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

s10.pdf - University of Toronto Scarborough Department of...

This preview shows page 1 out of 7 pages.

You've reached the end of your free preview.

Want to read all 7 pages?

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 ...
View Full Document

  • Summer '19
  • Trigraph, 2n

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