s10.pdf - University of Toronto Scarborough Department of Computer Mathematical Sciences MAT A32H Winter 2017 Solutions#10 1 Section 14.5 \u0012 \u0013 \u0012 \u0013\u0013 Z \u0012 Z

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Unformatted text preview: University of Toronto Scarborough Department of Computer & Mathematical Sciences MAT A32H Winter 2017 Solutions #10 1. Section 14.5     Z  Z 3 x2 5 1 5 9x2 + 5 dx = 3x + dx = + ln | x | + C. 2. 3x 3 x 2 3 Z x √ dx. Substitute u = x2 + 1 =⇒ du = 2x dx. 4. 4 2 + 1 x   Z Z x 1 2 1 4 3/4 −1/4 √ Now dx = + C = (x2 + 1)3/4 + C. u u du = 4 2 2 3 3 x2 + 1 Z 2 2x ex 2 2 6. dx. Substitute u = ex − 2 =⇒ du = ex (2x) dx. 2 x e −2 Z Z 2 2x ex 1 2 du = ln | u | + C = ln | ex − 2 | + C. Now dx = 2 x u e −2 Z Z Z  e(ln 5)t 5t ln 5 t t e dt = e(ln 5)t dt = 8. 5 dt = +C = + C. ln 5 ln 5  Z Z Z  11 (3x + 2)(x − 4) 3x2 − 10x − 8 long 3x − 1 − 12. dx = dx = dx = division x−3 x−3 x−3 3 x2 − x − 11 ln | x − 3 | + C. 2  Z Z  5 − 4 x2 4 long 18. dx = −x2 +3x−2 ln | 2x+3 |+ −2x + 3 − dx = division 3 + 2x 2x + 3 C.Z 5 es ds. Substitute u = 1 + 3es =⇒ du = 3 es ds. 20. 1Z + 3es Z 5 es 1 5 5 5 Now du = ln | u | + C = ln(1 + 3es ) + C (1 + 3es > ds = s 1 + 3e 3 u 3 3 1 > 0). √ Z p √ 1 1+ x √ 22. . Substitute u = 1 + x =⇒ du = √ dx. x 2 x  3/2  √ Z p Z √ √ 4 1+ x 2u √ + C = (1 + x)3/2 + C. Now u du = 2 =2 3 3 x Z 4x 2 4 dx. Substitute u = ln(2 x2 ) =⇒ du = dx = dx. 28. 2 2 x ln(2 x ) 2x x Z Z 4 1 Now dx = 2 du = 2 ln | u | + C = 2 ln | ln(2 x2 ) | + C. x ln(2 x2 ) u MATA32H 30. 34. 40. 42. 46. 48. 50. 56. 58. page 2 Solutions # 10 Z  a ax + b long dx + For a, b, c, d constants and c 6= 0, = cx + d c   division    Z  ad ax ad 1 ax 1 1 ad + b− dx = + b− ln | cx+ dx = b− c cx + d c c cx + d c c c ax bc − ad ln | cx + d | + C. d| + C = + c c2 √ Z Z (4x) 21 ln(1 + x2 ) 4x ln 1 + x2 dx = dx. Substitute u = ln(1+x2 ) =⇒ 1 + x2 1 + x2 2x du = dx. 2 Z Z1 + x √ 2 u2 [ln(1 + x2 )] ln 2 (1 + x2 ) 4x ln 1 + x2 dx = u du = +C = +C = + Now 1 + x2 2 2 2 C.Z ex − e−x dx. Substitute u = ex + e−x =⇒ du = ex − e−x dx. x −x e +e Z Z 1 ex − e−x dx = du = ln | u | + C = ln | ex + e−x | + C. Now x −x e +e u Z 2x 2x 2 dx. Substitute u = ln(x + 1) =⇒ du = dx. (x2 + 1) ln(x2 + 1) x2 + 1 Z Z 1 2x dx = du = ln | u | + C = ln | ln(x2 + 1) | + C. Now 2 2 (x + 1) ln(x + 1) u  Z  Z Z 1 1 1 1 dx = − x dx − dx. −x 2 x 8x + 1 e (8 + e−x )2 ) Z8x + 1 e (8 + e ) 1 1 First, dx = ln | 8x + 1 | + C1 . 8 Z Z8x + 1 e−x 1 dx = dx. Substitute u = 8+e−x =⇒ du = Second, −x )2 −x )2 ex (8 + e (8 + e Z Z 1 1 1 1 −x dx = − du = + C = + C2 . −e dx. Hence 2 x (8 + e−x )2 2 −x e u u 8 + e   Z 1 1 1 1 Now dx = ln | 8x + 1 | − − x + C. −x 2 8x + 1 e (8 + e ) 8 8 + e−x Z 3x ln x (1 + ln x) dx. Substitute u = x ln x =⇒ du = ln x + 1 dx. Z Z 3u 3x ln x x ln x Now 3 (1 + ln x) dx = 3u du = +C = + C. ln 3 ln 3 Z 1 7 dx. Substitute u = ln x =⇒ dx. π x (ln x) x Z Z 7 7 7 u−π+1 7 Now dx = du = + C = (ln x)1−π + C. x (ln x)π uπ −π + 1 1−π Z Z Z f (x)+ln(f ′ (x)) f (x) ln(f ′ (x)) e dx = e e dx = ef (x) f ′ (x) dx = ef (x) + C. Z Given that marginal revenue is 900 dr = , the revenue function is r = dq (2q + 3)3 MATA32H 62. 64. 68. page 3 900 −225 substitute dq + C. = 3 u=2q+3 (2q + 3) (2q + 3)2 −225 −225 + C =⇒ C = 25 =⇒ r(q) = + 25. Since q = 0 =⇒ 0 = r(0) = 2 9  (2q + 3)  r 9 −225 25 25 Since r = pq we have p = so p(q) = 1− is + = 2 q q(2q + 3) q q (2q + 3)2 the demand function. dc Given that the marginal cost function is = 4 e0.005 q , the cost function is d q Z Z dc 4 c = c(q) = dq = 4 e0.005 q dq = e0.005 q + K = 800 e0.005 q + K. dq 0.005 When q = 0, c = 2000 =⇒ 2000 = 800 + K =⇒ K = 1200. Hence the cost function is 800 e0.005 q + 1200. 1 1 dC = − √ , the Given that the marginal propensity to consume is 2 2 2 I Z Z d I I dC 1 1 dI = − consumption function is C = C(I) = dI = − √ dI 2 2 2 2I √ 2I + K. 2 √ √ 4 2I 3 3 3 3 I Since C(2) = we have = 1− +K =⇒ K = =⇒ C(I) = − + . 4 4 2 4 2 2 4 100 dc = 10 − , the cost function Given that the marginal cost function is q + 10  dq Z Z  dc 100 is c = c(q) = 10 − dq = dq = 10q − 100 ln | q + 10 | + K. Now dq q + 10 c 100 ln | q + 10 | K + . the average cost is c = = 10 − q q q 100 ln(110) K Since c(100) = 50 we have 50 = 10 − + =⇒ K = 100(40 + 100 100 ln 110) =⇒ c(q) = 10q − 100 ln | q + 10 | + 100(40 + ln 110). When q = 0, c(0) = −100 ln(10) + 100(40 + ln 110) ≈ 4239.8. The fixed cost are about \$ 4240. a dr = q , the revenue function is r = r(q) = Given that marginal revenue is dq e +b Z Z Z dr a a e−q dq = dq = dq. Substitute u = 1 + b e−q =⇒ du = dq eq + b 1 + b e−q −b e−q dq. Z Z a e−q −a −a −a 1 dq = du = ln | u | + C = ln | 1 + Now r = r(q) = −q 1 + be b u b b −a b e−q | + C = ln(1 + b e−q ), a, b constants, b > 0. If q = 0 then r = 0 so b a −a −a ln(1 + b) + C =⇒ C = ln(1 + b). Hence r(q) = ln(1 + b e−q ) + 0= b b b   a 1+b a . ln(1 + b) = ln b b 1 + b e−q r(q) = 60. Solutions # 10 Z dr dq = dq Z MATA32H r a Thus the demand function, p = = ln q bq 2. page 4 Solutions # 10   1+b . 1 + b e−q Section 15.1 Z 1 2. x e3x+1 dx. Put u = x, dv = e3x+1 dx =⇒ du = dx, v = e3x+1 . 3 Z Z 3x+1 3x+1 3x+1 x e 1 x e e e3x+1 Now x e3x+1 dx = − e3x+1 dx = − +C = (3x − 3 3 3 9 9 1) + C. Z ea x 4. x ea x dx. Put u = x, dv = ea x dx =⇒ du = dx, v = . a Z Z ea x x ea x ea x x ea x − dx = − 2 + C. Now x ea x dx = a a a a Z dx x3 6. x2 ln x dx. Put u = ln x, dv = x2 dx =⇒ du = , v= . x 3 Z Z  3  3 3 3 x ln x x ln x dx x x x3 Now x2 ln x dx = = − − +C = (3 ln(x) − 3 3 x 3 9 9 1) + C. Z Z t 8. dt = t e−t dt. u = t, dv = e−t dt =⇒ du = dt, v = −e−t . t e Z Z Z t −t −t dt = t e dt = −t e + e−t dt = −t e−t −e−t +C = −e−t (t+1)+C. Now et √ Z 12 x 1 + 4x dx √ 10. . Put u = 12x, dv = √ =⇒ du = 12 dx, v = . 2 1 + 4x √ Z √ Z 1 + 4x √ 1 + 4x 1 + 4x 12 x √ = 12 x − 12 dx = 6 x 1 + 4x−(1+4x)3/2 + Now 2 2 1 +√ 4x C = (2x − 1) 1 + 4x + C. Z 2 ln x 20. (ln x)2 dx. Put u = (ln x)2 , dv = dx =⇒ du = dx, v = x. x   Z Z Z 2 ln x 2 2 2 Now (ln x) dx = x (ln x) − x dx = x (ln x) − 2 ln x dx. x Z dx , v = x. For ln x dx, we put u = ln x, dv = dx =⇒ du = x  Z Z Continuing, we have (ln x)2 dx = x (ln x)2 − 2 ln x dx = x (ln x)2 − 2 x ln x −  Z dx = x (ln x)2 − 2 (x ln x − x) + C = x ((ln x)2 − 2 ln x + 2) + C. Z x ex dx −1 22. dx. Put u = x ex , dv = =⇒ du = (x + 1) ex , v = . 2 2 (x + 1) x +1  Z(x + 1) x Z xe x −x ex −x ex x Now dx = + e dx = +ex +C = ex 1 − + 2 2 2 (x + 1) (x + 1) (x + 1) x+1 ex C= + C. x+1 MATA32H Solutions # 10 Z page 5 2 x5 ex dx. Substitute z = x2 =⇒ dz = 2x dx. Z Z 1 5 x2 Now x e dx = z 2 ez dz. Put u = z 2 , dv = ez dz =⇒ du = 2z dz, v = 2 ez .   Z Z Z 1 2 z 1 z 2 z 5 x2 z e −2 z e dz . Put u = z, dv = z e dz = Hence x e dx = 2 2 ez dz =⇒ du = dz, Zv = ez .    Z Z 1 1 2 z 5 x2 2 z z z Continuing we have x e dx = z e dz = z e −2 z e − e dz = 2 2 2 ex 1 2 z z z (z e − 2z e + 2 e ) + C = (x4 − 2x2 + 2) + C. 2 2   i √ x d h 1 2 √ 1+ √ = 30. Using the hint, we have ln(x + x + 1) = 2+1 2+1 d x! x + x x √ 1 x2 + 1 + x 1 √ √ . =√ x + x2 + 1 x2 + 1 x2 + 1 Z √ √ Now ln(x + x2 + 1) dx. Put u = ln(x + x2 + 1), dv = dx =⇒ du = 28. √ 1 dx, v = x. Z √ √ x 2 2 √ Hence ln(x + x + 1) dx = x ln(x + x + 1) − dx = x ln(x + 2+1 x √ √ x2 + 1) − x2 + 1 + C. Z 36. f (x) ex dx. Put u = f (x), dv = ex dx =⇒ du = f ′ (x) dx, v = ex . Z Z x x Now integration by parts gives f (x) e dx = f (x) e − f ′ (x) ex dx. Hence Z Z x f (x) e dx + f ′ (x) ex dx = f (x) ex + C. 3. x2 + Z1 Section 14.6 "  2 !  2    2 2 2 2 2 2 2 2 (12 + 22 + + 2 + ··· + n = 8. (a) Sn = 2 n n n n n n 8 n(n + 1)(2n + 1) 4 (n + 1)(2n + 1) 3 2 + · · · + n2 ) = 3 = . n 6 3 n2    1 1 4 4 4 (n + 1)(2n + 1) 8 1+ 2+ = (2) = . lim (b) lim Sn = lim 2 n→∞ 3 n→∞ n→∞ 3 n n n 3 3 MATA32H 16. page 6 Solutions # 10 Z a b dx. 0 We put f (x) = b, 0 ≤ x ≤ a. n X a and Sn = f (k∆x)∆x = n k=1   n n X a X a a a = b = (b n) = a b. f k n n n n k=1 k=1 Z a Now b dx = lim Sn = lim a b = a b. n→∞ n→∞ 0 Z 4 18. (2x + 1) dx. We put f (x) = 2x + 1, a = Now ∆x = b a 1 3 4−1 = . Now 1, b = 4 and ∆x = y n n   n n X X 3k 3 Sn = = 9 f (1 + k∆x) ∆x = f 1+ n n k=1 k=1    n   n  3 X 3 X 3k 6k +1 = = 2 1+ 3+ n k=1 n n k=1 n 3     n X 3 6 6 n(n + 1) 3 3n + 3n + = k = 1 n n k=1 n n 2   1 . 9 + 9 (1) 1 + n    Z 4 1 = 18. (2x + 1) dx = lim Sn = lim 9 + 9 (1) 1 + Now n→∞ n→∞ n 1 Z 2 20. (x + 2) dx. We put f (x) = x + 2, a = 1 4 x 2−1 1 y 1, b = 2 and ∆x = = . Now n n   n n 4 X X k 1 Sn = = f (1 + k∆x) ∆x = f 1+ 3 n n k=1 k=1    n  n  k k 1 X 1 1 X 1+ +2 = 3+ = 3n+ x n k=1 n n k=1 n n 1 2     n 1 X n(n + 1) 1 1 1 3 n+ . = 3+ (1) 1+ k = n k=1 n 2n 2 n    Z 2 1 1 1 7 (x + 2) dx = lim Sn = lim 3 + (1) 1 + Now =3+ = . n→∞ n→∞ 2 n 2 2 1 2 if 0 ≤ x < 1 4 − 2x if 1 ≤ x < 2 . We note that f is continuous on 22. Given f (x) = 5x − 10 if 2 ≤ x ≤ 3 MATA32H page 7 Solutions # 10 [0, 3] and f (x) ≥ 0 ∀x ∈ [0, 3]. Z 3 f (x) dx = the 0 area Z 3 under y = f (x) from x = 0 to x = 3. Hence f (x) dx = A1 + A2 + A3 corresponding to the y 5 0 3 cases in the definition of f . Now A1 = 2 (rect5 angle), A2 = 1 (triangle) and A3 = (triangle). 2 Z 3 5 f (x) dx = A1 +A2 +A3 = 2+1+ = Therefore 2 0 11 . 2 2 1 2 3 x ...
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