( ) = f(x) 6 INTEGRATION. Things to remember: n + 1 EXERCISE A function F is an ANTIDERIVATIVE of f if F'(x) = f(x).

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1 6 INTEGRATION EXERCISE 6-1 Things to remember: 1. A function F is an ANTIDERIVATIVE of f if F() = f().. THEOREM ON ANTIDERIVATIVES If the derivatives of two functions are equal on an open interval (a, b), then the functions can differ by at most a constant. Symbolically: If F and G are differentiable functions on the interval (a, b) and F() = G() for all in (a, b), then F() = G() + k for some constant k.. The INDEFINITE INTEGRAL of f(), denoted! f()d, represents all antiderivatives of f() and is given by! f()d = F() + C where F() is any antiderivative of f() and C is an arbitrary constant. The symbol is called an INTEGRAL SIGN, the function f() is called the INTEGRAND, and C is called the CONSTANT OF INTEGRATION. 4. Indefinite integration and differentiation are reverse operations (ecept for the addition of the constant of integration). This is epressed symbolically by: ( ) = f()! (a) d d f()d (b)! F()d = F() + C 5. INDEFINITE INTEGRAL FORMULAS: (a)! n d = n+ 1 + C, n -1 n + 1 (b)! e d = e + C (c)! d = ln + C, 0 6. PROPERTIES OF INDEFINITE INTEGRALS: (a)! kf()d = k! f()d, k constant (b)! [f() ± g()] d =! f()d ±! g()d EXERCISE

2 1. d = 1 + C [Formula 5a] Check: d! 1 d + C =. 7 d = C [Formula 5a] Check: d! 1 d C = 7 5. d = d = [ + C] = + C Check: d [Formula 5a, Property 6a, d ( + C) = 0 = 1, replace C by C] 7. 5u - du = 5 u - du = 5u1 1 + C = -5u-1 + C Check: d du (-5u-1 + C) = 5u - [Formula 5a, Property 6a] 9. e dt = e dt = e t + C Check: d dt (e t + C) = e [Formula 5a, Property 6a] 11. z 49 dz = z C Check: d z 50 [Formula 5a] dz 50 + C = z49 1. π d = π d = π + C = π + C Check: d d (π + C) = π [Formula 5a, Property 6a] 15. 8u -1 du = 8 1 u du = 8 ln u + C Check: d du (8 ln u + C) = 8 1 u = 8u-1 [Formula 5c, Property 6a] / d = 15 1/ d = C = 15 + C = 10/ + C Check: d d (10/ + C) = 15 1/ [Formula 5a, Property 6a] 19. 7t -4/ dt = 7 t -4/ t!4 +1 dt = 7! C = 7t + C = -1t-1/ + C Check: d dt (-1t-1/ + C) = 7t -4/ [Formula 5a, Property 6a] 1. ( - )d = d - 1/ d = C = 1 - / + C Check: d 1 d + C( = - 1/ [Formula 5a, Property 6b] = - 58 CHAPTER 6 INTEGRATION

3 . dy d = 004 y = 00 4 d = 00 4 d = C = C 5. dp = 4-6 d P = (4-6)d = 4 d - 6 d = 4 d - 6 d 7. dy du = u5 - u - 1 = C = C y = (u 5 - u - 1)du = u 5 du - u du - 1 du 9. dy d = e + = u 5 du - u du - du = u6 6! u y = (e + )d = e d + d = e + + C 1. d dt = 5t u + C = u6 - u - u + C = (5t )dt = 5t -1 dt + 1 dt = 5 1 t dt + dt. True: f() = π, f () = 0 = k() 5. False: f() = True: f() = 5e, f () = 5e = f() = 5 ln t + t + C 9. False: f () = h(), g () = h(), (f() + g()) = f () + g () = h() 41. False: d d ( d) = d d + C = 4. The graphs in this set ARE NOT graphs from a family of antiderivative functions since the graphs are not vertical translations of each other. 45. The graphs in this set could be graphs from a family of antiderivative functions since they appear to be vertical translations of each other. EXERCISE

4 47. 5(1 - )d = 5 (1 - )d = 5 ( - )d = 5 d - 5 d = 5! 5 + C Check: 5! 5 + C = 5-5 = 5(1 - ) 49. ( + )( + )d = ( )d = 6 d + 5 d + 4 d = ! Check: C + C = C = = ( + )( + ) 51. du u = du u 1 = u -1/ du = u( ) C = u1 1 + C = u 1/ + C or u + C Check: (u 1/! + C) = 1 u-1/ = 1 u 1 = 1 u 5. d 4 = d = 1 4! Check:!! 8!! + C = + C! 8 + C = 1 8 (-)(-- ) = = u u du = 4 u + 1 du = 4 1 u du + 1 du = 4 ln u + u + C Check: d du (4 ln u + u + C) = 4 u + 1 = 4 + u u 57. (5e z + 4)dz = 5 e z dz + 4 dz = 5e z + 4z + C Check: (5e z + 4z + C) = 5e z ( d = d - d = d - - d = - + C = C Check: ( C) = - - = - 60 CHAPTER 6 INTEGRATION

5 ( d = 10 4 d d - d = 10 4 d d - d = !4!4 - + C = C Check: ( C) = = d = 1/ d + -1/ d = C = / + 4 1/ + C 1 Check: ( / + 4 1/! + C) = 1/! / 65. = 1/ + -1/ = + 4 ( d = / d d = 5 5-4!! = C 5 Check: 5 5 / + + C( = 5 / + (-) e! 4 d = e 4! 4 Check: 1 4 e = / = d = 1 4 e d - 4 d = 1 4 e C = 1 4 e - 8! 8 + C = 1 4 e z! z 69. z 4 dz = 1 z z ( dz z + C = 1 4 e - 4 = 1 z -4 dz + 5 z - dz - 1 z dz + C - 4 = 1 z!! + 5 z!! - ln z + C = -4z z- - ln z + C Check: d dz 4z 5 z lnz + C( = 1z z - - z = 1 + 5z z z 4 EXERCISE

6 ! d = 6 5 d - 1 d = ln + C = - 5 ln + C Check: 5! ln + C = = 6 5! 7. dy d = - y = ( - )d = d - d = - + C = - + C Given y(0) = 5: 5 = 0 - (0) + C. Hence, C = 5 and y = C() = 6-4 C() = (6-4)d = 6 d - 4 d = C = - + C Given C(0) = 000: 000 = (0 ) - (0 ) + C. Hence, C = 000 and C() = d dt = 0 t = 0 t dt = 0 t -1/ dt = 0 t1 1 + C = 40 t + C Given (1) = 40: 40 = C or 40 = 40 + C. Hence, C = 0 and = 40 t. 79. dy d = y = ( )d = - d + -1 d - d = + ln - + C =! + ln - + C Given y(1) = 0: 0 = ln C. Hence, C = and y = - + ln d dt = 4et - = (4e t - )dt = 4 e t dt - dt = 4e t - t + C Given (0) = 1: 1 = 4e 0 - (0) + C = 4 + C. Hence, C = - and = 4e t - t -. 6 CHAPTER 6 INTEGRATION

7 8. dy d = 4 - y = (4 - )d = 4 d - d = C = - + C Given y() = : = - + C. Hence, C = 1 and y = ! d = 4! d = d - - d = 87. 5! 4 d = 5 4! 4 d 89. e! 91. dm dt = t! 1 t M = t! 1 t = d - - d =! d = e - + C = C -!! + C = C d = e d - -1 d = e - ln + C dt = t t! 1 t dt = dt - t - dt = t - t + C = t + 1 t + C Given M(4) = 5: 5 = C or C = = 4. Hence, M = t + 1 t dy d = 5 + y = d = 1 1 ( = 5 / d + -1/ d = C = 5/ + / + C 5 Given y(1) = 0: 0 = 1 5/ + 1 / + C. Hence, C = -6 and y = 5/ + / p() = - 10 p() = 10 d = d = 0 + C = 10 + C Given p(1) = 0: 0 = C = 10 + C. Hence, C = 10 and p() = EXERCISE 6-1 6

8 97. d [ d! d] = [by 4(a)] 99. d d [ ]d = C = C 1 [by 4(b)] (C 1 = 1 + C is an arbitrary constant since C is arbitrary) 101. d! n +1 d n C = n 10. Assume > 0. Then = and ln = ln. Therefore, d d (ln + C) = d d (ln + C) = Assume f()d = F() + C 1 and g()d = G() + C. Then, d d (F() + C 1 ) = f(), d d (G() + C ) = g(), and d d (F() + C 1 + G() + C ) = d d (F() + C 1 ) + d d (G() + C ) = f() + g() C () = 1,000 C () = C ()d = 1,000 d = -1,000 - d Given C (100) = 5: 1, = -1,000 = 1,000 + C = 5 + C + C C = 15 Thus, C () = 1, Cost function: C() = C () = ,000 Fied costs: C(0) = 1, (A) The cost function increases from 0 to 8. The graph is concave downward from 0 to 4 and concave upward from 4 to 8. There is an inflection point at = 4. (B) C() = C()d = ( )d = d - 4 d + 5 d = K Since C(0) = 0, we have K = 0 and C() = C(4) = 4-1(4) + 5(4) + 0 = 114 thousand C(8) = 8-1(8) + 5(8) + 0 = 198 thousand 64 CHAPTER 6 INTEGRATION

9 (C) (D) Manufacturing plants are often inefficient at low and high levels of production S(t) = -5t / S(t) = S(t)dt = -5t / dt = -5 t / dt = -5 t5 5 + C = -15t5/ + C Given S(0) = 000: -15(0) 5/ + C = 000. Hence, C = 000 and S(t) = -15t 5/ Now, we want to find t such that S(t) = 800, that is: -15t 5/ = t 5/ = -100 t 5/ = 80 and t = 80 /5 14 Thus, the company should manufacture the computer for 14 months. 11. S(t) = -5t / - 70 S(t) = S(t)dt = (-5t / - 70)dt = -5 t / dt - 70dt = -5 t5 5-70t + C = -15t 5/ - 70t + C Given S(0) =,000 implies C =,000 and S(t) =,000-15t 5/ - 70t Graphing y 1 =,000-15t 5/ - 70t, y = 800 on 0 10, 0 y 1000, we see that the point of intersection is , y = 800. So we get t 8.9 months L() = g() = 400-1/ L() = g()d = 400-1/ d = 400-1/ d = C = / + C Given L(16) = 19,00: 19,00 = 4800(16) 1/ + C = 19,00 + C. Hence, C = 0 and L() = /. L(5) = 4800(5) 1/ = 4800(5) = 4,000 labor hours. EXERCISE