Mathematics 1274 SOLUTIONS TO SELECTED EXERCISES FROM MATH 1274 TEXTBOOK August 24, 2023 Minh Phuong Bui & Pichmony Anhaouy Langara College Mathematics and Statistics department panhaouy@langara.ca Contents 1 Integration 1.1 1.2 1.3 1.4 2 4.1 4.2 4.3 5 6 23 23 24 25 The Three-Dimensional Coordinate System . . . . . . . . . . . . . . . . . . . . Planes and Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Differential Equations 6.1 6.2 6.3 21 23 Continuous Income Streams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Consumers’ and Producers’ Surpluses . . . . . . . . . . . . . . . . . . . . . . . . Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Multivariate Calculus 5.1 5.2 5.3 5.4 5.5 5.6 7 8 12 13 14 18 19 21 Area Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . More Applications 1 2 4 5 7 Substitution Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Partial Fractions Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . Improper Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Applications of Integration 3.1 4 Antiderivatives and Indefinite Integration . . . . . . . . . . . . . . . . . . . . . . The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . Techniques of Integration 2.1 2.2 2.3 2.4 2.5 2.6 2.7 3 1 25 26 26 29 30 31 33 Graphical and Numerical Solutions to Differential Equations . . . . . . . Separable Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . First Order Linear Differential Equations . . . . . . . . . . . . . . . . . . . . . . 3 33 33 34 6.4 Modeling with Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . 36 Chapter 1 Integration 1.1 Antiderivatives and Indefinite Integration Z 9. x8 dx = x9 + C. 9 Z 11. dt = t + C. Z −1 1 dt = + C. 2 3t 3t Z √ 1 √ dx = 2 x + C. x 13. 15. Z 17. Z 5eθ dθ = 5eθ + C. Z 5t 5t dt = + C. 2 2 ln(5) 19. 21. Z 23. 2 3 (t + 3)(t − 2t) dt = Z eπ dx = eπ x + C. Z 1 dx = ln(|x|) + C. x 25. 27. sin θ dθ = − cos θ + C. Z (t5 + t3 − 6t) dt = t6 t4 + − 3t2 + C. 6 4 (a) What is the domain? The function is ln(|x|) so we need x > 0. The domain is (0, ∞). i 1 dh (b) ln(x) = dx x 1 (c) What is the domain of ln(−x) ? We need −x > 0 therefore x < 0. Domain of the function ln(−x) is (−∞, 0). (d) Find d of ln(−x). dx i dh 1 ln(−x) = − dx x  1 ln(−x), if x < 0 dx = (e) ln(x), if x > 0 x Z Z 0 29. f (x) = f (x) dx = sin(x) dx = − cos x + C. Z f (0) = 2 ⇒ − cos 0 + C = 2, ⇒ −1 + C = 2 ⇒ C = 3 ⇒ f (x) = − cos x + 3. Z Z 0 31. f (x) = f (x) dx = sec(x)2 dx = tan x + C. π π f ( ) = 5 ⇒ tan + C = 5, ⇒ 1 + C = 5 ⇒ C = 4 ⇒ f (x) = tan x + 4. 4 4 Z Z 33. f 0 (x) = f ”(x) dx = 5 dx = 5x + C. f 0 (0) = 7 ⇒ 5 ∗ 0 + C = 7, ⇒ 0 + C = 7 ⇒ C = 7 ⇒ f 0 (x) = 5x + 7. Z f (x) = Z 0 f (x) dx = (5x + 7) dx = 5x2 + 7x + C. 2 f (0) = 3 ⇒ 0 + 0 + C = 3, ⇒ 0 + C = 3 ⇒ C = 3 ⇒ f (x) = Z 0 35. f (x) = Z f ”(x) dx = 5x2 + 7x + 3. 2 5ex dx = 5ex + C. f 0 (0) = 3 ⇒ 5 ∗ e0 + C = 3, ⇒ 5 + C = 3 ⇒ C = −2 ⇒ f 0 (x) = 5ex − 2. Z f (x) = f 0 (x) dx = Z (5e x − 2) dx = 5ex − 2x + C. f (0) = 5 ⇒ 5 − 0 + C = 5, ⇒ 5 + C = 5 ⇒ C = 0 ⇒ f (x) = 5ex − 2x. 1.2 The Definite Integral Z 1 5. (a) 0 Math 1274 (−2x + 4) dx = (1 ∗ 2) + 1 ∗ (1 ∗ 2) = 2 + 1 = 3. 2 Minh Phuong Bui & Pichmony Anhaouy Solutions 2 Z 2 (−2x + 4) dx = (b) 0 Z 3 1 ∗ (2 ∗ 4) = 4. 2 0 (−2x + 4) dx = 4 − Z 3 Z 2 (c) (d) (−2x + 4) dx = 1 1 ∗ (1 ∗ 2) = 3. 2 Z 3 (−2x + 4) dx + 1 (−2x + 4) dx = 0. 2 Z 4 1 (−2x + 4) dx = − ∗ (2 ∗ 4) = −4. 2 2 (e) Z 1 (f) Z 1 (−6x + 12) dx = 3 0 0 Z 2 9. (a) f (x) dx = π22 = π. 4 f (x) dx = π22 = π. 4 f (x) dx = π22 = 2π. 2 0 Z 4 (b) 2 Z 4 (c) 0 Z 4 (d) 0 Z 3 19. 0 (−2x + 4) dx = 3 ∗ 3 = 9. 5f (x) dx = 5 ∗ 2 = 10. (f (x) − g(x)) dx = Z 3 0 f (x) dx − Z 3 0 g(x) dx = 7 − hZ 2 Z 3 g(x) dx + 0 i g(x) dx . 2 = 7 − (−3 + 5) = 5. Z 3 Z 3 Z 3 21. (af (x) + bg(x)) dx = 0 ⇒ a f (x) dx + b g(x) dx = 0. 0 0 0 ⇒ 7a + b(−3 + 5) = 0 ⇒ 7a + 2b = 0, for a = 2 ⇒ 14 + 2b = 0 ⇒ b = −7. Z 0 Z 5 Z 5 Z 5 (s(t) − r(t)) dt = − (s(t) − r(t)) dt = r(t) dt − s(t) dt. 23. 5 0 Z 5 = 0 Z 5 25. 0 r(t) dt − Z 3 0 Z 5 s(t) dt + 0 3 (ar(t) + bs(t)) dt = 0 ⇒ a  s(t) dt = 11 − (10 + 8) = −7. Z 5 Z 5 r(t) dt + b 0 choose a = 1 ⇒ 11 + 18b = 0 ⇒ b = Math 1274 0 0 s(t) dt = 0 ⇒ 11a + 18b = 0. −11 . 18 Minh Phuong Bui & Pichmony Anhaouy Solutions 3 1.3 5. Riemann Sums 4 X i2 = i=2 7. 4 X  4∗5∗9 n(n + 1)(2n + 1) −1= − 1 = 29. i2 − 1 = 6 6 i=1 2 X sin  iπ  2 i=−2 9. 6 X i=1 = sin(−π) + sin  −π  2 + sin(0) + sin π  2 + sin(π) = 0 − 1 + 0 + 1 + 0 = 0. −1i i = (−1)1 1 + (−1)2 2 + (−1)3 3 + (−1)4 4 + (−1)5 5 + (−1)6 6. = −1 + 2 − 3 + 4 − 5 + 6 = 3. 5 X 11. (−1)i cos(iπ) = (−1)0 cos(0) + (−1)1 cos(π) + (−1)2 cos(2π) + (−1)3 cos(3π)+. i=0 (−1)4 cos(4π) + (−1)5 cos(5π) = 6. Z 3 27. x2 dx. −3 ∆x = 3 − (−3) = 1. 6 x1 = −3 x2 = −2. x3 = −1 x4 = 0 f (x1 ) = 9 f (x2 ) = 4 f (x3 ) = 1. f (x4 ) = 0 f (x5 ) = 1 f (x6 ) = 4. Z 3 2 x dx = −3 6 X x5 = 1 x6 = 2. f (xi )∆x = 9 + 4 + 1 + 0 + 1 + 4 = 19. i=1 Z π 29. sin x dx. 0 ∆x = x1 = π−0 π = 6 6 π 6 x2 = xi = π 3 π 1 f (x1 ) = sin( ) = 6 2 Math 1274 iπ . 6 x3 = π 2 x4 = 2π 3 √ π 3 f (x2 ) = sin( ) = 3 2 x5 = 5π 6 x6 = π. π f (x3 ) = sin( ) = 1. 2 Minh Phuong Bui & Pichmony Anhaouy Solutions 4 √ 2π 3 f (x4 ) = sin( ) = 3 2 Z π 0 Z 3 35. −1 f (x5 ) = sin( 5π 1 )= 6 2 f (x6 ) = sin(π) = 0. √ √ √ π 3 3 1 (2 + 3)π sin x dx = + +1+ + +0 = . 2 2 2 2 6 6 1 (3x − 1) dx. ∆x = 3 − (−1) 4 = n n ⇒ xi = xi = −1 + 4i . n xi+1 = −1 + 4(i + 1) . n xi + xi+1 4i + 2 = −1 + . 2 n  4i + 2  12i + 6 ⇒ f (xi ) = f − 1 + = −4 + . n n ⇒ Z 3 −1 (3x − 1) dx = = −16 + Z 3 n = 100 ⇒ −1 lim 1.4 (3x − 1) dx = −16 + Z 3 n = 1000 ⇒ n→∞ i=1 n n n i=1 i=1 i=1 12i + 6  4 X 16 X 48i X 24 −4+ = + . − + n n n n2 n2 48 n(n + 1) 24 24n2 + 24n 24 + = −16 + + . 2 n 2 n n2 n n = 10 ⇒  n  X −1 2400 + 240 24 + = 12.8. 100 10 (3x − 1) dx = −16 + Z 3 −1 24 ∗ 1002 + 2400 24 + = 8.48. 1002 100 (3x − 1) dx = −16 + 24 ∗ 10002 + 24000 24 + = 8.048. 10002 1000 24n2 + 24n 24  − 16 + + = −16 + 24 = 8. n2 n The Fundamental Theorem of Calculus Z 3 5. 1 (3x2 − 2x + 1) dx =  3x3 3 − 3 3 2x2 = (x3 − x2 + x . +x 2 1 1 = (27 − 9 + 3) − (1 − 1 + 1) = 20. Z 1  x4 x6  1 1 1 1 1 7. (x3 − x5 ) dx = − = − − − = 0. 4 6 −1 4 6 4 6 −1 Z π 4 9. sec2 x dx = tan x 0 Math 1274 π 4 0 = tan π − tan 0 = 1. 4 Minh Phuong Bui & Pichmony Anhaouy Solutions 5 Z 1 5x dx = 11. −1 Z π 13. 0 5 − 15 5x 1 24 = = . ln 5 −1 ln 5 5 ln 5 π (2 cos x − 2 sin x) dx = (2 sin x + 2 cos x) . 0 = (2 sin π + 2 cos π) − (2 sin 0 + 2 cos 0) = −4. 3 Z 4√ Z 4 1 t 2 4 2 3 4 16 t 2 dt = 3 = t 2 = . 15. t dt = 3 0 3 0 0 0 2 Z 8 1 45 3 4 8 3 17. x 3 dx = x 3 = (16 − 1) = . 4 4 4 1 1 Z 2 Z 2 1  1 1 2 1 −2 x dx = − dx = = − 19. − 1 = . 2 x 1 2 2 1 1 x Z 1 1 1 1 1 21. x dx = x2 = − 0 = . 2 0 2 2 0 Z 1 1 1 1 23. x3 dx = x4 = . 4 0 4 0 Z 4 4 25. dx = x = 4 − (−4) = 8. −4 −4 Z 2 27. 0 dx = 0. −2 Z 2 31. 1 1 16 x2 dx = x3 = (8 + 8) = . 3 3 3 −2 Z 2 −2 x2 dx = f (c)(2 − (−2)) = 4f (c) ⇒ 4c2 = Z 16 √ x dx = 16 4 2 −2 ⇒ c2 = ⇒ c = √ , √ . 3 3 3 3 Z 16 1 2 3 1 128 x 2 dx = x 2 6 = . 3 3 0 0 0 √ √ 128 64 128 8 f (c)(16 − 0) = ⇒ 16 c = ⇒ c= ⇒c= . 3 3 3 9 Z π π 1 1 −1 2 35. y = sin x ⇒ Ave = sin x dx = (− cos x) = (cos π − cos 0) = . π−0 0 π π π 0 Z e e 1 1 1 1 1 1 39. g(t) = on [1, e] ⇒ Ave = dt = ln t = (ln e − ln 1) = . t e−1 1 t e−1 e − 1 e − 1 1 Z x3 +x Z x3 +x 1 d 1 1 d 3 3x2 + 1 0 53. F (x) = dt ⇒ F (x) = dt = 3 (x + x) = 3 . t dx 2 t x + x dx x +x 2 Z x2 Z 0 Z x2 55. F (x) = (t + 2) dt = (t + 2) dt + (t + 2) dt. 33. x x Math 1274 Z x2 0 Z 2 Z x i dh x =− (t + 2) dt + (t + 2) dt ⇒ F (x) = (t + 2) dt − (t + 2) dt . dx 0 0 0 0 d d = (x2 + 2) (x2 ) − (x + 2) (x) = (x2 + 2)2x − (x + 2) = 2x3 + 3x − 2. dx dx Z x 0 Minh Phuong Bui & Pichmony Anhaouy Solutions 6 Chapter 2 Techniques of Integration 2.1 Substitution Method 3 2 3. x − 5 = u ⇒ 3x dx = du ⇒ Z 1 1 u7 du = u8 + C = (x3 − 5)8 + C. 8 8 5. x2 + 1 = u ⇒ 2xdx = du ⇒ xdx = ⇒ 11. √ 1 1 u8 du = u9 + C = (x2 + 1)9 + C. 2 18 18 1 1 x = u ⇒ √ dx = du ⇒ √ dx = 2du. 2 x x ⇒ 13. Z du . 2 Z √ 2eu du = 2eu + C = 2e x + C. −1 1 + 1 = u ⇒ 2 dx = du. x x ⇒ Z −u du = 2 −1  1 −1 2 u +C = + 1 + C. 2 2 x 15. sin x = u ⇒ cos xdx = du. ⇒ Z 1 1 u2 du = u3 + C = sin3 x + C. 3 3 17. 4 − x = u ⇒ −dx = du. ⇒ Z − sec 2u du = − tan u + C = − tan(4 − x) + C. 19. tan x = u ⇒ sec2 xdx = du. ⇒ Z 1 1 u2 du = u3 + C = − tan3 x + C. 3 3 1 25. x3 = u ⇒ 3x2 dx = du ⇒ x2 dx = du. 3 7 ⇒ Z 1 u 1 1 3 e du = eu + C = ex + C. 3 3 3 31. ln x = u ⇒ 1 dx = du. x 1 1 u du = u2 + C = ln2 x + C. 2 2 Z 2 Z Z 2 x 3x 1 1 1 x + 3x + 1 dx = + + dx = x + 3 + dx = x2 + 3x + ln |x| + C. 35. x x x x x 2 Z 3 Z  Z x −1 1 3 1 2 2 2  2 37. dx = dx = x − x + x − dx. (x − x + 1) − x+1 x+1 3 2 x+1 ⇒ Z 1 1 = x3 − x2 + x − 2 ln |x + 1| + C. 3 2 Z x2 dx. 51. (x3 + 3)2 x3 + 3 = u ⇒ 3x2 dx = du ⇒ x2 dx = ⇒ Z 1 −1 −1 du = +C = + C. 3u2 3u 3(x3 + 3 53. 1 − x2 = u ⇒ −2xdx = du ⇒ xdx = ⇒ Z du . 3 −du . 2 p √ −1 √ du = − u + C = − 1 − x2 + C. 2 u 79. 1 − x2 = u ⇒ −2xdx = du. x = 0 ⇒ u = 1. x = 1 ⇒ u = 0. ⇒− 2.2 Z 0 u4 du = 1 −1 5 0 1 u = . 5 5 1 Integration by Parts Z 5. xe−x dx. u = x ⇒ du = dx. ⇒ Z Math 1274 e−x dx = dv ⇒ −e−x = v. xe−x dx = −xe−x + Z e−x dx = −xe−x − e−x = e−x (1 + x). Minh Phuong Bui & Pichmony Anhaouy Solutions 8 Z 7. x3 sin x dx. u = x3 ⇒ du = 3x2 dx. ⇒ Z 3 sin xdx = dv ⇒ − cos x = v. 3 x sin x dx = −x cos x + 3 u = x2 ⇒ du = 2xdx. ⇒ Z 3 Z 9. Z  3 Z 11. x3 ex dx. Z 3 x 3 x ex dx = dv ⇒ ex = v. x e dx = x e − 3 Z Z Z x2 ex dx. ex dx = dv ⇒ ex = v. Z   2 x x e dx = x e − 3 x e − 2 xex dx . 3 x 3 x ex dx = dv ⇒ ex = v.    x3 ex dx = x3 ex − 3 x2 ex − 2 xex − ex . ex sin x dx. ex dx = dv ⇒ ex = v. u = sin x ⇒ du = cos xdx. ⇒ Z x x e sin x dx = e sin x − u = cos x ⇒ du = − sin xdx. ⇒  x sin x dx .    x3 sin x dx = −x3 cos x + 3 x2 sin x − 2 sin x − x cos x . u = x ⇒ du = dx. ⇒ Z sin xdx = dv ⇒ − cos x = v. u = x2 ⇒ du = 2xdx. ⇒ 2 x sin x dx = −x cos x + 3 x sin x − 2 u = x3 ⇒ du = 3x2 dx. ⇒ x2 cos x dx. cos xdx = dv ⇒ sin x = v. u = x ⇒ du = dx. ⇒ Z Z Math 1274 Z ex cos x dx. ex dx = dv ⇒ ex = v. Z   ex sin x dx = ex sin x − ex cos x + ex sin x dx . Minh Phuong Bui & Pichmony Anhaouy Solutions 9 ⇒2 Z 13. Z ex sin x dx = ex sin x − ex cos x ⇒ 1 e2x dx = dv ⇒ ex = v. 2 Z 2x e 1 3 sin 3x dx = e2x sin 3x − 2 2 ⇒ ⇒ ⇒ Z 21. e2x cos 3x dx. Z 31 x 3 1 e cos 3x + e2x sin 3x dx = e2x sin 3x − 2 2 2 2 Z Z 1 3 9 e2x sin 3x dx = e2x sin 3x − ex cos 3x − 2 4 4  e2x sin 3x dx . 13 4 Z Z e2x sin 3x dx = Z  e2x sin 3x dx . 1 3 e2x sin 3x dx = e2x sin 3x − ex cos 3x 2 4 4 2x 3 e sin 3x − ex cos 3x. 26 13 (x − 2) ln x dx. u = ln x ⇒ du = ⇒ Z 1 e2x dx = dv ⇒ ex = v. 2 u = cos 3x ⇒ du = −3 sin 3xdx. ⇒  1  ex sin x dx = ex sin x − cos x . 2 e2x sin 3x dx. u = sin 3x ⇒ du = 3 cos 3xdx. ⇒ Z Z 1 dx. x 1 (x − 2) ln x dx = (x − 2)2 ln x − 2 1 1 = (x − 2)2 ln x − 2 2 1 1 = (x − 2)2 ln x − 2 2 Z 23. x ln x2 dx. Z u = x ⇒ du = 2xdx ⇒ w = ln u ⇒ dw = 1 du. u Z 1 1 (x − 2)2 dx. 2 x x2 − 4x + 4 dx. x Z  2 Math 1274 1 (x − 2)dx = dv ⇒ (x − 2)2 = v. 2 x−4+ Z 2 4 1 1 dx = (x − 2)2 ln x − x2 + 2x − 2 ln x + C. x 2 4 x ln x dx = Z 1 ln u du. 2 du = dv ⇒ u = v. Minh Phuong Bui & Pichmony Anhaouy Solutions 10 = 1 u ln u − 2 Z  1   u  1  du = u ln u − 1 = x2 ln x2 − 1 + C. u 2 2 Z π x sin x dx. 39. 0 u = x ⇒ du = dx. ⇒ = Z π 0  x sin x dx = −x cos x 4 u = x2 ⇒ du = 2xdx. cos x dx = x cos x 0 π 0 π + sin x . 0 Z π 4 −π 4 sin xdx = dv ⇒ − cos x = v. 2 2 x sin x dx = −x cos x + 2 u = x ⇒ du = dx. ⇒ 0 Z π + x2 sin x dx. −π 4 ⇒ π    − π cos π + 0 cos 0 + sin π − sin 0 = π. Z π 41. sin xdx = dv ⇒ − cos x = v. Z π 4 −π 4 Z ln √ Z π 4 −π 4 x cos x dx. cos xdx = dv ⇒ − sin x = v. x2 sin x dx =    π 4 − x2 cos x + 2 x sin x − cos x = 0. −π 4 2 2 xex dx. 43. 0 u = x2 ⇒ du = 2xdx. x=0⇒u=0 ⇒ Z ln √2 xe 0 Z 2 45. x2 x= 1 dx = 2 √ ln 2 ⇒ u = ln 2. Z ln 2 0 1 ln 2 1 eu du = eu = . 2 0 2 xe−2x dx. 1 u = x ⇒ du = dx. ⇒ Z 2 xe 1 =− −2x 1 e−2x dx = dv ⇒ − e−2x = v. 2 2 1 1 dx = − xe−2x + 2 2 1 Z 2 1 e−2x dx = −1 −2x 2 −1 −2x 2 xe − e . 2 4 1 1  1  1  −4 5 3 2e − e−2 − e−4 − e−2 = − e−4 + e−2 . 2 4 4 4 Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 11 Z π 47. 2 e2x cos x dx. − π2 u = cos x ⇒ − sin xdx = du. ⇒ Z π 2 − π2 e 2x 1 1 cos x dx = e2x cos x + 2 2 u = sin x ⇒ − cos xdx = du. ⇒ Z π 2 − π2 5 ⇒ 4 ⇒ ⇒ 2.3 Z π 2 − π2 Z π 2 − π2 Z π 2 − π2 Z π 2 − π2 e2x sin x dx. 1 e2x dx = dv ⇒ e2x = v. 2 1 1  1 2x 1 cos x dx = e2x cos x + e sin x − 2 2 2 2 Z π 2 − π2  e2x cos x dx . 1 1 e2x cos x dx = e2x cos x + e2x sin x. 2 4 e2x cos x dx = π 4 2x 4 2 e cos x + e2x sin x −π . 10 20 2 e2x cos x dx =  4 π e + e−π . 20 Trigonometric Integrals Z 5. e 2x 1 e2x dx = dv ⇒ − e2x = v. 2 sin3 x cos x dx. Z Let sin x = u ⇒ cos xdx = du ⇒ sin3 x cos x dx. Z 1 1 = u3 du = u4 + C = sin4 x + C. 4 4 Z Z Z 3 3 3 2 7. sin x cos x dx = sin x cos x cos x dx = sin3 x(1 − sin2 x) cos x dx. Z Z 3 2 Let sin x = u ⇒ cos xdx = du ⇒ sin x(1 − sin x) cos x dx = u3 (1 − u2 ) du. Z 1 1 1 1 = (u3 − u5 ) du = u4 − u6 + C = sin4 x − sin6 x + C. 4 6 4 6 Z Z Z 9. sin2 x cos7 x dx = sin2 x cos6 x cos x dx = sin2 x(1 − sin2 x)3 cos x dx. Z Let sin x = u ⇒ cos xdx = du ⇒ sin2 x(1 − sin2 x)3 cos x dx. Z Z Z = u2 (1 − u2 )3 du = u2 (1 − 3u2 + 3u4 − u6 ) du = u2 − 3u4 + 3u6 − u8 du. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 12 1 3 3 1 1 3 3 1 = u3 − u5 + u7 − u9 + C = sin3 x − sin5 x + sin7 x − sin9 x + C. 3 5 7 9 3 5 7 9 Z Z   1 1 1 1 11. sin 5x cos 3x dx = (sin 2x + sin 8x) dx = − cos 2x − cos 8x + C. 2 2 2 8 1 1 = − cos 2x − cos 8x + C. 4 16 Z 17. tan4 x sec2 x dx. Z Z Z 1 4 2 4 2 1−1 n = 2 ⇒ k = 1 ⇒ tan x sec x dx = u (1 + u ) du = u4 du = u5 + C. 5 1 = tan5 x + C. 5 Z π 27. sin x cos4 x dx. 0 Z Z Z 1 4 2 0 4 m = 1 ⇒ k = 0 ⇒ sin x cos x dx = − (1 − u ) u du = − u4 du = − u5 . 5 Z π π 1 2 ⇒ sin x cos4 x dx = − cos5 x = . 5 5 0 0 Z π Z π Z π 2 2 2 sin2 x cos7 x dx = sin2 x cos6 x cos x dx = sin2 x(1 − sin2 x)3 cos x dx. 29. − π2 − π2 − π2 Let u = sin x ⇒ du = cos xdx. Z 1 1 3 3 1 1 32 ⇒ u2 (1 − u2 )3 du = u3 − u5 + u7 − u9 = . 3 5 7 9 315 −1 −1 Z π Z π 2 2 cos x cos 2x dx = cos x(1 − 2 sin2 x) dx. 31. − π2 − π2 Let u = sin x ⇒ du = cos xdx. Z π Z 1  2 2 2 1 2 ⇒ = . cos x(1 − 2 sin x) dx = (1 − 2u2 ) du = u − u3 3 3 −1 −π −1 2 2.4 Trigonometric Substitutions Z p p 1 p 1 x2 + 1 dx = x x2 + 1 + ln |x + x2 + 1| + C. 2 2 Z p 1 p 1 x 7. 1 − x2 dx = x 1 − x2 + arctan √ + C. 2 2 1 − x2 Z p p 1 p 1 x2 − 1 dx = x x2 − 1 − ln |x + x2 − 1| + C. 9. 2 2 Z p Z p p 2x p 2 1 11. 4x2 + 1 dx = (2x)2 + 1 dx = 4x + 1 + ln |2x + 4x2 + 1| + C. 2 2 Z p p 1 p 1 13. (4x)2 − 1 dx = 4x (4x)2 − 1 − ln |4x + (4x)2 − 1| + C. 2 2 Z Z 3 1 x √ 15. dx = 3 q √ dx = arcsin √ + C. 7 7 − x2 ( 7)2 − x2 5. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 13 Z 16. √ Z √ 5 p 1 q dx = 5 ln |x + x2 − 8| + C. √ x2 − ( 8)2 Z x2 − 8 5 − x2 dx = 5 q√ ( 5)2 − x2 1 1 1 p x  2 − arcsin √ dx = − 5 − x + C. 7x2 7 x2 7 x 5 Z 1p Z 1p 1 xp 1 2 27. 1 − x dx = 12 − x2 dx = 1 − x2 + arcsin(x) . 2 2 −1 −1 −1 25. Z dx = 1p  π 1√ 1 1 1 − 1 + arcsin(1) − 1 − −12 + arcsin(−1) = . 2 2 2 2 2 Z π Z π 2 sin(x) cos8 (x) 2 − 1 2 sin2 (x) cos7 (x) dx = − + cos7 (x) dx. 29. π π 2 + 7 2 + 7 − − = 2 2 sin(x) cos8 (x) 1  1 6 = + cos6 (x) sin(x) + 9 9 7 7 Z π 2 − π2  cos5 (x) dx . 61 4 sin(x) cos8 (x) 1  1 + cos6 (x) sin(x) + cos4 (x) sin(x) + = 9 9 7 7 5 5 Z π 2 − π2 sin(x) cos8 (x) 1 6 1 4 + cos6 (x) sin(x) + cos4 (x) sin(x) + 9 63 63 5 5 Z π sin(x) cos8 (x) 1 6 24 = + cos6 (x) sin(x) + cos4 (x) sin(x) + 9 63 315 315 Z π = cos3 (x) dx 2 − π2  .  cos3 (x) dx . 2 − π2 cos3 (x) dx. sin(x) cos8 (x) 1 6 24  1 + cos6 (x) sin(x) + cos4 (x) sin(x) + cos2 (x) sin(x)+ 9 63 315 315 3 Z π  2 2 cos(x) dx . 3 −π = 2 =  sin(x) cos8 (x) 9 + 1 6 24  1 cos6 (x) sin(x) + cos4 (x) sin(x) + cos2 (x) sin(x)+ 63 315 315 3  π 2 48  −π  96 32 π 2 = sin(x) sin( ) − sin( ) = = . π 3 945 2 2 945 315 −2 Z 1p 1 √ xp 9 x 1 31. 32 − x2 dx = 9 − x2 + arcsin( ) = 9 arcsin + 8. 2 2 3 −1 3 −1 2.5 Partial Fractions Decomposition Z 7. 7x + 7 dx = 2 x + 3x − 10 Math 1274 Z 7x + 7 dx. (x + 5)(x − 2) Minh Phuong Bui & Pichmony Anhaouy Solutions 14 A B + dx = (x + 5) (x − 2) Z = Z A(x − 2) + B(x + 5) dx. (x + 5)(x − 2) ⇒ A(x − 2) + B(x + 5) = 7x + 7 ⇒ x(A + B) + (5B − 2A) = 7x + 7.  A+B =7 ⇒ 5B − 2A = 7  A=4 . B=3 Z 4 3 7x + 7 dx = + dx = 4 ln |x + 5| + 3 ln |x − 2| + C. ⇒ 2 x + 3x − 10 (x + 5) (x − 2) Z Z −4 1 −4 9. dx = dx. 2 3x − 12 3 (x − 2)(x + 2) Z −4 = 3 Z A B −4 + dx = (x − 2) (x + 2) 3 Z A(x + 2) + B(x − 2) dx. (x + 2)(x − 2) ⇒ A(x + 2) + B(x − 2) = 1 ⇒ x(A + B) + (2A − 2B) = 1.  ⇒ A+B =0 ⇒ 2A − 2B = 1  A = 14 . B = − 14 −4 4 dx = − 3x2 − 12 3 Z Z 1 1 − dx. 4(x − 2) 4(x + 2) 1 1 = − ln |x − 2| + ln |x + 2| + C. 3 3 Z  Z A B C  −12x2 − x + 33 dx = + + dx. 13. (x − 1)(x + 3)(3 − 2x) (x − 1) x + 3 3 − 2x ⇒ A(x + 3)(3 − 2x) + B(x − 1)(3 − 2x) + C(x − 1)(x + 3) = −12x2 − x + 33. ⇒ x2 (−2A − 2B + C) + x(−3A + 5B + 2C) + 9A − 3B − 3C = −12x2 − x + 33.    −2A − 2B + C = −12  A=5 −3A + 5B + 2C = −1 ⇒ B=2 .   9A − 3B − 3C = 33 C=2 Z  A B C  + + dx = (x − 1) x + 3 3 − 2x Z  5 2 2  + + dx. (x − 1) x + 3 3 − 2x = 5 ln |x − 1| + 2 ln |x + 3| − ln |3 − 2x| + C. Z 15. x2 + x + 1 dx = x2 + x − 2 Math 1274 Z x2 + x − 2 + 3 dx = x2 + x − 2 Z  1+  3 dx. x2 + x − 2 Minh Phuong Bui & Pichmony Anhaouy Solutions 15 Z = Z 1 dx + =x+ Z  3 dx = x + 2 x +x−2 Z 3 dx. (x − 1)(x + 2) B  A + dx (x − 1) (x + 2) ⇒ A(x + 2) + B(x − 1) = 3 ⇒ x(A + B) + (2A − B) = 3.  A+B =0 ⇒ 2A − B = 3 ⇒x+ Z   A=1 . B = −1 A B  + dx = x + (x − 1) (x + 2) Z  1 −1  + dx. (x − 1) (x + 2) = x + ln |x − 1| − ln |x + 2| + C. Z Z Z 2x2 − 4x + 6 2(x2 − 2x + 3) 17. dx = dx = 2 dx = 2x + C. x2 − 2x + 3 x2 − 2x + 3 Z Z  Z A Bx + C  6x2 + 8x − 4 dx = = + dx. 21. (x − 3)(x2 + 6x + 10) x − 3 x2 + 6x + 10 ⇒ A(x2 + 6x + 10) + (Bx + C)(x − 3) = 3. ⇒ x2 (A + B) + x(6A − 3B + C) + (10A − 3C) = 6x2 + 8x − 4.    A=2  A+B =6 B=4 . 6A − 3B + C = 8 ⇒   C=8 10A − 3C = −4 ⇒ Z Z  A Bx + C  + 2 dx = x − 3 x + 6x + 10 2 dx + x−3 Z = Z 4x + 8 dx = 2 x + 6x + 10 2 dx + 2 x−3 Z Z  2 dx + x−3 Z 2x + 6 dx − 2 x + 6x + 10 2 = 2 ln |x − 3| + 2 ln |x + 6x + 10| − 4 2 4x + 8  + 2 dx =. x − 3 x + 6x + 10 Z Z 4x + 12 − 4 dx. x2 + 6x + 10 4 Z x2 + 6x + 10 dx. 1 dx. (x + 3)2 + 1 = 2 ln |x − 3| + 2 ln |x2 + 6x + 10| − 4 arctan (x + 3) + C. Z Z Z  x2 − 20x − 69 A Bx + C  23. dx = = + dx. (x − 7)(x2 + 2x + 17) x − 7 x2 + 2x + 17 ⇒ A(x2 + 2x + 17) + (Bx + C)(x − 7) = 3. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 16 ⇒ x2 (A + B) + x(2A − 7B + C) + (17A − 7C) = x2 − 20x − 69.    A+B =1  A = −2 2A − 7B + C = −20 ⇒ B=3 .   17A − 7C = −69 C=5 ⇒ Z  Bx + C  A + 2 dx = x − 7 x + 2x + 17 = −2 ln |x − 7| + = −2 ln |x − 7| + 3 −2 3x + 5  + 2 dx. x − 7 x + 2x + 17 3x + 3 + 2 dx. x2 + 2x + 17 Z = −2 ln |x − 7| + 3 Z  Z Z 2 (x + 1) dx + 2 x + 2x + 1 + 16 Z (x + 1) dx + 2 (x + 1)2 + 42 1 dx. (x + 1)2 + 42 Z x2 + 2x + 1 + 16 dx. x+1 3 ln |x2 + 2x + 17| + arctan ( ) + C. 2 4 Z Z Z  6x2 + 45x + 121 A Bx + C  25. dx = = + dx. (x + 2)(x2 + 10x + 27) x + 2 x2 + 10x + 27 = −2 ln |x − 7| + ⇒ A(x2 + 10x + 27) + (Bx + C)(x + 2) = 6x2 + 45x + 121. ⇒ x2 (A + B) + x(10A + 2B + C) + (27A + 2C) = 6x2 + 45x + 121.    A=5  A+B =6 B=1 . 10A + 2B + C = 45 ⇒   C = −7 27A + 2C = 121 ⇒ ⇒ Z  A Bx + C  dx = + 2 x + 2 x + 10x + 27 Z   5 x−7 dx. + 2 x + 2 x + 10x + 27 Z  A Bx + C  + 2 dx = x + 2 x + 10x + 27 Z   x−7 5 + 2 dx. x + 2 x + 10x + 27 = 5 ln |x + 2| + = 5 ln |x + 2| + = 5 ln |x + 2| + Math 1274 Z x + 5 − 12 x2 + 10x + 25 + 2 Z dx. x+5 dx − 12 (x + 5)2 + 2 Z 1 dx. (x + 5)2 + 2 x + 5 √ 1 + C. ln |x2 + 10x + 27| − 6 2 arctan √ 2 2 Minh Phuong Bui & Pichmony Anhaouy Solutions 17 14x + 6 dx = (3x + 2)(x + 4) Z 27. Z = Z  A B  + dx. 3x + 2 x + 4 ⇒ A(x + 4) + B(3x + 2) = 14x + 6. ⇒ x(A + 3B) + (4A + 2B) = 14x + 6.  A + 3B = 14 ⇒ 4A + 2B = 6 Z 5 0  A = −1 . B=5 B  A + dx = 3x + 2 x + 4 Z 5 0 5  −1 + dx. 3x + 2 x + 4  9  1  17  5 1 = − ln |3x + 2| + 5 ln |x + 4| = 5 ln − ln . 3 4 3 2 0 Z 1 Z 1 x x+1−1 29. dx = dx. 2 + 2x + 1) 2 + 2x + 1) (x + 1)(x (x + 1)(x 0 0 Z 1 x+1 dx − (x + 1)(x2 + 2x + 1) = 0 Z 1 1 dx − (x + 1)2 = 0 2.6 Z 1 0 Z 1 0 1 dx. (x + 1)(x2 + 2x + 1)  1 1 1 1 1 dx = − + = . (x + 1)3 x + 1 2(x + 1)2 0 8 Improper Integration Z ∞ 7. 5−2x e Z t dx = lim t→∞ 0 0 e5−2x dx = lim  t→∞ t 1 − e5−2x . 2 0 e5 1 lim (e5−2t − e5 ) = . 2 t→∞ 2 Z 0 Z 0 1 1 11. 2x dx = lim 2x dx = lim (20 − 2t ) = . t→−∞ t ln 2 t→−∞ ln 2 −∞ =− Z ∞ 13. x dx = 2 −∞ x + 1 Z 0 x dx + 2 −∞ x + 1 Z ∞ 0 x dx = lim 2 t→−∞ x +1 Z 0 t x dx + lim 2 t→∞ x +1 Z t x 0 x2 + 1 dx. 0 t 1 1 ln |x2 + 1| + lim ln |x2 + 1| = ∞ t→−∞ 2 t→∞ 2 t 0 Z ∞ Z t  1  1 1 1 t 15. dx = lim dx = lim = lim − 1 = −1. t→∞ t − 1 t→∞ 2 (x − 1)2 t→∞ x − 1 2 (x − 1)2 2 = lim Z 1 19. 1 dx = −1 x Math 1274 Z 0 1 dx + −1 x Z 1 0 1 dx = lim t→0 x Z t 1 dx + lim t→0 −1 x Z 1 t 1 dx. x Minh Phuong Bui & Pichmony Anhaouy Solutions 18 = lim ln |x| t→0 Z ∞ xe 23. −x t −1 1 + lim ln |x| t→0 Z t dx = lim xe t→∞ 0 0 t = ∞. −x  dx = lim xe t→∞ −x t 0 Z t +  e−x dx . 0    t = lim te−t − e−t + e0 = 1. = lim xe−x − e−x t→∞ Z ∞ 25. xe t→∞ 0 −x2 Z 0 dx = −∞ xe −x2 Z ∞ dx + −∞ 0 1 1 2 2 0 2 t xe−x dx = lim − e−x + lim − e−x . t→−∞ 2 t→∞ 2 t 0 1 1 = − (1 − 0)) − (0 − 1) = 0. 2 2 Z Z 1 Z 1  1 1 1 x dx . x2 ln x − 27. x ln x dx = lim x ln x dx = lim t→0 t t→0 2 2 t 0 1 1 1 =− . x2 ln x − x2 t→0 2 4 4 t Z ∞ Z t 33. e−x cos x dx = lim e−x cos x dx. = lim 1 t→∞ 0 0 Z t e −x 0 −x cos x dx = −e = −e−x cos x − ⇒ ⇒ 2.7  cos x − − e−x sin x + Z t e−x sin x dx. 0 Z t  e−x cos x dx . 0 Z t t 1 e−x cos x dx = e−x (sin x − cos x) . 2 0 0 Z ∞ e−x cos x dx = 0  t 1 1 lim e−x (sin x − cos x) = . t→∞ 2 2 0 Numerical Integration 5. (a): Z 10 5x dx. 0 10 − 0 = 2.5. Here, f (x1 ) = f (2.5) = 12.5, f (x2 ) = f (5) = 25, f (x3 ) = f (7.5) = 37.5, f (x4 ) = 4 f (10) = 50. Z 10 2.5 5x dx = (0 + 2(12.5 + 25 + 37.5) + 50) = 250. 2 0 ∆x = 5. (c): Z 10 0 5 10 5x dx = x2 = 250. 2 0 Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 19 11. (a): Z 3p 9 − x2 dx. −3 √ 3 − (−3) 3 3 ∆x = = 1.5. Here, f (x0 ) = f (−3) = 0, f (x1 ) = f (−1.5) = , f (x2 ) = f (0) = 9, f (x3 ) = 4√ 2 3 3 f (1.5) = , f (x4 ) = f (3) = 0. 2 √ √ Z 3p √ 1.5 3 3 3 3 9 2 9 − x dx = (0 + 2( +3+ ) + 0) = (1 + 3). 2 2 2 2 −3 11. (c): Z 3p xp 9 9 x 3 = π. 9 − x2 dx = 9 − x2 + arcsin 2 2 3 −3 2 −3 Z 1 2 13. ex dx. −1 4 1 2 1 1 − (−1) 1 = . f (x0 ) = f (−1) = ef (x1 ) = f (− ) = e 9 , f (x2 ) = f (− ) = e 9 , f (x3 ) = f (0) = 6 3 3 3 1 4 1 2 e0 = 1, f (x4 ) = f ( ) = e 9 , f (x5 ) = f ( ) = e 9 , f (x6 ) = f (1) = e1 = e. 3 3 Z 1 4 1 1 2 e2 dx = (e + 2(2e 9 + 2e 9 + 1) + e) = 3.0241. 6 −1 Z π 15. x sin x dx. 0 √ π−0 π π π π 3π ∆x = = . Here, f (x0 ) = f (0) = 0; f (x1 ) = f ( ) = , f (x2 ) = f ( ) = , f (x3 ) = 6 6 6 12 3 6 5π π π 2π π 5π f ( ) = , f (x4 ) = f ( ) = √ , f (x5 ) = f ( ) = , f (x6 ) = f (π) = 0. 2 2 3 6 12 3 √ Z π π  3π π 5π  π π x sin x dx = 0+2 + + +√ + + 0 = 3.0695. 12 12 6 2 3 12 0 ∆x = 21. (a): Z 4 1 √ dx. x 1 3 3 3 f ” = x−2.5 < f” < < 1 ⇒ M = 1. 4 128 4 (4 − 1)3 ET = < 0.0001 ⇒ n = 150. 12n2 23. (a): Z 5 x4 dx. 0 f ” = 12x2 ET = 300 Math 1274 0 < f ” < 300 (5 − 0)3 12n2 < 0.0001 ⇒ M = 300. ⇒ n = 5591. Minh Phuong Bui & Pichmony Anhaouy Solutions 20 Chapter 3 Applications of Integration 3.1 Area Between Curves Z 1 5. A = −1 3 2 (−3x + 3x + 2) − (x + x − 1) dx = Z 1 −1 (−3x3 − x2 + 2x + 3) dx. 1 1 16 3 = . − x4 − x3 + x2 + 3x 4 3 3 −1 Z π Z π π 7. A = (sin x + 1) − (sin x) dx = 1 dx = x = π. =  0 Z 5π 4 9. A = 0 0 π 4 5π √ 4 (sin x − cos x) dx = (cos x + sin x) π = 2 2. 4 11. 2x2 + 5x − 3 = x2 + 4x − 1 ⇒ x2 + x − 2 = 0 ⇒ x = 1, −2. Thus, Z 1 A= −2 2 2 (2x + 5x − 3) − (x + x − 2) dx = Z 1 −2 (x2 + x − 2) dx = 1 1 1 9 x3 + x2 − 2x = . 3 2 2 −2 π π 2x = 0 ⇒ x = − , 0, . Due to symmetry property, we take integral for the positive part π 2 2 π (from 0 to ), and double the result. 2 13. sin x − A=2 Z π 2 0 15. x − √  2x  x2  π2 π sin x − dx = 2 − cos x − =2− . π π 0 2 x=0⇒ √ √ x( x − 1) = 0 ⇒ x = 0, 1. Z 1 2 √ 1 1 1 ( x − x) dx = x1.5 − x2 = . 3 2 6 0 0 Z 1 Z 2 √ 1  1  19. A = x + x dx + (−2x + 3) + x dx 2 2 0 1 A= = 2 1 1  1 2 5 x1.5 + x2 + − x2 + 3x + x2 = . 3 4 4 3 0 1 21 1 1 21. − y + 1 = y 2 ⇒ −y + 2 = y 2 ⇒ y 2 + y − 2 = 0 ⇒ y = −2, 1. So, 2 2 A= Z 1 −1 1 2 y dy + 2 Z 1 −2  1 9 1  1 1 1 1 1 + − y2 + y − y3 = . − y + 1 − y 2 dy = y 3 2 2 6 −1 4 6 4 −2 23. The line goes through (1,1) and (2,3) is y − 1 = 3−1 (x − 1) ⇒ y = 2(x − 1) + 1 ⇒ y = 2x − 1. 2−1 The line goes through (1,1) and (3,3) is y − 1 = 3−1 (x − 1) ⇒ y = (x − 1) + 1 ⇒ y = x. 3−1 The line goes through (2,3) and (3,3) is y − 3 = 3−3 (x − 3) ⇒ y = 3. 3−2 Intersection points are (1.5, 1.5), (3, 3) and (2, 3). Therefore, A= Z 2 1 = 1 2 Math 1274  (2x − 1) − x dx + x2 − x Z 3 2 (3 − x) dx = Z 2 1.5 (x − 1) dx + Z 3 2 (3 − x) dx.  1 3 + 3x − x2 = 1. 2 1 2 2 Minh Phuong Bui & Pichmony Anhaouy Solutions 22 Chapter 4 More Applications 4.1 Continuous Income Streams Z ∞ s  1 e−5r − e−rs e−5r = 200 · lim 1. P V = 200 · e−rt dt = 200 · lim − e−rt = 200 · . s→∞ s→∞ r  r r 5  5 −5r   1−e 1 − e−5r e−5r 1000 · = P V ⇒ 1000 · = 200 · ⇒ 5 − 5e−5r = e−5r ⇒ 6e−5r = 5 ⇒ e−5r = r r r 5 6 1 6 ≈ 3.64%. ⇒ e5r = ⇒ r = ln 6 5 5 5 40 2000e−0.09t 3. The future value is F V = f (t) dt = (2000 + 400t)e dt = e + −0.09 0 0 0 40  −0.09t Z Z e−0.09t ert rt 3.6 te − e = 2, 371, 230. Note that we use the fact that e = + c and tert = −0.09 −0.092 0 r tert ert − 2 + c when evaluate this integral. r r The answer is $2,371,230, which in current dollars assuming a 4% rate of inflation is $478,743 - enough to Z Z 40 Z 40 0.09(40−t) 3.6  40 live on for a good number of years. The amount of principal deposited is (2000+400t)dt = 400, 000, 0 which leaves roughly two million of interest - the lion’s share. Note that even the first $2000 deposited becomes $73,200. Z 7 5. The present value of saving is 10, 000e−0.10t dt = · · · = $50, 340. The present value for maintenance 0 Z 7 and repair is (1000 + 200t)e−0.10t dt = · · · = $8149.60. The present value of the scrap for $1000 in 0 7 years is 1000e−0.10·7 = $496.60. So, the maximum amount that we should be willing to pay now is $50, 340 + $8149.60 + $496.60 = $42, 687. 4.2 1. Consumers’ and Producers’ Surpluses a) D(x) = S(x) ⇒ 100 − 0.05x = 10 + 0.1x ⇒ 0.15x = 90 ⇒ x = 600 ⇒ p = 100 − 0.05(600) = 70. So, the equilibrium quantity and price is (70, 600). Z 600 h Z 600 h i i b) CS = D(x) − 70 dx = 30 − 0.05x dx = 9000 (dollars/unit), and Z0 600 h Z 0600 h i i PS = 70 − S(x) dx = 60 − 0.1x dx = 18, 000 (dollars/unit) 0 0 23 3. a) D(x) = S(x) ⇒ e9−x = ex+3 ⇒ 9 − x = x + 3 ⇒ 2x = 6 ⇒ x = 3 ⇒ p = e6 ≈ 403. So, the equilibrium quantity and price is (403, 3). Z 3h Z 3h i i b) CS = D(x) − 403 dx = e9−x − 403 dx = 45, 432 (dollars/unit), and Z0 3 h Z 03 h i i PS = 403 − S(x) dx = 403 − ex+3 dx = ... (dollars/unit). 0 5. 0 a) D(x) = S(x) ⇒ 20e−x = 5ex ⇒ e2x = 4 ⇒ 2x = ln 4 = 2 ln 2 ⇒ x = ln 2 ⇒ p = 5eln 2 = 10. So, the equilibrium quantity and price is (10, ln 2). Z ln 2 h Z ln 2 h i i b) CS = D(x) − 10 dx = 20e−x − 10 dx = 3.07 (dollars/unit), and Z0 ln 2 h Z 0ln 2 h i i PS = 10 − S(x) dx = 10 − 5ex dx = 1.93 (dollars/unit). 0 4.3 0 Probabilities Z ∞  1 s e−5r e−5r − e−rs 1. P V = 200 · e−rt dt = 200 · lim − e−rt = 200 · . = 200 · lim s→∞ s→∞ r  r r 5  5 −5r   1−e 1 − e−5r e−5r 1000 · = P V ⇒ 1000 · = 200 · ⇒ 5 − 5e−5r = e−5r ⇒ 6e−5r = 5 ⇒ e−5r = r r r 5 6 1 6 ⇒ e5r = ⇒ r = ln ≈ 3.64%. 6 5 5 5 40  −0.09t 40 2000e−0.09t e−0.09t 3.6 te f (t) dt = (2000 + 400t)e dt = e +e − 3. F V = −0.09 −0.09 −0.092 0 0 0 0 Z Z rt rt rt e e te = 2, 371, 230. Note that we use the fact that ert = + c and tert = − 2 + c when evaluate r r r this integral. Z 40 Z 40 0.09(40−t) 3.6  The answer is $2,371,230, which in current dollars assuming a 4% rate of inflation is $478,743 - enough to Z 40 live on for a good number of years. The amount of principal deposited is (2000+400t)dt = 400, 000, 0 which leaves roughly two million of interest - the lion’s share. Note that even the first $2000 deposited becomes $73,200. Z 7 5. The present value of saving is 10, 000e−0.10t dt = · · · = $50, 340. The present value for maintenance 0 Z 7 and repair is (1000 + 200t)e−0.10t dt = · · · = $8149.60. The present value of the scrap for $1000 in 0 7 years is 1000e−0.10·7 = $496.60. So, the maximum amount that we should be willing to pay now is $50, 340 + $8149.60 + $496.60 = $42, 687. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 24 Chapter 5 Multivariate Calculus 5.1 The Three-Dimensional Coordinate System 1. Position now (−3, 4, 1); distance travelled is 3+4+1 = 8 units; distance from origin = √ 26 ≈ 5.0990 units. p (−3)2 + 42 + 12 = z ( 3, 0, 1) ( 3, 4, 1) 2 -3 (0, 0, 1) 1 -2 ( 3, 4, 0) (0, 4, 1) -1 0 1 2 3 4 y (0, 4, 0) x 3. a) Midpoint P1,2 between P1 and P2 = b) Take (1+6t, 2−5t, 1+3t) t=0 1 + 7 2 − 3 1 + 4  1 5 , , = 4, − , . 2 2 2 2 2 = (1, 2, 1) ≡ P1 ; take (1+6t, 2−5t, 1+3t) (7, −3, 4) ≡ P2 . Notice, P1,2 = (1 + 6t, 2 − 5t, 1 + 3t) t=1/2 t=1 = (1+6, 2?5, 1+3) = . c) If Q(2, 7, −5) lies on the line through P1 and P2 , then there is a common t-value that will give Q. Start with x-coordinate: 2 = 1 + 6t ⇒ t = 1/6; next y-coordinate: 7 = 2 − 5t ⇒ t = −1. Different t-values for each coordinate means Q is not collinear with P1 and P2 . 5. x2 + y 2 + z 2 + x + y + z = 0 ⇒ x2 + x + y 2 + y + z 2 + z = 0 ⇒ x2 + 2(1/2)x + (1/2)2 + y 2 + 2(1/2)y + 2 2 2 2 2 (1/2)2 + z 2 + 2(1/2)z + (1/2) √ = 3(1/2) ⇒ (x + 1/2) + (y + 1/2) + (z + 1/2) = 3/4. Centre: (−1/2, −1/2, −1/2); radius: 3/2. 7. Plane x = 8: 64 + y 2 + z 2 = 100 ⇒ y 2 + z 2 = 36. Circle in yz-plane with : centre (0, 0); radius: 6. 9. Plane z = 10: x2 + y 2 + 100 = 100 ⇒ x2 + y 2 = 0. Circle (point) in xy-plane with: centre (0, 0); radius: 0. 25 5.2 1. Planes and Surfaces a) TRUE: planes’ orientation is determined by their normals (relative size and sign of each variable’s coordinate); so if two normal are each parallel to a third normal, then they must be parallel to each other; so the planes are parallel. Without normals terminology: if two planes are parallel, then the coordinates for each variable are in a fixed ratio (planes x + y + z = d1 and −77x − 77y − 77z = d2 are parallel); therefore, if if planes #1 and #2 are each parallel to plane #3, then the ratio of coordinates for the three variables of #1 to those of #2 will also be in a fixed ratio. b) FALSE: consider planes x + y + z = d1 and y = d2 : they are both perpendicular to the plane x − z = d3 but are not parallel to each other. Notice, however, that the converse is TRUE: two parallel planes will be perpendicular to the same planes. c) FALSE: A plane parallel to a line means that the plane is parallel to a plane through the line (the plane can be moved - translated - so it contains the line); two non-parallel planes can certainly pass through the same line. For instance, planes x + y + z = d1 and y = d2 are both parallel to the line, passing through the origin, x = z, but as was argued in part (b), those planes are not parallel. d) TRUE: The quick answer is again using normal: the each plane’s normal is perpendicular to that plane; so if two planes are perpendicular to one particular line, then those planes’ normal must be parallel to the line, and so parallel to each other; and so the planes must be parallel. 3. The plane 5x − 13y + z = −7 passes through all three points - you can check by substituting each point. 5. A point on the line of intersection of the planes must satisfy both planes’ equations. That is, if the point is (p, q, r), then p + q + r = 6 and p − q + r = 0; subtracting these equations gives, q = 3 and so p + r = 3; pick two p-values, say p = 0 and p = 1; The associated points are (0, 3, 3) and (1, 3, 2). 7. a) The domain of f is all (x, y) such that 1 + x + y > 0; that is, the line y = (1 + x) is the excluded boundary, and y-values should be larger than (1 + x); as shown in figure below, where the feasible region is shaded and its diagonal edge is not included.   b) The range of ln (1 + x + y) is R, all real numbers. 5.3 Functions of Several Variables 3. z = y 1/2 − x1/2 ⇒ y = (z + x)2 , with x, y > 0. So for z = 0, 1, 2: z = 0 ⇒ y = x, x > 0; 2 2   z = 1 ⇒ y = 1 + x1/2 ; and z = 2 ⇒ y = 2 + x1/2 . The contours are in ascending order: the lowest (and thinnest) is for z = 0; the highest (and thickest) is for z = 2. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 26   5. z = xey ⇒ y = ln (z/x) = ln z − ln x, when x and z are positive, and ln (−z) − ln (−x) when both are negative. For z = 1 ⇒ y = − ln x; z = −1 ⇒ y = − ln (−x); z = e ⇒ y = 1 − ln x; and z = e ⇒ y = 1 − ln (−x). The curves appear in symmetric pairs (reflected about the y-axis); the positive z-values, the first and third functions, are the thin curves on the right, positive side; the negative z-values, the second and fourth functions, are the thicker curves on the left, negative x-axis side; the lower curves are for the first two functions, with |z| = 1; and the upper curves are for the second two functions, with |z| = e.   7. yz-traces: x = 0 ⇒ z = (2y)2 = 4y 2 ; x = 1 ⇒ z = (2y − 1)2 . These yz-plane graphs. .   xz-traces: y = 0 ⇒ z = (−x)2 = x2 , y = 1/2 ⇒ z = (1 − x)2 . These xz-plane graphs. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 27   9. yz-traces: x = 0 ⇒ z = −y, the lower, thinnest line (passing through origin); x = 1 ⇒ z = 1 − y, middle line; x = −2 ⇒ z = 4 − y, upper, thickest line. These yz-plane graphs. .   xz-traces: y = 0 ⇒ z = x2 , the middle, thinnest line (tangent to x-axis at origin); y = 1 ⇒ z = x2 − 1, the lowest line; y = −2 ⇒ z = x2 + 2, the top, thickest line (with z intercept 2). These xz-plane graphs.   11. yz-traces: x = 0 ⇒ z = 2(ln y − ln 3), the lowest, thinnest line with the z-axis its vertical asymptote  y2  ; x = 1 ⇒ z = ln 1 + , the top line tangent to the y-axis at the origin; x = 1/3 ⇒ z = 9 2 ln (1 + y ) − 2 ln 3, the thickest line crossing the z-axis near z = −2. These yz-plane graphs. . xz-traces: y = 0 ⇒ z = 2 ln x, the lowest, thinnest line with the z-axis its vertical asymptote ; Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 28   y = 1 ⇒ z = ln (x2 + 1/9), the thickest line crossing the z-axis near z = 2; y = 3 ⇒ z = ln (x2 + 1) is the highest line tangent to the x-axis at the origin. These xz-plane graphs.   5.4 Partial Derivatives 2  y 1/3 2  27 1/3 2 3 · ⇒ fx (8, 27) = 100 · · = 100 · · = 100. That is, the MPL is 3 x 3 8 3 2 $100.00 additional production for each additional hour labour used. 1. fx (x, y) = 100 · 3. a) fx (0, 0) = 0 [looks maybe < 0]; b) fy (0, 0) = 0; c) fx (−2, 0) < 0; d) fy (−2, 0) > 0 [looks maybe = 0]. 5. fx = −72x; fy = −128y; fxx = −72; f xy = 0; fyy = −128. Thus, we have D = (−72)(−128)−(0)2 = 210 · 32 = 9216(> 0).  y y 1 y     1 2 7. fx = , fy = ln x, fxx = − 2 , fxy = , fyy = 0. Thus, we have D = − 2 · 0 − = x x x x x 1 − 2 < 0. x 9. fx = yex+y , fy = (1 + y)ex+y , fxx = yex+y , fxy = (1 + y)ex+y , fyy = (2 + y)ex+y . Thus, we have      2 D = yex+y · (2 + y)ex+y − (1 + y)ex+y = e2(x+y) (−1) < 0. Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 29 y −x −2xy x2 − y 2 2xy = = = . Thus, we have , f , f , f , fyy = 2 y xx xy x2 + y 2 x2 + y 2 (x2 + y 2 )2 (x2 + y 2 )2 (x + y 2 )2  −2xy     x2 − y 2 2 1 2xy D= =− 2 · − < 0. (x2 + y 2 )2 (x2 + y 2 )2 (x2 + y 2 )2 (x + y 2 )2 11. fx = 13. 15. 5.5 ∂z ∂z ∂z ∂z = f 0 (xy) · y ⇒ = 5, and = f 0 (xy) · x ⇒ = 5. Now, zxx = f 00 (xy) · y 2 , zxy = ∂x ∂x 1,1 ∂y ∂y 1,1  2 f 0 (xy) + f 00 (xy) · xy, zyy = f 00 (xy) · x2 . So, D(x, y) = − f 0 (xy) − 2xyf 0 (xy)f 00 (xy) ⇒ D(1, 1) = 5. a) f (x, y) = axα y β ⇒ fx = aαxα−1 y β , fy = aβxα y β−1 , fxx = aα(α − 1)xα−2 y β , fxy = aαβxα−1 y β−1 , fyy = aβ(β − 1)xα y β−2 .     2   b) D(x, y) = aα(α−1)xα−2 y β · aβ(β −1)xα y β−2 − aαβxα−1 y β−1 = a2 αβx2(α−1) y 2(β−1 1−  α − β = a2 x2(α−1) y 2(β−1) {1 − (α + β)}. When, α + β = 1, D(x, y) = 0 for all (x, y) ∈ R2 . Maxima and Minima     1. • at 0, 0, f (0, 0) = 0 , T (x, y) = x + y. 0, 0, f (0, 0) = 0 , T (x, y) = f (0, 0) + fx (0, 0)(x − 0) + fy (0, 0)(y − 0) = x + y.   1 1 1 • at 1, 0, f (1, 0) = ln 2 , T (x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (0, 0)(y − 0) = ln 2 − + x + y. 2 2 2   1 1 1 at 1, 0, f (1, 0) = ln 2 , T (x, y) = f (1, 0) + fx (1, 0)(x − 1) + fy (0, 0)(y − 0) = ln 2 − + x + y. 2 2 2     2 1 1 • at 1, 1, f (1, 1) = ln 3 , T (x, y) = ln 3 − + x + y. at 1, 1, f (1, 1) = ln 3 , T (x, y) = 3 3 3 2 1 1 f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1) = ln 3 − + x + y. 3 3 3     • at 1, −1, f (1, −1) = 0 , T (x, y) = x + y. at 1, −1, f (1, −1) = 0 , T (x, y) = f (1, −1) + fx (1, −1)(x − 1) + fy (1, −1)(y + 1) = x + y.   3. fx = 2x + 2, fy = 2y − 2. So, at 1, 1, f (1, 1) = 2 , we have T (x, y) = f (1, 1) + fx (1, 1)(x − 1) + fy (1, 1)(y − 1) = 2 + 4(x − 1) + 0(y − 1) = −2 + 4x. Thus, f (0.9, 1.1) ≈ T (0.9, 1.1) = −2 + 4(0.9) = 1.6. T (x, y) = −2 + 4x. f (0.9, 1.1) ≈ T (0.9, 1.1) = −2 + 4(0.9) = 1.6. 5. a) D = (1)(1) − (−2)2 = −3 < 0. So, it is a saddle point. b) D = (−1)(−2) − (1)2 = 1 > 0, fxx = −1 < 0. So, it is a local maximum. c) D = (−1)(1) − (0)2 = −1 < 0. So, it is a saddle point. 2 2 7. fx = 3x2 + 2x = x(3x + 2) = 0 ⇒ x = 0, − , fy = −3y 2 + 2y = y(−3y + 2) = 0 ⇒ y = 0, . So, we 3  2   2  3 2 2  have four points: (0, 0), − , 0 , 0, , − , . Now, fxx = 6x + 2, fxy = 0, 3 3 3 3 fyy = −6y + 2 ⇒ D(x, y) = (6x + 2)(−6y + 2) − (0)2 = 4(3x + 1)(1 + 3y). Then, we have • D(0, 0) = 4 > 0, fxx (0, 0) = 2 > 0 ⇒ the point is a local minimum.  2  • D − , 0 = −4 < 0 ⇒ the point is a saddle point. 3  2 • D 0, = −4 < 0 ⇒ the point is a saddle point. 3  2 2  2 2 • D − , = 4 > 0, fxx − , < 0 ⇒ the point is a local maximum. 3 3 3 3 Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 30 1 x2 y − 1 1 8 xy 2 + 8 9. fx = − 2 + y = = 0 ⇒ y = , f = x + = = 0 ⇒ xy 2 = −8 y x x2 x2 y2 y2  1  1 1 1 ⇒ x · 2 = −8 ⇒ x3 = − ⇒ x = − ⇒ y = 4. So, we have a point: − , 4 . x 8 2 2  2  16  2 16 32 Now, fxx = 3 , fxy = 1, fyy = − 3 ⇒ D(x, y) = − 3 − (1)2 = − 3 3 − 1. Then, 3 x y x y  x1   1y  we have D − , 4 = 3 > 0, fxx − , 4 = −16 < 0 ⇒ the point is a local maximum. 2 2 11. fx = 3x2 − 6y = 0 ⇒ x2 = 2y, fy = −6x + 3y 2 = 0 ⇒ 2x = y 2 . From here, we get  y 2 2 = 2y ⇒ 2 y 4 = 8y ⇒ y 4 − 8y = 0 ⇒ y(y 3 − 8) =⇒ y = 0, y = 2 ⇒ x = 0, 2 as well. So, we have two points: (0, 0), (2, 2). Now, fxx = 6x, fxy = −6, fyy = 6y ⇒ D(x, y) = (6x)(6y) − (−6)2 = 36(xy − 1). Then, we have • D(0, 0) = −36 < 0 ⇒ the point is a saddle point.     • D 2, 2 = 108 > 0, fxx 2, 2 = 12 > 0 ⇒ the point is a local minimum. 13. fx = 6y − 2x = 0 ⇒ 3y = x, fy = 6x − 18y 2 = 0 ⇒ x = 3y 2 . Solving these two equations gives y = y 2 ⇒ y = 0, 1 ⇒ x = 0, 3. So, we have two points: (0, 0), (3, 1). Now, fxx = −2, fxy = 6, fyy = −36y ⇒ D(x, y) = (−2)(−36y) − (6)2 = 36(2y − 1). Then, we have • D(0, 0) = −36 < 0 ⇒ the point is a saddle point. • D(3, 1) = 180 > 0, fxx (3, 1) = −2 < 0 ⇒ the point is a local maximum. 5.6 Lagrange Multipliers 1. Here, g = x2 + y 2 − 1. fx = λgx ⇒ 2x = λ(2x) ⇒ x = 0 or λ = 1, but x 6= 1. fy = λgy ⇒ 1 = √ λ(2y). 1 1 3 . If x = 0 we have y 2 = 1 ⇒ y = ±1, and λ = ± . But, if x 6= 0, then λ = 1 ⇒ y = ⇒ x = ± 2 2 2  √3 1  So, we have four points: (0, 1), (0, −1), ± , . Thus, 2 2 1 • f (0, 1) = 1, corresponding to λ = . 2 1 • f (0, −1) = −1, which is a minimum, corresponding to λ = − . 2  √3 1  5 • f ± , = , which is a maximum, corresponding to λ = 1. 2 2 4 3. Here, g = 3x + 5y − 47. fx = λgx ⇒ 2(x − 1) = 3λ. fy = λgy ⇒ 2(y − 2) = 5λ. This gives 2 2 λ = (x − 1) = (y − 2) ⇒ −5x + 3y = 1. Together with 3x + 5y = 47, we can solve for x and y to 3 3 be x = 4, y = 7. Thus, f (4, 7) = 30, which is a minimum, corresponding to λ = 2. 5. 1 1 1 √ 1 √ , = , corresponding to λ = 1. Here, g = x + y − 1. fx = λgx ⇒ 1 = λ · √ , x > 0. 4 4 2 2 x √ 1 √ fy = λgy ⇒ 2 = λ · √ , y > 0. This gives λ = 2 x = 2 y ⇒ x = y. Together with 2 y 1 1 1 √ 1 1 √ x + y = 1, we can solve for x and y to be x = , y = . Thus, f , = , corresponding 4 4 4 4 2 to λ = 1. a) f Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 31 1 1 √ 1 √ = , but f (1, 0) = f (0, 1) = 1. From x + y = 1, we can solve for y to be , 4 4 2  √ 2 √ 1  1 y = (1 − x) ⇒ y 0 = 2(1 − x) · − √ = 1 − x−1/2 ⇒ y 00 = x−3/2 > 0. 2 2 x √ 1 √ c) At this point, contour z = is a tangent to the constraint x + y = 1. Points (0, 1) and (1, 0) 2 √ √ are corner point solutions (since x + y = 1 meets implicit constraint constraints y ≥ 0 and x ≥ 0. b) f Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 32 Chapter 6 Differential Equations 6.1 Graphical and Numerical Solutions to Differential Equations 1. No Selected Questions. 6.2 Separable Differential Equations 5. y 0 + 1 − y 2 = 0 ⇒ y 0 = y 2 − 1 ⇒ ⇒ dy = dx ⇒ y2 − 1 1 ⇒ 2 ⇒ Z 1 dy = y2 − 1 1 1 dy − (y − 1) 2 Z Z dx ⇒ 1 dy = (y + 1) Z 1 dy = (y − 1)(y + 1) Z dx. Z dx. 1 1 ln |y − 1| + ln |y + 1| = x + C 2 2 7. xy 0 = 4y ⇒ ⇒ Z dy = y 2 − 1. dx 4y dy dx dy = ⇒ = ⇒ dx x 4y x Z 1 dy = 4y Z 1 dx. x 1 ln |y| = ln |x| + C 4 9. ex yy 0 = e−y + e−2x−y ⇒ ex yy 0 = e−y + ⇒ yy 0 ey = e−2x ⇒ ex yy 0 ey = 1 + e−2x . ey 1 + e−2x ⇒ yey dy = (e−x + e−3x )dx. ex ⇒ yey − ey = −e−x − 3e−3x + C. sin x dy sin x 13. y = ⇒ = ⇒ cos ydy = sin xdx ⇒ cos y dx cos y 0 ⇒ sin y = − cos x + C ⇒ sin Z Z cos y dy = pi = cos 0 + C ⇒ C = 2 2 33 sin x dx ⇒ sin y = − cos x + 2. 15. y 0 = dy 2x 2xdx 2x ⇒ = ⇒ ydy = 2 ⇒ y + x2 y dx y(x2 + 1) x +1 Z Z y dy = 2x dx x2 + 1 1 1 ⇒ y 2 = ln (x2 + 1) + C ⇒ (−4)2 = ln (0 + 1) + C ⇒ C = 8. 2 2 1 ⇒ y 2 − ln (x2 + 1) = 8. 2 x ln (x2 + 1) 17. y = ⇒ (y − 1)dy = x ln (x2 + 1)dx ⇒ y−1 0 Z (y − 1) dy = Z x ln (x2 + 1) dx.  1 1 2 ⇒ y2 − y = (x + 1) ln(x2 + 1) − (x2 + 1) + C. 2 2 1 1 1 ⇒ 22 − 2 = ((0 + 1) ln(0 + 1) − (0 + 1)) + C ⇒ C = . 2 2 2  1 1 1 2 ⇒ y2 − y = (x + 1) ln(x2 + 1) − (x2 + 1) + . 2 2 2 Z Z 1 1 0 2 2 2 19. y = (cos x)(cos 2y) ⇒ dy = (cos x)dx ⇒ dy = (cos2 x) dx. cos2 2y cos2 2y 1 1 ⇒ tan(2y) = 2 2 ⇒ Z (cos 2x + 1) dx = 1 1 sin 2x + x + C. 4 2 1 1 1 tan 0 = sin 0 + 0 + C ⇒ C = 0. 2 4 2 ⇒ 2 tan 2y = 2x + sin 2x. 6.3 First Order Linear Differential Equations dy 1. y = 2y − 3 ⇒ = dx ⇒ 2y − 3 0 Z (2y − 3) dy = Z dx ⇒ 1 ln(2y − 3) = x + C. 2 1 y 0 = 2y − 3 ⇒ ln(2y − 3) = 2x + C ⇒ 2y − 3 = Ce2x ⇒ y = (Ce2x + 3). 2 ⇒y= 3 + Ce2x . 2 3. x2 y 0 − xy = 1 ⇒ y 0 − ⇒ R 1 1 1 1 y = 2 ⇒ µ = e − x dx ⇒ µ = e− ln x = . x x x d 1  1 1 y y = ⇒ = dx x x x2 x Math 1274 Z x−3 dx ⇒ y 1 1 = − x−2 + C ⇒ y = − + Cx. x 2 2x Minh Phuong Bui & Pichmony Anhaouy Solutions 34 5. (cos2 x sin x)y 0 + (cos3 x)y = 1 ⇒ y 0 + µ=e R cos x sin x dx cos x 1 y= . 2 sin x cos x sin x = eln | sin x| = | sin x|. d sin x 1 ⇒ (y sin x) = = ⇒ y sin x = 2 dx cos x sin x cos2 x Z 1 dx = tan x + C. cos2 x ⇒ y = sec x + C csc x. 7. x3 y 0 − 3x3 y = x4 e2x ⇒ y 0 − 3y = xe2x ⇒ µ = e− d (ye−3x ) = xe−x ⇒ ye−3x = ⇒ dx Z R 3 dx = e−3x . xe−x dx ⇒ ye−3x = −xe−x − e−x + C. ⇒ y = −xe2x − e2x + Ce3x = Ce3x − (x + 1)e2x . R 9. y 0 = y + 2xex ⇒ y 0 − y = 2xex ⇒ µ = e −1 dx = e−x d ⇒ (ye−x ) = 2x ⇒ ye−x = dx Z 2x dx ⇒ ye−x = x2 + C ⇒ y = x2 ex + Cex . y(0) = 2 ⇒ 2 = 0e0 + Ce0 ⇒ C = 2 ⇒ y = x2 ex + 2ex . 11. xy 0 + (x + 2)y = x ⇒ y 0 + R x+2 ⇒µ=e x dx R x+2 y = 1. x 2 = e 1+ x dx = ex+2lnx = x2 ex . d (yx2 ex ) = x2 ex ⇒ yx2 ex = ⇒ dx 2 x 2 x  x ⇒ yx e = x e − 2 xe − ⇒y =1− Z Z R x+2 Z 2xex dx.    ex dx ⇒ yx2 ex = x2 ex − 2 xex − ex + C. C 2 2 e ⇒ C = −e ⇒ y = 1 − + 2 − 2 x . e x x x e x+1 dx x+2 2xe−x y= . x+1 x+1 = ex+ln |x+1| = (x + 1)ex . d (y(x + 1)ex ) = 2x ⇒ y(x + 1)ex = dx Math 1274 2 x x e dx ⇒ yx e = x e − 13. (x + 1)y 0 + (x + 2)y = 2xe−x ⇒ y 0 + ⇒ 2 x 2 2 C + 2 + 2 x. x x x e y(1) = 0 ⇒ 0 = 1 − 2 + 2 + ⇒µ=e 2 x Z 2x dx = x2 + C ⇒ y = x2 + C . (x + 1)ex Minh Phuong Bui & Pichmony Anhaouy Solutions 35 y(0) = 1 ⇒ 1 = C x2 + 1 ⇒C=1⇒y= . 1 (x + 1)ex 15. (x2 − 1)y 0 + 2y = (x + 1)2 ⇒ y 0 + R ⇒µ=e 2 dx x2 −1 R =e 2 dx (x−1)(x+1) x − 1 d  x − 1 y =1⇒y = dx x + 1 x+1 y(0) = 2 ⇒ 2 = 1. =e R x+1 . x−1 y= 1 1 ( (x−1) − (x+1) ) dx = eln |x−1|−ln |x+1| = eln |x−1| . eln |x+1| x−1 . x+1 = 6.4 2 x2 − 1 Z dx = x + C ⇒ y = (x + C)(x + 1) . x−1 C (x − 2)(x + 1) ⇒ C = −2 ⇒ y = . −1 x−1 Modeling with Differential Equations dy dy = k(10 − y) ⇒ = kdx ⇒ dx 10 − y Z 1 dy = 10 − y Z k dx. ⇒ − ln(10 − y) = kx + D ⇒ (10 − y)−1 = ekx+D . ⇒ 1 1 = Eekx where E = eD ⇒ 10 − y = e−kx . 10 − y E ⇒ y = 10 − 1 −kx e ⇒ y = 10 + Ce−kx . E 3. Let y be the number of students who heard the rumor at any time so (250-y) is the number of students who have not heard the rumor. dy dy = ky(250 − y) ⇒ = dt ⇒ dt ky(250 − y) 1 ⇒ 250 Z  1  1 + dy = y 250 − y Z k dt ⇒ ⇒ ln y − ln(250 − y) = 250kt + D ⇒ ln y(0) = 1 ⇒ ln dy dy = y(250 − y) Z  Z 1  1 + dy = y 250 − y k dt. Z 250k dt.  y = 250kt + D. 250 − y 1 = D ⇒ D = − ln(240). 240 y(3) = 75 ⇒ ln Math 1274  Z  75  = 750k − ln(240). 250 − 75 Minh Phuong Bui & Pichmony Anhaouy Solutions 36 ⇒k= ln   75 ∗ 240   720  1 1 ln = ln = 6.18x10−3 . 750 175 750 7  1  720  y = ln t − ln(240) 250 − y 3 7 80 percents of the students have heard the rumor so y = 200. ⇒ ln  200  1  720  = ln t − ln(240) ⇒ t = 4.45 days. 250 − 200 3 7 7. Let T denote the temperature of the object at any time t. ⇒ dT = k(T − (60 + 20e−0.25t )) ⇒ T 0 − kT = −k(60 + 20e−0.25t ). dt R ⇒ µ = e −k dt = e−kt ⇒ d(T e−kt ) = −ke−kt (60 + 20e−0.25t )dt. ⇒ T e−kt = Z ⇒ T = 60 + −ke−kt (60 + 20e−0.25t ) dt = 60e−kt + 20 e−t(k+0.25) + C. k + 0.25 20 e−0.25t + Cekt . k + 0.25 T (0) = 100 and T (20) = 80 ⇒ T = 60 − 3.69858e−0.25t + 43.69858e−0.0390169t . 11. Let y denote the amount of salt in the tank at anytime t. Rate in = 1 ∗ 4 = 4(g/min). To calculate rate out, we need to calculate the volume at anytime t, y 3y volume = 1 + (4 − 3)t = 1 + t ⇒ Rate out = 1+t ∗ 3 = 1+t R 3 dy 3y 3 =4− ⇒ y0 + y = 4 ⇒ µ = e 1+t dt = (1 + t)3 . dt 1+t 1+t ⇒ d(y(1 + t)3 ) = 4(1 + t)3 dt ⇒ y(1 + t)3 = Z 4(1 + t)3 dt = (1 + t)4 + C. ⇒ y = (1 + t) + C We have y(0) = 2 ⇒ 2 = 1 + C ⇒ C = 1. (1 + t)3 ⇒ y = (1 + t) + 1 ⇒ y(10) = 11.00075. (1 + t)3 Math 1274 Minh Phuong Bui & Pichmony Anhaouy Solutions 37