: sin(2θ) = 2 sin(θ) cos(θ) = 2 · √ 1 x 1 ·√ 2 1+x 1 + x2 Overall, the solution is: 2x 1 + x2 NOTE: The expression sin(2 tan−1 (x)) is not a simplified expression in x, so is not a complete answer to the problem. √ 4. Same kind of problem as the previous one, but t + 2 = 3 sec(θ), and the result was cos(2θ) (TYPO: The question should read “convert this expression back to t”). tan−1 (x) + SOLUTION: We get the triangle from the substitution, and we need to recall the identity for cos(2θ), of which there are several possibilities. If you’ve forgotten them, hopefully you recall the half angle formula: cos2 (θ) = 1 + cos(2θ) 2 ⇒ cos(2θ) = 2 cos2 (θ) − 1 Now subsititute the appropriate expression in for cos(θ): √ !2 3 6 −1= −1 2 t+2 (t + 2)2 5. (a) The substitution would be x = 2 sin(θ), so that the integrand becomes: 8 sin2 (θ) cos(θ) 8 sin2 (θ) cos(θ) 4 sin2 (θ) 2 cos(θ) = = = tan2 (θ) (4 − x2 )3/2 8 cos3 (θ) (4(1 − sin2 (θ))3/2 Therefore, we get Z Z x2 dx = tan2 (θ) dθ 2 3/2 (4 − x ) (b) For the second integral, normally we would first check to see if the quadratic is reducible (it is not; b2 − 4ac < 0). However, in this case, we’re told to make the trig substitution- “Complete the square” in the denominator: x x x = 2 = 2 x + 2x + 5 (x + 2x + 1) + 4 (x + 1)2 + 4 We take x + 1 = 2 tan(θ) so that the integrand becomes: Z (2 tan(θ) − 1)2 sec2 (θ) dθ Z (2 tan(θ) − 1) dθ = 4 tan2 (θ) + 4 2 6. Find the length of the arc of the curve y = x3/2 from the point (1, 1) to (4, 8). SOLUTION: Note that the given y−values are not necessary; we only need 1 ≤ x ≤ 4. Compute the integrand for the arc length formula: s q 1 + (y 0 )2 = q 1 + ((3/2)x1/2 )2 = 9 1√ 1+ x= 4 + 9x 4 2 Now integrate from 1 to 4:  1Z 4√ 1 1 Z 40 1/2 1  3/2 4 + 9x dx = · u du = 40 − 143/2 2 1 2 9 14 27 2 7. Show that Z xf 00 (x) dx = xf 0 (x) − f (x) This is integration by parts: + x f 00 (x) − 1 f 0 (x) + 0 f (x) ⇒ xf 0 (x) − f (x) 8. True or False? (And give a short reason) (a) If f is continuous on [0, ∞) and 1∞ f (x) dx converges, then SOLUTION: This is true, since we can write: R Z ∞ f (x) dx = Z 1 0 f (x) dx + 0 Z ∞ R∞ 0 f (x) dx converges. f (x) dx 1 The first part of the sum (on the interval [0, 1]) exists because we’re told that f is continuous on [0, ∞), and therefore f is continuous on [0, 1] so that the Fundamental Theorem of Calculus (part 1) is satisfied there. (b) To find sin2 (x) cos5 (x) dx, rewrite the integrand as sin2 (x)(1 − sin2 (x))2 cos(x) SOLUTION: That is true; then let u = sin(x) and du = cos(x) dx. R (c) Integration by parts is the integral version of the Product Rule for derivatives. SOLUTION: That is true. We showed it in class, but you could also start with the product rule, then integrate both sides: 0 0 (f g) = f g + f g 0 → f (x)g(x) = Z 0 f (x)g(x) dx + Z f (x)g 0 (x) dx dx, start by completing the square in the denominator. (d) To find x22x−3 −3x+5 SOLUTION: False- Start by looking for an obvious u, du substitution- In this case, u = x2 − 3x + 5 and du = 2x − 3 dx Side Note: Would it be wrong to complete the square? You would get to the same answer, but it would take you significantly more time... R 3 dx, start by completing the square in the denominator. (e) To find x2 −3x+5 SOLUTION: False. Start by checking that you cannot factor the denominator- In this case, we cannot, so then continue by completing the square. R 3 (f) To find x2 −4x+3 dx, start by completing the square in the denominator. SOLUTION: False. Start by checking the denominator- In this case, we can factor it, so we should do that and use partial fractions. R (g) u, du substitution is the integral version of the Chain Rule. SOLUTION: True. We showed it in class, and gives you some good insight into when to use it. 9. Does the following integral converge or diverge? Z ∞ 1 3 2 + sin(x) √ dx x SOLUTION: Since −1 ≤ sin(x) ≤ 1, then 1 ≤ 2 + sin(x) ≤ 3, so that 1 2 + sin(x) √ 0≤ √ ≤ x x Further, Z ∞ 1 √ √ x−1/2 dx = lim (2 x|t1 = lim 2 t − 2 → ∞ t→∞ t→∞ Therefore, by the comparison theorem, the original integral also diverges. 10. The region under the curve of y = cos2 (x) (0 ≤ x ≤ π/2) is rotated about the x−axis. Find the volume of the resulting solid. SOLUTION: Using disks, the integral is an even power of cosine, which means we use the half angle identity to integrate it: Z π/2 4 π cos (x) dx = π 0 Z π/2 0 1 + cos(2x) 2 !2 dx = πZ 1 + 2 cos(2x) + cos2 (2x) dx 4 Use the half angle identity again to get: πZ πZ 3 1 2 1 + 2 cos(2x) + cos (2x) dx = + 2 cos(2x) + cos(4x) dx 4 4 2 2 Finishing it off, we should get that the volume is 3π 2 /16. 11. Does the integral converge or diverge? If it converges, evaluate it. (a) Z ∞ te−st dt 0 (s is a constant- state any conditions on s for the integral to converge.) SOLUTION: First we’ll take care of the integration. Use integration by parts, we get the following (I’ve put it into a single fraction, but that is not necessary): e−st + t − 1 (−1/s)e−st + 0 (1/s2 )e−st ⇒ (st + 1)e−st − s2 T = lim − 0 T →∞ (sT + 1)e−sT 1 + 2 2 s s For the limit, we can factor out the s2 (it’s constant with respect to T ), and we get a fraction on which we can use l’Hospital’s rule: −1 sT + 1 −1 s lim = 2 lim sT = 0 2 sT s T →∞ e s T →∞ se The previous steps were valid as long as s > 0 (otherwise, e−sT would diverge to −∞). Overall then, the integral converges to 1/s2 . Z 4 dx √ (b) 1 x−1 SOLUTION: Rewriting the integrand as (x − 1)−1/2 , we see that the antiderivative √ 1/2 is 2(x − 1) = 2 x − 1. Therefore, Z 4 1 √ 4  √  √ √ √ dx = lim+ 2 x − 1 = lim+ 2 3 − 2 t − 1 = 2 3 t→1 t→1 x−1 t 4 (c) Z ∞ ln(x) dx x 3 SOLUTION: The integral diverges. We can use the comparison theorem with 1/x. That is, since 1 < ln(x) for x > e, then for x > 3, we have: 1 ln x < x x and 3∞ 1/x dx diverges. ALTERNATIVE SOLUTION: You could perform the integration, and show that the limit diverges. In this case, if we take care of the integrand first: R Z ln(x) dx x u = ln(x) du = (1/x) dx Z 1 1 u du = u2 = (ln(x))2 2 2 Now we can take the limit: Z ∞ 3 ln(x) 1 dx = lim (ln(x))2 t→∞ 2 x t 3 This limit diverges to infinity. Z ∞ x (d) dx −∞ x2 + 1 SOLUTION: Rewrite the integral using a convenient number: Z 0 −∞ Z ∞ x x dx + dx 2 2 x +1 x +1 0 For each, we can use u = x2 + 1 and 12 du = x dx Z x 1Z 1 1 dx = du = ln(x2 + 1) 2 x +1 2 u 2 This expression diverges if x → ∞ and when x → −∞, so the integral diverges. 12. Evaluate using any method, unless specified below: (a) Z 4 dx (4 + x2 )3/2 SOLUTION: Trig substitution is the most direct choice. Let x = 2 tan(θ). Then 4 + x2 = 4 + 4 tan2 (θ) = 4 sec2 (θ) and dx = 2 sec2 (θ) dθ Substituting these in, we get: Z 8 sec2 (θ) dθ Z = cos(θ) dθ = sin(θ) + C 8 sec3 (θ) Use the reference triangle to convert this answer back to x: x √ +C 4 + x2 5 (b) Z tan3 (x) sec2 (x) dx SOLUTION: This is a trig integral- Try to reserve something to pull off a u, du substitution. In this case,Rreserve sec2 (x) so that u = tan(x) and du = sec2 (x) dx, and the integral becomes u3 du. 1 tan4 (x) + C 4 Z 3x + 2 3x + 2 dx = dx (c) 2 x + 6x + 8 (x + 2)(x + 4) SOLUTION: Since the denominator factors, use partial fractions. Here is the final answer: Z = (d) 5 2 − dx = 5 ln |x + 4| − 2 ln |x + 2| + C x+4 x+2 Z t2 cos(t3 − 2) dt sin2 (t3 − 2) SOLUTION: Look for the u, du substitution first. In this case, we do have what we need, if we let u = sin(t3 − 2). Then the integral becomes Z 1 Z −2 1 u du = − csc(t3 − 2) + C 3 3 (e) Z q cos5 (x) sin(x) dx SOLUTION: Look for a substitution first. Looks like we can reserve one of the cosines for the du term, and make u = sin(x): Z q cos4 (x) sin(x) [cos(x) dx] = √ Z q (1 − sin2 (x))2 sin(x) [cos(x) dx] = 2 4 2 u1/2 − 2u5/2 + u9/2 du = u3/2 − u7/2 + u11/2 3 7 11 To finish up the problem, back substitute the x. Z x (f) dx 2 x +4 SOLUTION: Straight u, du substitution: 12 ln |x2 + 4| + C. Z (g) 2 2 (1 − u ) u du = Z dx 1 − 6x − x2 SOLUTION: We’ll need to complete the square in the denominator, then probably do a trig substitution. To complete the square, notice that Z √ √ 2 1 − 6x − x2 = 1 − (x2 + 6x+ ) = 10 − (x + 3)2 = 10 − (x + 3)2 √ I can make the substitution: x + 3 = 10 sin(θ) so that the denominator becomes: q 10 − 10 sin2 (θ) = 6 √ 10 cos(θ) √ 10 cos(θ) dθ: Z Z √ dx 10 cos(θ) dθ √ √ =θ+C = 2 1 − 6x − x 10 cos(θ) and don’t forget the dx term: dx = Convert back to x to get ! −1 sin (h) x+3 √ +C 10 x−1 dx x2 + 3 SOLUTION: It might be easiest to separate these into two integrals, or you could do a trig substitution. Separating we get: Z Z Z Z x−1 x 1 dx = dx − dx 2 2 2 x +3 x +3 x +3 The first integral is set up for u, du substitution. For the second integral, factor 3 from the denominator so that we can do a different u, du substitution: Z 1 Z 1 1Z 1 1 dx =√ dx = du = √ tan−1 (u)  2 2 2 x +3 3 3 u +1 3 √x +1 3 Put the two together: (i) Z 1 2 ln |x2 + 3| + √1 3 tan−1  √x 3  +C sin2 (3t) dt SOLUTION: Use the half angle identity: Z (j) 1Z 1 1 sin (3t) dt = 1 − cos(6t) dt = t − sin(6t) + C 2 2 12 2 3x − 2 dx (x2 + 2)2 SOLUTION: We could break this into two, then use u, du substitution on one and trig substitution on the other, or we can just go for the trig substitution gusto from the start! √ Let x = 2 tan(θ) and make the necessary substitutions to get: √ √ √ √ Z Z Z 3x − 2 (3 2 tan(θ) − 2)( 2 sec2 (θ)) 2 (3 2 tan(θ) − 2) = dθ = dθ (x2 + 2)2 4 sec4 (θ) 4 sec2 (θ) Z Continuing to simplify, √ Z √ Z 3Z 2 3Z 2 2 sin(θ) cos(θ) dθ− cos (θ) dθ = sin(θ) cos(θ) dθ− (1+cos(2θ)) dθ 2 2 2 4 These can now each be evaluated to get: √ √ √ √ 2 2 2 2 3 2 3 2 sin (θ) − θ− sin(2θ) = sin (θ) − θ− sin(θ) cos(θ) 4 4 8 4 4 4 7 Finally, back substitute x using a triangle (which is why we converted sin(2θ) in the previous answer). Unsimplified, the answer is: 3 4 (k) x √ x2 + 2 !2 NOTE: If you evaluate answer... Z √ √ ! 2 x 2 x 2 −1 √ − √ √ tan − · +C 4 4 x2 + 2 x2 + 2 2 √ R sin(θ) cos(θ) dθ = − 12 cos2 (θ), you get a slightly different sin−1 (x) dx Use integration by parts + sin−1 (x) 1 1 − √1−x x 2 ⇒ x sin−1 (x) − Let u = 1 − x2 , du = −2x dx to finish: x sin−1 (x) + Z √ (l) x3 x2 + 4 dx Z x √ dx 1 − x2 √ 1 − x2 + C Substitution: u = x2 + 4, du = 2x dx, and x2 = u − 4. Then Z √ 1Z 1 Z 3/2 1/2 2 x x + 4 dx = (u − 4)u du = u − 4u1/2 du 2 2 3 (and continue...) 4 1 2 (x + 4)5/2 − (x2 + 4)3/2 + C 5 3 (m) Z √ 2x − x2 dx Rq Rq Complete the square first: −(x2 − 2x + 1) + 1 dx = 1 − (x − 1)2 dx Use a trig substitution: sin(θ) = x − 1 and cos(θ) dθ = dx. The integral becomes the following, which we can evaluate using either the half angle formulas or your table of formulas: Z 1 1 cos2 (θ) dθ = cos(θ) sin(θ) + θ 2 2 Use the reference triangle to convert back to x: √ 1 (sin−1 (x − 1) + (x − 1) 2x − x2 ) + C 2 (n) Z √ t ln(t) dt Integration by parts: √ + ln(t) t 2 3/2 − 1/t 3 t (o) Z ⇒ 2 3/2 2 Z 1/2 2 4 t ln(t) − t dt = t3/2 ln(t) − t3/2 + C 3 3 3 9 3x − 1 dx (x + 2)(x − 3) 8 By partial fractions, 3x − 1 A B = + (x + 2)(x − 3) x+2 x−3 ⇒ 3x − 1 = A(x − 3) + B(x + 2) Substitute x = 3 to get A = 7/5 and substitute x = −2 to get B = 8/5. Then the integral becomes: (p) Z 7Z 1 8Z 1 7 8 3x − 1 dx = dx + dx = ln |x + 2| + ln |x − 3| + C (x + 2)(x − 3) 5 x+2 5 x−3 5 5 Z ln(y 2 + 9) dy SOLUTION: Just like the regular log, we can integrate by parts + ln(y 2 + 9) 1 2y − y y 2 +9 ⇒ y ln(y 2 + 9) − 2 Z y2 dy y2 + 9 For the integral in y, we can use trig substitution: y = 3 tan(θ) so that y 2 + 9 = 9(tan2 (θ) + 1) = 9 sec2 (θ) and dy = 3 sec2 (θ): Z Z Z y2 9 tan2 (θ)(3 sec2 (θ)) dy = dθ = 3 tan2 (θ) dθ y2 + 9 9 sec2 (θ) Now, use the formulas that will be handed out (about half way down the page) to get that Z 3 tan2 (θ)dθ = 3 (tan(θ) − θ) Convert back to y so that: −2 Z y y y2 dy = −6 · + 6 tan−1 2 y +9 3 3   Put it all together: y ln(y 2 + 9) − 2y + 6 tan−1 (y/3) + C (q) sin3 (x) dx cos4 (x) Retain one sin(x) to go with dx, and set up the substitution u = cos(x) du = − sin(x) dx: Z Z − (1 − u2 )u−4 du = − (r) Z Z u−4 − u−2 du = 1 sec3 (x) − sec(x) + C 3 e−x sin(2x) dx Integrate by parts twice to get the same integral on both sides, + sin(2x) e−x − 2 cos(2x) −e−x + −4 sin(2x) e−x 9 Therefore, we have: Z −x e −x sin(2x) dx = −e (sin(2x) + 2 cos(x)) − 4 and Z (s) Z e−x sin(2x) dx 1 e−x sin(2x) dx = − e−x (sin(2x) + 2 cos(x)) + C 5 w dw w+5 SOLUTION: After some trial and error, we might take √ u= w+5 Z √ We’ll need to solve this for w and dw to make the substitution: w = u2 − 5 ⇒ dw = 2u du Therefore, Z √ Z w 2 (u2 − 5)2u du = u3 − 10u + C = dw = u 3 w+5 2 (w + 5)3/2 − 10(w + 5)1/2 + C 3 (t) Z y 2 e−3y dy SOLUTION: Integration by parts using a table + y2 e−3y − 2y (−1/3)e−3y + 2 (1/9)e−3y − 0 (−1/27)e−3y Then just write out the answer. Notice that we can factor out −e−3y to get: −e −3y  1 2 2 2 y + y+ +C 3 9 27  10