(x! 4) (x! 4)10 + C + C. 2 e2x dx = 1 2 (1 + e 2x ) 3 2e 2x dx. # 8 '(4)(1 + e 2x ) 3 e 2x (2) = e 2x (1 + e 2x ) 3 & dx = 1

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1 33. x(x - 4) 9 Let u = x - 4, then du = and x = u + 4. x(x - 4) 9 = (u + 4)u 9 du = (u 0 + 4u 9 )du = u + 4u0 0 = (x! 4) (x! 4)0 (x " 4) (x " 4)0 ( '( = ()(x - 4)0 () (0)(x - 4)9 () = (x - 4) 9 [(x - 4) + 4] = x(x - 4) Let u = + e 2x, then du = 2e 2x. e 2x ( + e 2x ) 3 = ( + e 2x ) e2x = 2 ( + e 2x ) 3 2e 2x = 2 u3 du = 2 u 4 4 = 8 ( + e2x ) 4 " 8 ( + e2x ) 4 ' = " 8 '(4)( + e 2x ) 3 e 2x (2) = e 2x ( + e 2x ) Let u = 4 + 2x + x 2, then du = (2 + 2x) = 2( + x). + x 4 + 2x + x 2 = + x 4 + 2x + x 2 2( + x) = + x x + x 2 2(2 + x) = 2 u du = 2 ln u = 2 ln 4 + 2x + x2 " 2 ln4 + 2x + x2 ' = " 2 ' 4 + 2x + x 2 (2 + 2x) = + x 4 + 2x + x Let u = x 4 + 2x 2 +, then du = (4x 3 + 4x) = 4(x 3 + x). x 3 + x (x 4 + 2x 2 + ) 4 = (x 4 + 2x 2 + ) (x3 + x) = 4 (x 4 + 2x 2 + ) -4 4(x 3 + x) = 4 u -4 du = 4 u!3!3 =!u!3 2 = "(x4 + 2x 2 + ) "3 2 " 2 (x4 + 2x 2 + ) "3 ' ( = " (("3)(x 4 + 2x 2 + ) -4 (4x 3 + 4x) 2' = (x 4 + 2x 2 + ) -4 (x 3 + x) EXERCISE

2 4. (A) Differentiate F(x) = ln 2x - 3 to see if you get the integrand f(x) = 2x! 3 (B) Wrong; d [ln 2x - 3 ] = 2x! 3 (2) = 2 2x! 3 2x! 3 (C) Let u = 2x - 3, then du = 2 2x! 3 = 2x! = 2 2x! 3 2 = 2 u du = 2 ln u = 2 ln 2x ln2x " 3 ' ( = 2 2x! 3 2 = 2x! (A) Differentiate F(x) = e x4 to see if you get the integrand f(x) = x 3 e x4. (B) Wrong; d [ex4 + c] = e x4 (4x 3 ) = 4x 3 e x4 x 3 e x4 (C) Let u = x 4, then du = 4x 3 x 3 e x4 = 4 4 x3 e x4 = 4 4x 3 e x4 = 4 e u du = 4 eu = 4 ex4 " 4 ex 4 ' = 4 ex4 (4x 3 ) = x 3 e x4 45. (A) Differentiate F(x) = (x2! 2) 2 3x f(x) = 2(x 2-2) 2 (B) Wrong; d (x 2 " 2) 2 3x to see if you get the integrand ( '( = 3x! 2(x2 " 2)(2x) " (x 2 " 2) 2! 3 9x 2 = (x2! 2)[9x 2 + 6] 9x 2 = 9x4! 2x 2! 2 9x 2 = 3x4! 4x 2! 4 3x 2 2(x 2-2) CHAPTER 6 INTEGRATION

3 (C) 2(x 2-2) 2 = 2(x 4-4x 2 + 4) = 2 5 x 5 " 4 3 x3 + 4x ' ( = 2 5 x5-8 3 x3 + 8x 2 5 x 5 " 8 3 x3 + 8x ' ( = 2x4-8x = 2[x 4-4x 2 + 4] = 2(x 2-2) Let u = 3x 2 + 7, then du = 6x. x 3x = (3x 2 + 7) /2 x = (3x 2 + 7) /2 6 6 x = u /2 du = 6 6 u = 9 (3x2 + 7) 3/2 " 9 (3x2 + 7) 3 2 ' =! 3 9 " 2 (3x2 + 7) /2 (6x) = x(3x 2 + 7) /2 49. x(x 3 + 2) 2 = x(x 6 + 4x 3 + 4) = (x 7 + 4x 4 + 4x) 5. x 2 (x 3 + 2) 2 = x x5 + 2x 2 " x x 5 + 2x 2 ' Let u = x 3 + 2, then du = 3x 2. x 2 (x 3 + 2) 2 = x 2 (x 3 + 2) 2 3x 2 ' = x7 + 4x 4 + 4x = x(x 6 + 4x 3 + 4) = x(x 3 + 2) 2 3 = 3 (x 3 + 2) 2 3x 2 = 3 u2 du = 3 u 3 3 = 9 u3 = 9 (x3 + 2) 3 " 9 (x3 + 2) 3 ' = 9 (3)(x3 + 2) 2 (3x 2 ) = x 2 (x 3 + 2) Let u = 2x 4 + 3, then du = 8x 3. x 3 2x = (2x 4 + 3) -/2 x 3 = (2x 4 + 3) -/2 8 8 x3 = u -/2 du = 8 8 u 2 2 = 4 (2x4 + 3) /2 " 4 (2x4 + 3) 2 ' = " '(2x 4 + 3) -/2 (8x 3 ) = 4 2 x 3 (2x 4 + 3) 2 EXERCISE

4 55. Let u = ln x, then du = x. (ln x)3 = u 3 du = u4 (ln x)4 = x 4 4 " (ln x) 4 ' 4 ' = 4 (4)(ln x)3 x 57. Let u =! x = -x-, then du = x 2. x 2 e-/x = e u du = e u = e -/x [e-/x ] = e -/x " ' = 59. = 7t2 (t 3 + 5) 6 x 2 x 2 e-/x = (ln x)3 x Let u = t 3 + 5, then du = 3t 2. x = 7t 2 (t 3 + 5) 6 = 7t 2 (t 3 + 5) 6 = 7(t 3 + 5) t2 = 7 3 u 6 du = 7 3 u 7 7 = 3 (t3 + 5) 7 6. dy = 3t t 2 " 4 Let u = t 2-4, then du = 2t. 3t y = (t 2! 4) 2 = 3 (t 2-4) -/2 t = 3(t 2-4) -/2 2 2 t 63. dp = ex + e!x (e x! e!x ) 2 = 3 2 u-/2 du = 3 2 u 2 Let u = e x - e -x, then du = (e x + e -x ). 2 = 3(t2-4) /2 p = ex + e!x (e x! e!x ) 2 = (e x - e -x ) -2 (e x + e -x ) = u -2 du 65. Let v = au, then dv = a du. e au du = e au a = u!! = -(ex - e -x ) - a du = a e au a du = a e v dv = a ev = a eau " du a eau ' + a eau (a) = e au 374 CHAPTER 6 INTEGRATION

5 67. p'(x) =!6000 (3x + 50) 2 Let u = 3x + 50, then du = 3.!6000 p(x) = (3x + 50) 2 = (3x + 50) -2 = -6000(3x + 50) Given p(50) = 4: = (3! ) 4 = C = 0 Thus, p(x) = x = -2000u -2 u! du = -2000! = x + 50 Now, 2.50 = x (3x + 50) = x + 25 = x = 875 x = 250 Thus, the demand is 250 bottles when the price is C'(x) = x +, x > 0! 500 C(x) = 2 + " x + = (u = x +, du = ) x + = 2x ln(x + ) Now, C(0) = Thus, C(x) = 2x ln(x + ) The average cost is: C (x) = ln(x + ) + x x and C (000) = ln(00) = ln(00) or 7.45 per pair of shoes 7. S'(t) = 0-0e -0.t, 0 t 24 (A) S(t) = (0-0e -0.t ) = 0-0e -0.t = 0t - 0!0. e-0.t = 0t + 00e -0.t Given S(0) = 0: e 0 = 0 00 = 0 C = -00 Total sales at time t: S(t) = 0t + 00e -0.t 00, 0 t 24. EXERCISE

6 (B) S(2) = 0(2) + 00e -0.(2) - 00 = e Total estimated sales for the first twelve months: 50 million. (C) On a graphing utility, solve 0t + 00e -0.t - 00 = 00 or 0t + 00e -0.t = 200 The result is: t 8.4 months. " Q(t) = R(t) = t ' = 00 t = 00 ln(t + ) + 5t Given Q(0) = 0: 0 = 00 ln() + 0 Thus, C = 0 and Q(t) = 00 ln(t + ) + 5t, 0 t 20. Q(9) = 00 ln(9 + ) + 5(9) = 00 ln thousand barrels. 75. W(t) = W(t) = 0.2e 0.t = e 0.t (0.) = 2e 0.t Given W(0) = 2: 2 = 2e 0. Thus, C = 0 and W(t) = 2e 0.t. The weight of the culture after 8 hours is given by: W(8) = 2e 0.(8) = 2e grams. 77. dn = t + t 2, 0 t 0 (A) To find the minimum value of dn, calculate d " dn ' = d2 N 2 = -( + t2 )(2000)! 2000t (2t) ( + t 2 ) 2 = [ " t2 ] ( + t 2 ) 2 = critical value: t = Now dn t=0 = 0 dn t= = -,000 "2000( " t)( + t) ( + t 2 ) 2 dn t=0 =!20, Thus, the minimum value of dn is -,000 bacteria/ml per day. 376 CHAPTER 6 INTEGRATION

7 (B) N =!2,000t + t 2 Let u = + t 2, then du = 2t N =!2,000t + t 2 = -,000 2t + t 2 = -,000 u du = -,000 ln u = -,000 ln( + t 2 ) Given N(0) = 5,000: 5,000 = -,000 ln() = C (ln = 0) Thus, C = 5,000 and N(t) = 5,000 -,000 ln( + t 2 ) Now, N(0) = 5,000 -,000 ln( ) = 5,000 -,000 ln(0) 385 bacteria/ml (C) Set N(t) =,000 and solve for t:,000 = 5,000 -,000 ln( + t 2 ) ln( + t 2 ) = 4 + t 2 = e 4 t 2 = e 4 - t = e 4 " 7.32 days 79. N'(t) = 6e -0.t, 0 t 5 N(t) = N(t) = 6e -0.t = 6e -0.t = 6!0. e -0.t (-0.) = -60e -0.t Given N(0) = 40: 40 = -60e 0 Hence, C = 00 and N(t) = 00-60e -0.t, 0 t 5. The number of words per minute after completing the course is: N(5) = 00-60e -0.(5) = 00-60e words per minute. 8. de = 5000(t + )-3/2, t 0 Let u = t +, then du = E = 5000(t + ) -3/2 = 5000(t + ) -3/2 = 5000u -3/2 du = 5000 u! 2! 2 = -0,000(t + )-/2 = "0,000 t + Given E(0) = 2000: 2000 = "0,000 Hence, C = 2,000 and E(t) = 2,000-0,000 t +. EXERCISE

8 The projected enrollment 5 years from now is: E(5) = 2,000-0,000 0,000 0,000 = 2,000 - = 2, = 9500 students EXERCISE 6-3 Things to remember:. A DIFFERENTIAL EQUATION is an equation that involves an unknown function and one or more of its derivatives. The ORDER of a differential equation is the order of the highest derivative of the unknown function. 2. A SLOPE FIELD for a first-order differential equation is obtained by drawing tangent line segments determined by the equation at each point in a grid. 3. EXPONENTIAL GROWTH LAW If dq = rq and Q(0) = Q 0, then Q(t) = Q 0 ert, where Q 0 = Amount at t = 0 r = Relative growth rate (expressed as a decimal) t = Time Q = Quantity at time t 4. COMPARISON OF EXPONENTIAL GROWTH PHENOMENA DESCRIPTION MODEL SOLUTION GRAPH USES Unlimited growth: Rate of growth is proportional to the amount present Exponential decay: Rate of growth is proportional to the amount present Limited growth: Rate of growth is proportional to the difference between the amount present and a fixed limit dy = ky k, t > 0 y(0) = c dy = -ky k, t > 0 y(0) = c dy = k(m - y) k, t > 0 y(0) = 0 y = ce kt y = ce -kt y = M( - e -kt ) c c M y 0 y 0 y 0 t t t Short-term population growth (people, bacteria, etc.) Growth of money at continuous compound interest Price-supply curves Depletion of natural resources Radioactive decay Light absorption in water Price-demand curves Atmospheric pressure (t is altitude) Sales fads (e.g., skateboards) Depreciation of equipment Company growth Learning Logistic growth: Rate of growth is proportional to the amount present and to the difference between the amount present and a fixed limit dy = ky(m - y) k, t > 0 y(0) = M!+!c y = M!-!ce -kmt M y 0 t Long-term population growth Epidemics Sales of new products Rumor spread Company growth 378 CHAPTER 6 INTEGRATION