s9.pdf - University of Toronto Scarborough Department of Computer Mathematical Sciences MAT A32H Winter 2017 Solutions#9 1 Section 14.2 Z 1 dx = ln | x

s9.pdf - University of Toronto Scarborough Department of...

This preview shows page 1 out of 5 pages.

You've reached the end of your free preview.

Want to read all 5 pages?

Unformatted text preview: University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT A32H Winter 2017 Solutions #9 1. Section 14.2 Z 1 dx = ln | x | + C. 2. x  24+1  Z Z 5 x2 5 x2 5 x 24 24 +C = +C = + C. 4. 5 x dx = 5 x dx = 5 24 + 1 25 5     Z Z −3 1 1 z −3+1 1 z −2 −1 z −3 dz = z dz = +C = +C = + C. 6. 3 3 3 −3 + 1 3 −2 6 z2 Z Z 7 7 7 x−5/4 −14 −9/4 10. dx = + C. x dx = −5 + C = 9/4 2x 2 2 5 x5/4 4 Z Z Z Z 7 r6 4 r3 5 2 5 2 + + r + C. 12. (7r + 4r + 1) dr = 7 r dr + 4 r dr + 1 dr = 6 3  2 Z Z Z Z w 2 2 14. (5 − 2w − 6w ) dw = 5 dw − 2 w dw − 6 w dw = 5 w − 2 − 2  3 w 6 + C = 5 w − w2 − 2 w3 + C. 3 Z Z Z Z Z t3 t5 t7 2 4 6 2 4 6 16. (1 + t + t + t ) dt = dt + t dt + t dt + t dt = t + + + + C. 3 5 7   3    Z Z Z 2 2 8 4 8 2 x 8 x5 2x 2 4 − x dx = x dx − x dx = − +C = 20. 7 3 7 3 7 3 3 5 8 x5 2 x3 = + C. 21 15  3  2 Z Z Z Z x x x 2 x 2 x 22. (e +3x +2x) dx = e dx+3 x dx+2 x dx = e +3 +2 + 3 2 C = ex + x3 + x2 + C.   Z Z Z −4 −4 −4 x−2 2 −4 −3 dx = dx = + C. x dx = + C = 28. (3 x)3 27 x3 27 27 −2 27 x2  Z  Z Z Z √ 1 u4/3 1/3 −1/2 1/3 3 u+ √ 40. du = (u + u ) du = u du + u−1/2 du = + 4 u 3 3 u4/3 u1/2 1/2 + 2 u + C. +C = 1 4 2 MATA32H page 2 Solutions # 9 x6 x4 x (x + 5x + 2) dx = (x5 + 5x4 + 2x3 ) dx = + x5 + + C. 6 2 Z Z z3 2 + 2 z 2 + 4z + C. 44. (z + 2) dz = (z 2 + 4z + 4) dz = 3    Z  Z 4 2 1 1 x3 x − 5x2 + 2x 2 x −5+ dx = − 5x + 2 ln | x | +C = dx = 50. 5 x2 5 x 5 3 2 x3 − x + ln | x | + C. 15 5 Z 42. Z 54. (a) (b) 3 2 Z We need a function F (x) such that F (x) dx = x ex + C. Now Z d d (x ex + C) =⇒ F (x) = ex + x ex = ex (1 + x). F (x) dx = dx dx Z Assume that G(x) is any function such that G(x) dx = x ex + C. Now d (x ex + C) = ex (1 + x) = F (x) =⇒ F is unique – there is only G(x) = dx one. 2. Section 14.3 19 Given y ′ = x2 − x with y(3) = , we wish to find y. 2 Z Z x2 19 x3 − + C. Since y(3) = , we have y = y ′ dx = (x2 − x) dx = 3 2 2 19 33 32 19 9 x3 x2 = y(3) = − + C =⇒ C = − 9 + = 5 =⇒ y = − + 5. 2 3 2 2 2 3 2 ′ 2 4. Given Z y = −x Z+ 2x with y(2) = 1, we3wish to find y. −x y = y ′ dx = (−x2 + 2x) dx = + x2 + C. Since y(2) = 1 we have 3 −8 −1 −x3 1 1 = y(2) = + 4 + C =⇒ C = =⇒ y = + x2 − =⇒ y(1) = 3 3 3 3 1 1 −1 +1− = . 3 3 3 ′′ 6. Given y = x + Z1 with y ′ (0) = 0 and y(0) = 5, we wish to find y. Z x2 y′ = y ′′ dx = (x + 1) dx = + x + C1 . Since y ′ (0) = 0, we have 0 = 2 x2 02 + 0 + C1 =⇒ C1 = 0 =⇒ y ′ = + x. y ′ (0) = 2Z 2  Z  2 x3 x2 x + x dx = + + C2 . Since y(0) = 5, we have Now y = y ′ dx = 2 6 2 03 02 x3 x2 5 = y(0) = + + C2 =⇒ C2 = 5 =⇒ y = + + 5. 6 2 6 2 dr 12. Given the marginal revenue function = r ′ (q) = 5000 − 3(2q + 2q 2 ), the Z Zd q 2. revenue function is r(q) = r ′ (q) dq = (5000 − 3(2q + 2q 2 ) dq = 5000q − 3q 2 − MATA32H Solutions # 9 page 3 3 q4 3 q4 +C. Since q = 0 =⇒ r = 0 we have C = 0. Hence r(q) = 5000q −3q 2 − . 2 2 4 3q 5000q − 3q 2 − r 2 so the demand function is = Recall that r = p q so p = q q 3 q3 p = p(q) = 5000 − 3 q − . 2 dc = c ′ (q) = 0.000204 q 2 −0.046 q+6, the cost 16. Given the marginal cost function Z Z dq function is c(q) = c ′ (q) dq = (0.000204 q 2 − 0.046 q + 6) dq = 0.000068 q 3 − 0.023 q 2 + 6 q + K. Since the fixed costs are $15,000, we have 15000 = c(0) = 0.000068 03 − 0.023 02 + 6 0 + K = 0 + K =⇒ K = 15000 =⇒ c(q) = 0.000068 q 3 − 0.023 q 2 + 6 q + 15000. When q = 200 we have c(200) = 0.000068 (200)3 −0.023 (200)2 +6 (200)+15000 = 15824. dr 20. Given the marginal revenue function = r ′ (q) = 100 − 3q 2 , the revenue d q Z Z (100 − 3q 2 ) dq = 100q − q 3 + C. Now q = 0 =⇒ r r = 0 =⇒ C = 0 =⇒ r(q) = 100q − q 3 . Since r = pq =⇒ p = , the demand q function is p = p(q) = 100 − q 2 . Now q = 5 =⇒ p(5) = 100 − 25 = 75. Recall that the (point) elasticity of p p p p/q p q q q demand is η = = ′ . Here we have η = ′ = = . When d p/d q p (q) p (q) −2q −2 q 2 75 −3 75 = = . q = 5 we have η = 2 −2 (5) −50 2 function is r(q) = 3. r ′ (q) dq = Section 14.4 Z 2. 15 (x + 2)4 dx. Substitute u = x + 2 =⇒ du = dx.  5 Z Z u 4 4 Now 15 (x + 2) dx = 15 u du = 15 + C = 3(x + 2)5 + C. 5 Z 4. (4x+3)(2x2 +3x+1) dx. Substitute u = 2x2 +3x+1 =⇒ du = (4x+3) dx. Z Z u2 (2x2 + 3x + 1)2 2 Now (4x + 3)(2x + 3x + 1) dx = u du = +C = + C. 2 2 Z Z 4x dx = (2x2 − 7)−10 4x dx. Substitute u = 2x2 − 7 =⇒ du = 8. (2x2 − 7)10 4x dx.Z Z u−9 (2x2 − 7)−9 −1 4x −10 dx = u du = +C = +C = + Now 2 10 (2x − 7) −9 −9 9 (2x2 − 7)9 C. MATA32H page 4 Solutions # 9 Z 1 √ dx = (x − 5)−1/2 dx. Substitute u = x − 5 =⇒ du = dx. 10. Z x−5 Z √ 1 √ Now dx = u−1/2 du = 2 u1/2 + C = 2 (x − 5)1/2 + C = 2 x − 5 + C. x−5 Z 12. x2 (3x3 + 7)3 dx. Substitute u = 3x3 + 7 =⇒ du = 9 x2 dx =⇒ x2 dx = Z 1 du. 9 Z   Z 1 (3 x3 + 7)4 1 u4 3 Now x (3x + 7) dx = +C = + C. u du = 9 9 4 36 Z √ 14. x 3 + 5 x2 dx. Substitute u = 3 + 5 x2 =⇒ du = 10x dx =⇒ x dx = 1 du. 10 Z 18. 20. 26. 28. 44. Now Z 2 x √ 3 3 1 3 + 5 x2 dx = 10 Z √ 1 u du = 10  2 3/2 u 3  (3 + 5x2 )3/2 +C = + C. 15 5 e3t+7 dt. Substitute u = 3t + 7 =⇒ du = 3 dt. Z Z 5 5 5 3t+7 Now 5 e dt = eu du = eu + C = e3t+7 + C. 3 3 3 Z 3 −3 w2 e−w dw. Substitute u = −w3 =⇒ du = −3w2 dw. Z Z 3 2 −w3 Now −3 w e dw = eu du = eu + C = e−w + C. Z 12x2 + 4x + 2 dx. Substitute u = x + x2 + 2x3 =⇒ du = (1 + 2x + 6x2 ) dx. 2 + 2x3 x + x Z Z 12x2 + 4x + 2 1 Now dx = 2 du = 2 ln | u | + C = 2 ln | x + x2 + 2x3 | + C. 2 3 x + x + 2x u Z 2 6x − 6x dx. Substitute u = 1 − 3x2 + x3 =⇒ du = (−6x + 6x2 ) dx. 2 + 2x3 1 − 3x Z Z 6x2 − 6x 1 Now dx = du = ln | u | + C = ln | 1 − 3x2 + 2x3 | + C. 2 3 1 − 3x + 2x u −1/3 Z  Z 3 2 x2 + x + 1 3 r x + x + 3x dx = (x2 + x + 1) dx. Substitute 2 3 3 x3 + x2 + 3x 2 3 2 3 u = x + x + 3x =⇒ du = (3x2 + 3x + 3) dx.   Z 2 2 Z x +x+1 1 1 3 2/3 −1/3 r + C = Now dx = u u du = 3 3 2 3 3 x3 + x2 + 3x 2  2/3 3 1 x3 + x2 + 3x + C. 2 2 MATA32H 52. 60. 70. 78. 82. Solutions # 9 Z page 5 (6t2 + 4t) (t3 + t2 + 1)6 dt. Substitute u = t3 + t2 + 1 =⇒ du = (3t2 + 2t) dt.  7 Z Z 2 (t3 + t2 + 1)7 u 2 3 2 6 6 +C = +C. Now (6t +4t) (t +t +1) dt = 2 u du = 2 7 7 Z Z e2x e−2x x −x 2 2x −2x (e − e ) dx = (e − 2 + e ) dx = − 2x − + C. 2 2  Z  Z Z 3 1 1 dx = 2 + dx + (x − 1)−2 dx = 3 ln | x − 1 | − x − 1 (x − 1)2 x−1 1 + C. (x − 1)−1 + C = 3 ln | x − 1 | − x−1 r Z 1 −1 2 1 + 9 dt. Substitute u = + 9 =⇒ du = 2 dt. 2 t tr t t  3/2   3/2 Z Z √ 2 1 2u −4 1 Now u du = −2 +C = + 9 dt = −2 +9 + C. t2 t 3 3 t Z Z Z x 2x 1 1 x ′ ′ , then y = y dx = dx = dx = ln | x2 + If y = 2 2 2 x +6 x +6 2 x +6 2 1 2 2 6 | + C = ln(x + 6) + C (x + 6 ≥ 6 > 0). 2 1 1 Since y(1) = 0, we have 0 = ln(1 + 6) + C =⇒ C = − ln 7 =⇒ y = 2     22 1 1 x +6 2 ln(x + 6) − ln 7 = ln . 2 2 7 r x2 + 6 is also a correct answer. Note that y = ln 7 ...
View Full Document

  • Summer '19
  • Trigraph, r6, z dz

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
A+ icon
Ask Expert Tutors