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

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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 ...
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