CHAPTER SEVEN. Solutions for Section x x t t4 4. ) + 4x = 7. 6( x4 3x4

Transcription

1 CHAPTER SEVEN 7. SOLUTIONS 6 Solutions for Section t t + t t t ( 4 4 ) + 4 = q We break the antierivative into two terms. Since y is an antierivative of y an y 4 /4 is an antierivative of y, an antierivative of y y is. + 8( 4 4 ) = Antierivative F() = C. 4. 6( z ) + 7 = Since ( z) = z /, an antierivative of ( z) is 7. ln z t t 6 t 9. z y 5. + ln y 5. F() = ( 5 5 ) + C = C. ln + C. e t = e t. y y4 4. z (/)+ (/) + = 5 z5/.

2 64 Chapter Seven /SOLUTIONS 4. sin t 5. G(t) = 5t + sin t + C 6. G(θ) = cos θ sin θ + C 7. f() =, so F() = + C. F() = implies that + C =, so C =. Thus F() = is the only possibility. 8. f() = 7, so F() = 7 + C. F() = implies that 7 + C =, so C =. Thus F() = 7 / is the only possibility. 9. f() = , so F() = C. F() = implies that C =. Thus F() = is the only possibility.. f() =, so F() = +C. F() = implies that +C =, so C =. Thus F() = is the only possibility.. f() = /, so F() = / + C. F() = implies that / + C =, so C =. Thus F() = / is the only possibility.. Since (e ) = e, we take F() = e + C. Now F() = e + C = + C =, so an C = F() = e C 4. + C C. t 6. + C. 7. ( + ) ( + ) = + C. Another way to work the problem is to epan ( + ) to + + as follows: ( + ) = ( + + ) = C These two answers are the same, since ( + ) plus a constant. ( + ) = + + C = + C (t + 5t + ) t = t + 5 t + t + C = = + + +, which is + +, 4. 5e z + C 4. (t + 6t ) t = t t + C = t4 4 + t + C C w / w = w/ / + C = w/ + C C

3 7. SOLUTIONS ln t + t + C 46. e t + C / + C 48. ( ) = C 49. e C 5. + ln + C sin θ + C e r r = er + C 5. cos t + C 54. 5e.4q q = e 4 + C 56. e 8sin + C 57. sin + 7cos + C 58. cos() + C (.4 e.4q ) + C = 65e.4q + C 59. We use the substitution w = + 4, w = : cos( + 4) = cos w w = sin w + C = sin( + 4) + C. 6. sin() + C 6. 4cos() + C 6. 6cos() + sin(5) + C 6. An antierivative is F() = 5 + C. Since F() = 5, we have 5 = + C, so C = 5. The answer is F() = An antierivative is F() = + + C. Since F() = 5, we have 5 = + C, so C = 5. The answer is F() = / An antierivative is F() = 4cos() + C. Since F() = 5, we have 5 = 4cos + C = 4 + C, so C = 9. The answer is F() = 4cos() An antierivative is F() = e + C. Since F() = 5, we have 5 = e + C = + C, so C =. The answer is F() = e The marginal cost, MC, is given by ifferentiating the total cost function, C, with respect to q so C q = MC. Therefore, C = MC q (q = + 4q + 6 ) q = q + q + 6q + D,

4 66 Chapter Seven /SOLUTIONS where D is a constant. We can check this by noting C q = ( q + q + 6q + D ) = q + 4q + 6 = MC. q The fie costs are given to be so C = when q =, thus D =. The total cost function is C = q + q + 6q (a) The marginal revenue, MR, is given by ifferentiating the total revenue function, R, with respect to q so R q = MR. Therefore, R = = MR q ( 4q) q = q q + C. We can check this by noting R q = ( q q + C ) = 4q = MR. q When no goos are prouce the total revenue is zero so C = an the total revenue is R = q q. (b) The total revenue, R, is given by pq where p is the price, so the eman curve is p = R q = q. Solutions for Section 7.. We use the substitution w = +, w =. ( + ) 4 = w 4 w = w5 5 + C = 5 ( + ) 5 + C.. We use the substitution w = +, w =. + = w w = ln w + C = ln( + ) + C.. We use the substitution w = +, w =. ( + ) = ( 4 ( + )4 + C) = ( + ). 4. We use the substitution w = + 9, w = : ( + 9) 6 = w w = w4 4 + C = 4 ( + )4 + C. w 6 w = w C = 4 ( + 9) 7 + C.

5 7. SOLUTIONS We use the substitution w = q +, w = qq. qe q + q = e w w = e w + C = e q + + C. 6. We use the substitution w = 5t +, w = 5t. 5e 5t+ t = e w w = e w + C = e 5t+ + C. t (e5t+ + C) = 5e 5t+. 7. Make the substitution w = t, w = t t. The general antierivative is te t t = (/)e t + C. 8. We use the substitution w =, w =. e = ( e + C) = ( e ) = e. 9. We use the substitution w = t, w = t t. e w w = e w + C = e + C. t (t ) t = (t ) (t t) = w ( w ) = w + C = (t ) + C. t [ (t ) + C] = (t ) (t ) = t (t ).. We use the substitution w = +, w = 6. ( + ) = w ( 6 w) = 6 (w ) + C = 8 ( + ) + C. [ 8 ( + ) + C] = 8 [( + ) (6 )] = ( + ).. We use the substitution w = 4, w =. ( 4) 7/ = ( 4) 7/ () = w 7/ w = ( ) 9 w9/ + C = 9 ( 4) 9/ + C. ( 9 ( 4) 9/ + C) = ( 9 ) 9 ( 4) 7/ = ( 4) 7/.. We use the substitution w = +, w =. ( + ) = w ( w) = w + C = 6 ( + ) + C. [ 6 ( + ) + C] = [ ( + ) () ] = ( + ). 6. We use the substitution w = 4, w =. = w = w + C = 4 + C. 4 w ( 4 + C) = 4 = 4.

6 68 Chapter Seven /SOLUTIONS 4. We use the substitution w = y + 5, w = y, to get y y + 5 = w w = ln w + C = ln y C. (ln y C) = y y We use the substitution w =, w =. cos( ) = cos(w)w = 4sin(w) + C = 4sin( ) + C. 6. We use the substitution w = t 7, w = t. (t 7) 7 t = [ ] t 48 (t 7)74 + C 7. In this case, it seems easier not to substitute. w 7 w = = (t 7)7 () = (t 7) 7. ()(74) w74 + C = 48 (t 7)74 + C. ( + ) = ( ) = C. [ ] C = = ( + ). 8. In this case, it seems easier not to substitute. y ( + y) y = y (y + y + ) y = (y 4 + y + y )y = y5 5 + y4 + y + C. ( ) y 5 y 5 + y4 + y + C = y 4 + y + y = y (y + ). 9. We use the substitution w = t, w = t. sin( t)t = (cos( t) + C) = sin( t)( ) = sin( t). t. We use the substitution w = cos θ + 5, w = sin θ θ. sin(w)w = ( cos(w)) + C = cos( t) + C. sin θ(cos θ + 5) 7 θ = w 7 w = 8 w8 + C = 8 (cos θ + 5)8 + C. θ [ 8 (cos θ + 5)8 + C ] = 8 8(cos θ + 5)7 ( sin θ) = sin θ(cos θ + 5) 7

7 7. SOLUTIONS 69. We use the substitution w = cos t, w = sin t t. cost w sin t t = w = w + C = 9 (cost) + C. [ ] t 9 (cos t) + C = 9 (cos t) ( sin t) = cos t sin t.. We use the substitution w = + t, w = 6t t. t + t t = w ( 6 w) = 6 ln w + C = 6 ln( + t ) + C. (We can rop [ the absolute value signs since + t > for all t). t 6 ln( + t ) + C] = 6 + t (6t) = t + t.. We use the substitution w = sin θ, w = cos θ θ. sin 6 θ cos θ θ = [ ] sin 7 θ + C = sin 6 θ cos θ. θ 7 4. We use the substitution w = +, w =, to get w 6 w = w7 7 + C = sin7 θ + C. 7 e + = e w w = ew + C = + e + C. ( + e + C) = + e = e We use the substitution w = sin, w = cos : sin cos = w w = w + C = (sin ) + C. 6. We use the substitution w = sin α, w = cos α α. sin αcos α α = ( ) sin 4 α + C = α 4 4 4sin α cos α = sin αcos α. 7. We use the substitution w = 4, w = 8. sin(4 ) = 8 w w = w4 4 + C = sin4 α + C. 4 sin(w)w = 8 cos(w) + C = 8 cos(4 ) + C. 8. We use the substitution w = 4, w =. e 4 = e w w = ew + C = e 4 + C. ( e 4 + C) = e 4 = e 4.

8 7 Chapter Seven /SOLUTIONS 9. We use the substitution w =, w = 6. e = e w w = 6 6 ew + C = 6 e + C. ( 6 e + C) = 6 e 6 = e.. We use the substitution w = + 4, w = = w / w = w / 6 6 / + C = 9 ( + 4) / + C.. We use the substitution w = 5q + 8, w = qq. q 5q + 8 q = w w = ln w + C = ln(5q + 8) + C. ( q ln(5q + 8) + C) = 5q + 8 q = q 5q We use the substitution w = ln z, w = z. z (lnz) z = z [ ] (ln z) + C = z (ln z) z w w = w (ln z) =. z + C = (ln z) + C.. We use the substitution w = y + 4, w = y y. y y + 4 y = w w = ln w + C = ln(y + 4) + C. (We can rop [ the absolute value signs since y + 4 for all y.) y ln(y + 4) + C] = y + 4 y = y y We use the substitution w = e t + t, w = (e t + ) t. e t + e t + t t = w w = ln w + C = ln et + t + C. t (ln et + t + C) = et + e t + t. 5. We use the substitution w = y, w = y y. e y y = e w w = e w + C = e y + C. y y (e y + C) = e y y y = e. y 6. We use the substitution w =, w = cos =. (sin + C) = cos ( cos w( w) = sin w + C = sin + C. ) = cos. 7. We use the substitution w = + e, w = ( + e ). + e w = + e w = w + C = + e + C. ( + e + C) = ( + e ) ( + e ) = + e. + e

9 7. SOLUTIONS 7 8. We use the substitution w = e t +, w = e t t: e t e t + t = w w = ln w + C = ln(et + ) + C. 9. We use the substitution w = + + 9, w = ( + ). ( + ) = w w = ln w + C = ln( + + 9) + C. (We can rop the absolute value signs, since = ( + ) + 8 > for all.) [ ln( + + 9)] = ( + ) = We use the substitution w = e + e, w = (e e ). e e w e + e = w = ln w + C = ln(e + e ) + C. (We can rop the absolute value signs since e + e > for all ). [ln(e + e ) + C] = e + e (e e ). 4. (a) This integral can be evaluate using integration by substitution. We use w =, w =. sin = sin(w)w = cos(w) + C = cos( ) + C. (b) This integral cannot be evaluate using a simple integration by substitution. (c) This integral cannot be evaluate using a simple integration by substitution. () This integral can be evaluate using integration by substitution. We use w = +, w =. ( + ) = w w = ( w ) + C = ( + ) + C. (e) This integral cannot be evaluate using a simple integration by substitution. (f) This integral can be evaluate using integration by substitution. We use w = + cos, w = sin. sin + cos = w = ln w + C = ln + cos + C. w 4. (a) (i) Multiplying out gives 4. (a) (ii) Substituting w = + 5, so w =, gives ( + + 5) = C. ( + 5) = (b) The results of the two calculations are not the same since ( + 5) w w = w + C = ( + 5) + C = C. However they iffer only by a constant, 5/, as guarantee by the Funamental Theorem of Calculus. 4( + ) = (b) If w = +, then w =. (4 + 4) = C. 4( + ) = + C. w w = w + C = ( + ) + C. (c) The epressions from parts (a) an (b) look ifferent, but they are both correct. Note that ( + ) + C = C. In other wors, the epressions from parts (a) an (b) iffer only by a constant, so they are both correct antierivatives.

10 7 Chapter Seven /SOLUTIONS Solutions for Section 7.. Since F () = 5, we use F() = 5. By the Funamental Theorem, we have 5 = 5 = 5() 5() = 5 5 =.. Since F () = 6, we use F() =. By the Funamental Theorem, we have = = 4 = 48 = 48.. Since F () = +, we use F() = +. By the Funamental Theorem, we have ( + ) = ( + ) = ( + ) ( + ) = 4 = If f(t) = t + 4t +, then F(t) = t + t + t. By the Funamental Theorem, we have (t + 4t + ) t = (t + t + t) = + ( ) + () =. 5. Since F (t) = /t = t, we take F(t) = t = /t. Then t = F() F() t = ( ) =. 6. Since F () = = /, we use F() = / =. By the Funamental Theorem, we have 4 = 7. Since F () =, we take F() =. Then 5 4 = 4 = 4 =. = F(5) F() = 5 = If F (t) = t, then F(t) = t4. By the Funamental Theorem, we have 4 t t = F() F() = t4 4 = = If F () = 6, then F() =. By the Funamental Theorem, we have 6 = = (7) () = 54 = 5.

11 7. SOLUTIONS 7. Since F (t) = 5t, we take F(t) = 5 4 t4. Then 5t t = F() F() = 5 4 (4 ) 5 4 (4 ) = = Since F () =, we take F() = / / = /. Then 9 4 = F(9) F(4) = 9/ 4/ = 7 8 = 8.. Since F (y) = y + y 4, we take F(y) = y + y5 5. Then (y + y 4 ) y = F() F() ( ) ( ) = = + 5 = Since F (t) = /(t), we take F(t) = ln t. Then t = F() F() t = ln ln = ln ( π ) = ( 4 4 π ) 5 = π If f(t) = e.t, then F(t) = 5e.t. (This can be verifie by observing that Funamental Theorem, we have e.t t = ( 5e.t ) = 5(e. ) ( 5)() = 5 5e..96. t ( 5e.t ) = e.t.) By the 6. e = e = e.47.

12 74 Chapter Seven /SOLUTIONS 7. If F (t) = cos t, we can take F(t) = sin t, so cos t t = sin t 8. Since sin( ) = sin, we can simplify the answer an write π/4 (sin t + cos t)t = ( cos t + sin t) π/4 = cos t t = sin ( + = sin sin( ). ) ( + ) =. 9. If f(t) = e.5t, then F(t) = e.5t (you can check this by observing that t (e.5t ) = e.5t ). By the Funamental Theorem, we have e.5t t = e.5t = e.5 e = (e.5 ).. If f(q) = 6q + 4, then F(q) = q + 4q. By the Funamental Theorem, we have (6q + 4) q = (q + 4q) = () + 4() ( + ) = 6.. (a) We substitute w = +, w =. = + = w= = w w = ln w w= (b) We substitute w = cos, w = sin. = ln. = π 4 = w= / sin cos = w w w= / = ln w = ln = ln.. We substitute w = +, so w =. ( + ) = w w = w + C = 6 ( + ) + C. Using the Funamental Theorem, we have ( + ) = 6 ( + ) = = = 4 6 = 6.. Let w = +, then w =. When =, w = an when =, w =. Thus we have + = w = ln w w = ln ln = ln.

13 7. SOLUTIONS Let w = t, then w = tt so tt = w. When t =, w = an when t =, w =. Thus we have 5. We substitute w = t +, so w = t. t = t + te t t = e w ( w ) = e w w = e w = e ( e ) = e. w = w Using the Funamental Theorem, we have t = t + t + w / w = w / + C = t + + C. = 4 = 4 =. 6. We have Area = 4 = 4 = 4 = 64 =. 7. The integral which represents the area uner this curve is Area = (6 + ). Since ( + ) = 6 +, we can evaluate the efinite integral: (6 + ) = ( + ) = ( ) + (() + ) = 6 + = We have Average value = ( ) ( + ) = + = We see in Figure 7. that the average value of / 4. for f() looks right. ( + ) =. f() = Figure One antierivative of f() = e.5 is F() = e.5. Thus, the efinite integral of f() on the interval is e.5 = F() F() = e.5.

14 76 Chapter Seven /SOLUTIONS The average value of a function on a given interval is the efinite integral over that interval ivie by the length of the interval: ( ) ( ) ( ) Average value = e.5 = e.5 = (e.5 e ).. From the graph of y = e.5 in Figure 7. we see that an average value of. on the interval oes make sense. 5 y y = e.5. Figure 7.. Since y = = ( )( + ), the graph crosses the ais at the three points shown in Figure 7.. The two regions have the same area (by symmetry). Since the graph is below the ais for < <, we have ( ( Area = ) ) = [ 4 4 y ] ( = 4 ) =. y = Figure 7.. We have Area = b b 4 = = b. We fin the value of b making the area equal to 4: 4 = b 4 = b = b b =.

15 7. SOLUTIONS 77. The area uner f() = 8 between = an = b is given by b (8). Using the Funamental Theorem to evaluate the integral: b Area = 4 = 4b 4. Since the area is 9, we have Since b is larger than, we have b = 7.. We have Area = We fin the value of b making the area equal to : 4b 4 = 9 b 4b = 96 b = 49 b = ±7. = b = b. = b = b b = () / = (a) At time t =, the rate of oil leakage = r() = 5 thousan liters/minute. At t = 6, rate = r(6) = 5.6 thousan liters/minute. (b) To fin the amount of oil leake uring the first hour, we integrate the rate from t = to t = 6: 6 5. (a) In, we have P = 6.e. = 6.9 billion people. In, we have P = 6.e. = 7.8 billion people. (b) We have ( Oil leake = 5e.t t = 5 ) 6. e.t = 5e. + 5e = 747 thousan liters. Average population = 6.e.t t = = ( ) 6.. (e. e ) = e.t The average population of the worl between an is preicte to be 6.5 billion people. 6. (a) The graph of y = e is in Figure 7.4. The integral is shae. e represents the entire area uner the curve, which f() = e 5 5 Figure 7.4 (b) Using a calculator or computer, we see that e =.494, e =.764, e =.77, 5 e =.77 5

16 78 Chapter Seven /SOLUTIONS (c) From part (b), we see that as we eten the limits of integration, the area appears to get closer an closer to about.77. We estimate that e = Figure 7.5 shows the graphs of y = / an y = /. We see that is larger, since the area uner / is larger than the area uner /. y / / Figure (a) An antierivative of F () = ( is F() = since have: b b = = b +.. (b) Taking a limit, we have Since the limit is, we know that So the improper integral converges to : 9. (a) Using a calculator or computer, we get 7 e t t =.4988 e t t = lim ( ) b b + = + =. b lim =. b =. ( ) = ). So by the Funamental Theorem we 5 e t t = e t t = The values of these integrals are getting closer to.5. A reasonable guess is that the improper integral converges to.5. (b) Since e t is an antierivative of e t, we have b (c) Since e b = /e b, we have Therefore, e t t = e t b = e b ( ) e = e b +. e b as b, so e b =. eb lim b b So the improper integral converges to / =.5: ( e t t = lim b e b + ) = + =. e t t =.

17 7. SOLUTIONS (a) Evaluating the integrals with a calculator gives (b) The results of part (a) suggest that 5 e / = 6.4 e / = e / = e / =. e / 4. (a) The total number of people that get sick is the integral of the rate. The epiemic starts at t =. Since the rate is positive for all t, we use for the upper limit of integration. Total number getting sick = ( te.5t ) t (b) The graph of r = te.5t is shown in Figure 7.6. The shae area represents the total number of people who get sick. r Figure 7.6 t 4. Since y = ( ) is positive for an y =, when =,, the area is given by Area = ( ) = ( 4 ) = =. 4. Since y = only when = an =, the area lies between these limits an is given by Area = = ( ) = =. 44. (a) Since v(t) = 6/5 t is never, the car never stops. (b) For time t, (c) Evaluating b 6 t for b =,5, gives 5t t = 5.6 t ( + ) = Distance travele = 6 5 t t. 6 5 t = 5.7 t ( + 4 ) 6 t = 5.7, 5t so the integral appears to converge to 5.7; so we estimate the istance travele to be 5.4 miles.

18 8 Chapter Seven /SOLUTIONS 45. (a) No, it is not reache since Thus, the total number of rabbits is. (b) Yes, since Total number of rabbits = t =. t t t oes not converge to a finite value, which means that infinitely many rabbits coul be prouce, an therefore million is certainly reache. (c) Yes, since t t oes not converge to a finite value. 46. (a) In the first case, we are given that R = wigets/year. So we have R = e.5t. To etermine the total number sol, we nee to integrate this rate over the time perio from to. Therefore In the secon case, Total number of wigets sol = e.5t t =, wigets. Total number of wigets sol = (b) We want to etermine T such that Trying a few values of T, we get T Similarly, in the secon case, we want T so that we get T 5,,e.5t t =.5 billionwigets. e.5t t =,. T 6.7 years. 5,,e.5t t = T 6.7 years., 5,, So the half way mark is reache at the same time regarless of the initial rate. (c) Since half the wigets are sol in the last years of the ecae, if each wiget is epecte to last at least.5 years, their claim coul easily be true. Solutions for Section 7.4. Let u = t an v = e 5t, so u = an v = 5 e5t. Then te 5t t = 5 te5t 5 e5t t = 5 te5t 5 e5t + C.. Let u = p an v = e (.)p, u =. Thus, v = e (.)p p = e (.)p. With this choice of u an v, integration by parts gives: pe (.)p p = p( e (.)p ) ( e (.)p ) p = pe (.)p + e (.)p p = pe (.)p e (.)p + C.

19 7.4 SOLUTIONS 8. Let u = z +, v = e z. Thus, v = ez an u =. Integrating by parts, we get: (z + )e z z = (z + ) ez ez z = (z + )ez 4 ez + C = 4 (z + )ez + C. 4. Let u = ln y, v = y. Then, v = y an u =. Integrating by parts, we get: y y lny y = y ln y y y y = y ln y y y = y ln y 4 y + C. 5. Let u = ln an v =, so u = an v = 4 4. Then ln = 4 4 ln 4 4 = ln C. 6. Let u = ln 5q, v = q 5. Then v = 6 q6 an u =. Integrating by parts, we get: q q 5 ln 5q q = 6 q6 ln 5q (5 5q ) 6 q6 q = 6 q6 ln 5q 6 q6 + C. 7. Let u = y an v = (y + ) /, so u = an v = (y + )/ : y y + y = y(y + )/ (y + )/ y = y(y + )/ 4 5 (y + )5/ + C. 8. Let u = t + an v = + t, so u = an v = ( + 9 t)/. Then (t + ) + t t = 9 (t + )( + t)/ ( + t) / t 9 = 9 (t + )( + t)/ 4 5 ( + t)5/ + C. 9. Let u = z, v = e z. Thus v = e z an u =. Integration by parts gives: ze z z = ze z ( e z ) z = ze z e z + C = (z + )e z + C.

20 8 Chapter Seven /SOLUTIONS. Let u = ln, v =. Then v = an u =. Integrating by parts, we get: ln = ln ( ) = ln + C.. Let u = y an v = 5 y, so u = an v = (5 y) /. y y = y(5 y) / + (5 y) / y = y(5 y) / 4 5 y (5 y)/ + C. t + 7 t. t = t + 7 (5 t) / t. 5 t 5 t To calculate the first integral, we use integration by parts. Let u = t an v = 5 t, so u = an v = (5 t) /. Then t t = t(5 t) / + (5 t) / t = t(5 t) / 4 5 t (5 t)/ + C. We can calculate the secon integral irectly: 7 (5 t) / = 4(5 t) / + C. Thus t t t = t(5 t) / 4 (5 t)/ 4(5 t) / + C.. Let u = t, v = sin t. Thus, v = cos t an u =. With this choice of u an v, integration by parts gives: 4. 5 cos = (cos + sin ) 5 tsin tt = tcos t ( cos t) t = tcos t + sin t + C. = cos 5 + 5sin 5 cos sin Let u = t an v = e 5t, so u = t an v = 5 e5t. Then t e 5t t = 5 t e 5t 5 te 5t t. Using Problem, we have t e 5t t = 5 t e 5t 5 ( 5 te5t 5 e5t ) + C = 5 t e 5t 5 te5t + 5 e5t + C. 6. Let u = (ln t) an v =, so u = ln t an v = t. Then t 7. (We use the fact that 5 ln tt = (tln t t) (ln t) t = t(lnt) ln tt = t(ln t) t lnt + t + C. ln = ln + C, a result which can be erive using integration by parts.) 5 = 5ln We use integration by parts. Let u = z an v = e z, so u = an v = e z. Then ze z z = ze z + = e + ( e z ) = e e z z

21 9.. 5 ( tln t t = t lnt ) 4 t = 9 ln ln( + t)t = (( + t) ln( + t) ( + t)) = 6ln (a) We evaluate this integral using the substitution w = +. (b) We evaluate this integral using the substitution w =. (c) We evaluate this integral using the substitution w = +. () We evaluate this integral using the substitution w = +. (e) This integral can be evaluate using integration by parts with u = ln, v =. (f) This integral can be evaluate using integration by parts with u = ln, v =. 7.4 SOLUTIONS 8. A calculator gives ln =.86. An antierivative of ln is ln, so the Funamental Theorem of Calculus gives ln = (ln ) = ln. Since ln =.86, the value from the Funamental Theorem agrees with the numerical answer.. Using integration by parts with u = e t, v = t, so u = e t an v =, we have Area = te t t = te t = ( te t e t ) 4. Since ln( ) = ln an ln = ln + C, we have Area = = (ln( ) ln ) = e t t = e e + = e. ln = (ln ) (ln ln ) = ln (ln ) = ln. 5. Since the graph of f(t) = ln(t ) is above the graph of g(t) = ln(t ) for t >, we have ( ) Area = (ln(t t ) ln(t )) t = ln t = ln(t + ) t. t We can cancel the factor of (t ) in the last step above because the integral is over t, where (t ) is not zero. We use ln = ln with the substitution = t +. The limits t =, t = become =, = 4, respectively. Thus 4 4 Area = ln(t + ) t = ln = (ln ) = 4ln 4 4 (ln ) = 4 ln4 ln. 6. Since f () =, integration by parts tells us that f()g () = f()g() f ()g() = f()g() f()g() g().

22 84 Chapter Seven /SOLUTIONS We can use left an right Riemann Sums with = to approimate g() : Left sum g() + g() + 4 g(4) + 6 g(6) + 8 g(8) = ((.) + (.) + 4(4.) + 6(5.5) + 8(5.9)) = 5.6. Right sum g() + 4 g(4) + 6 g(6) + 8 g(8) + g() = ((.) + 4(4.) + 6(5.5) + 8(5.9) + (6.)) = 7.6. A goo estimate for the integral is the average of the left an right sums, so Substituting values for f an g, we have g() = f()g () = f()g() f()g() g() (6.) (.) (66.6) = We have Bioavailability = 5te.t t. We first use integration by parts to evaluate the inefinite integral of this function. Let u = 5t an v = e.t t, so u = 5t an v = 5e.t. Then, Thus, 5te.t t = (5t)( 5e.t ) = 75te.t te.t t = ( 75te.t 75e.t ) ( 5e.t )(5t) The bioavailability of the rug over this time interval is 45.7 (ng/ml)-hours. e.t t = 75te. 75e.t + C. = = (a) We know that E = r, so the total energy E use in the first T hours is given by E = t integration by parts. Let u = t, v = e at. Then u =, v = a e at. E = T te at t = t T a e at = a Te at + a T T ( ) a e at t e at t = a Te at + a ( e at ). T te at t. We use (b) ( ) T lim E = T a lim + ) ( lim. T e at a T e at Since a >, the secon limit on the right han sie in the above epression is. In the first limit, although both the numerator an the enominator go to infinity, the enominator e at goes to infinity more quickly than T oes. So in the en the enominator e at T is much greater than the numerator T. Hence lim =. (You can check this by T eat graphing y = T on a calculator or computer for some values of a.) Thus lim e E = at T a.

23 7.5 SOLUTIONS We integrate by parts. Since we know what the answer is suppose to be, it s easier to choose u an v. Let u = n an v = e, so u = n n an v = e. Then n e = n e n n e. Solutions for Section 7.5. By the Funamental Theorem, Thus, our table is as follows: F() = F() + F (t) t = 5.5 =.5 F() = F() + F (t) t =.5.5 = F() = F() + F (t) t =.5 =.5 4 F(4) = F() + F (t) t = = 5 F(5) = F(4) + F (t) t 4 = +.5 =.5 Table 7. t 4 5 F(t) First, we observe that g is increasing when g is positive, which is when < < 4. g is ecreasing when g is negative, which is when 4 < < 6. Therefore, = 4 is a local maimum. Table 7. shows the area between the curve an the -ais for the intervals,, etc. It also shows the corresponing change in the value of g. These changes are use to compute the values of g using the Funamental Theorem of Calculus: Since g() =, Similarly, g() g() = g () =. g() =. g() g() = g () = g() = g() + =.

24 86 Chapter Seven /SOLUTIONS Continuing in this way gives the values of g in Table 7.. Table 7. b Interval Area Total change in g = g () a / / 4 / / 4 5 / / 5 6 / / Table 7. g() / / 5/ 4 5 5/ 6 Notice: the graph of g will be a straight line from to because g is horizontal there. Furthermore, the tangent line will be horizontal at = 4, = an = 6. The maimum is at (4,). See Figure 7.7. y Figure 7.7. Since F() =, F(b) = b f(t) t. For each b we etermine F(b) graphically as follows: F() = F() = F() + Area of rectangle = + = F() = F() + Area of triangle ( ) = +.5 =.5 F() = F() + Negative of area of triangle =.5.5 = F(4) = F() + Negative of area of rectangle = = F(5) = F(4) + Negative of area of rectangle = = F(6) = F(5) + Negative of area of triangle =.5 =.5 The graph of F(t), for t 6, is shown in Figure F(t) Figure 7.8 t 4. (a) The value of the integral is negative since the area below the -ais is greater than the area above the -ais. We count boes: The area below the -ais inclues approimately.5 boes an each bo has area ()() =, so 5 f(). The area above the -ais inclues approimately boes, each of area, so So we have 7 f() = f() + f() f() + 4 = 9.

25 7.5 SOLUTIONS 87 (b) By the Funamental Theorem of Calculus, we have so, 5. See Figure F(7) F() = f() 7 F(7) = F() + f() = 5 + ( 9) = 6. F() = F() = Figure See Figure 7. F() = F() = Figure See Figure 7.. F() = F() = Figure See Figure 7. F() = Figure 7. F() =

26 88 Chapter Seven /SOLUTIONS 9. (a) The function f() is increasing when f () is positive, so f() is increasing for < or >. The function f() is ecreasing when f () is negative, so f() is ecreasing for < <. Since f() is increasing to the left of =, ecreasing between = an =, an increasing to the right of =, the function f() has a local maimum at = an a local minimum at =. (b) See Figure 7.. f() 4 4 Figure 7.. (a) The function f() is increasing when f () is positive, so f() is increasing for < < or >. The function f() is ecreasing when f () is negative, so f() is ecreasing for <. Since f() is ecreasing to the left of = an increasing to the right of =, the function has a local minimum at =. Since f() is increasing on both sies of =, it has neither a local maimum nor a local minimum at that point. (b) See Figure 7.4. f() Figure 7.4. (a) f() is increasing when f () is positive. f () is positive when < < 5. So f() is increasing when < < 5. f() is ecreasing when f () is negative. f () is negative when < or > 5. So f() is ecreasing when < or > 5. A function has a local minimum at a point when its erivative is zero at that point, an when it ecreases immeiately before an increases immeiately after. f () =, f ecreases to the left of, an f increases immeiately after, therefore f() has a local minimum at =. A function has a local maimum at a point when its erivative is zero at that point, an when it increases immeiately before an ecreases immeiately after. f (5) =, f increases before 5, an f ecreases after 5. Therefore f() has a local maimum at = 5. (b) Since we o not know any areas or vertical values, we can only sketch a rough graph. We start with the minimum an the maimum, then connect the graph between them. The graph coul be more or less steep an further above or below the -ais. See Figure 7.5.

27 7.5 SOLUTIONS 89 f() Figure 7.5. (a) The function f is increasing where f is positive, so f is increasing for < or >. The function f is ecreasing where f is negative, so f is ecreasing for < <. The function f has critical points at =,,. The point = is a local maimum (because f is increasing to the left of = an ecreasing to the right of = ). The point = is a local minimum (because f ecreases to the left of = an increases to the right). The point = is neither a local maimum nor a local minimum, since f() is ecreasing on both sies. (b) See Figure 7.6. f() Figure 7.6. Since the rate at which the leaf grows is proportional to the rate of photosynthesis, the slope of the size graph is proportional to the given graph. Thus, if S(t) is the size of the leaf an p(t) is the rate of photosynthesis S (t) = kp(t) for some positive k. We plot the antierivative of p(t) to get the graph of S(t) in Figure 7.7. (Since no scale is given on the vertical ais, we can imagine k =.) The size of the leaf may be represente by its area, or perhaps by its weight. size of leaf S(t) r (ays) Figure For every number b, the Funamental Theorem tells us that b F () = F(b) F() = F(b) = F(b).

28 9 Chapter Seven /SOLUTIONS Therefore, the values of F(), F(), F(), an F(4) are values of efinite integrals. The efinite integral is equal to the area of the regions uner the graph above the -ais minus the area of the regions below the -ais above the graph. Let A, A, A, A 4 be the areas shown in Figure 7.8. The region between = an = lies above the -ais, so F() is positive, an we have F() = F () = A. The region between = an = also lies entirely above the -ais, so F() is positive, an we have F() = F () = A + A. We see that F() > F(). The region between = an = inclues parts above an below the -ais. We have F() = F () = (A + A ) A. Since the area A is approimately the same as the area A, we have F() F(). Finally, we see that F(4) = 4 F () = (A + A ) (A + A 4). Since the area A + A appears to be larger than the area A + A 4, we see that F(4) is positive, but smaller than the others. The largest value is F() an the smallest value is F(4). None of the numbers is negative. Area = A F () Area = A 4 Area = A Area = A4 Figure (a) The total volume emptie must increase with time an cannot ecrease. The smooth graph (I) that is always increasing is therefore the volume emptie from the blaer. The jagge graph (II) that increases then ecreases to zero is the flow rate. (b) The total change in volume is the integral of the flow rate. Thus, the graph giving total change (I) shows an antierivative of the rate of change in graph (II). 6. We can start by fining four points on the graph of F(). The first one is given: F() =. By the Funamental Theorem of Calculus, F(6) = F() + 6 F (). The value of this integral is 7 (the area is 7, but the graph lies below the -ais), so F(6) = 7 = 4. Similarly, F() = F() =, an F(8) = F(6) + 4 =. We sketch a graph of F() by connecting these points, as shown in Figure 7.9. (, ) (, ) F() (8, ) 4 8 (6, 4) Figure 7.9

29 7. We see that F ecreases when <.5 or > 4.67, because F is negative there. F increases when.5 < < 4.67, because F is positive there. So F has a local minimum at =.5. F has a local maimum at = We have F() = 4. Since F is negative between an.5, the Funamental Theorem of Calculus gives us Similarly F(.5) F() = F(4.67) = F(.5) + F(6) = F(4.67) +.5 F () = 4 F(.5) = 4 4 = F () = + 5 = 5. F () = 5 5 = A graph of F is in Figure 7.. The local maimum is (4.67, 5) an the local minimum is (.5, ). 7.5 SOLUTIONS 9 4 F() Figure The areas given enable us to calculate the changes in the function F as we move along the t-ais. Areas above the ais count positively an areas below the ais count negatively. We know that F() =, so Thus, Similarly, an A graph is shown in Figure 7.. F() F() = F (t)t = Area uner F t F() = F() + 5 = + 5 = 8. F(5) F() = 5 F (t)t = 6 F(5) = F() 6 = 8 6 = 8 F(6) = F(5) F (t)t = 8 + =. = 5

30 9 Chapter Seven /SOLUTIONS (, 8) (, ) F(t) (6, ) t (5, 8) Figure The critical points are at (, 5), (, ), (4,), an (5,5). A graph is given in Figure 7.. y (, ) G(t) 5 (5, 5) (4, ) 5 (, 5) 4 5 t Figure 7.. (a) Critical points of F() are the zeros of f: = an =. (b) F() has a local minimum at = an a local maimum at =. (c) See Figure F() Figure 7. Notice that the graph coul also be above or below the -ais at =.. (a) Critical points of F() are =, = an =. (b) F() has a local minimum at =, a local maimum at =, an a local minimum at =. (c) See Figure 7.4.

31 7.5 SOLUTIONS 9 4 F(). By the Funamental Theorem, Figure 7.4 f() f() = f (), Since f () is negative for, this integral must be negative an so f() < f().. First rewrite each of the quantities in terms of f, since we have the graph of f. If A an A are the positive areas shown in Figure 7.5: so Since Area A > Area A, an therefore f() f() = f(4) f() = f(4) f() = A < f() f() < y 4 4 f (t)t = A f (t)t = A A + A f A + A (t) t = < A A + A A < < A f(4) f() < f(4) f(). 4 A A y = f () Figure See Figure 7.6. a b Figure 7.6 f() f(b) f(a)

32 94 Chapter Seven /SOLUTIONS 5. See Figure 7.7. Slope= f(b) f(a) b a f() f(b) f(a) a Figure 7.7 b 6. See Figure 7.8. f() a b Figure See Figure 7.9. Note that we are using the interpretation of the efinite integral as the length of the interval times the average value of the function on that interval, which we evelope in Section 6.. f() a b Figure 7.9 F(b) F(a) b a Solutions for Chapter 7 Review. t + 4 t t5. t + 7t + t F(z) = e z + z + C 5. P(r) = πr + C 6. P(y) = ln y + y / + y + C 7. t 8. cos t

33 SOLUTIONS to Review Problems For Chapter Seven Antierivative G() = ( + )4 + + C = + C 4. We use substitution with w = + an w =. Then f() = ( + ) = w w4 ( + )4 w = + C = + C t + 7t + C. + C C e.5t + C = e.5t + C t t + C 6. Since f() = t + t + C. 8. p + ln p + C ln + C.. 5 sin + cos + C. ln π cos + C. + e + C = +, the inefinite integral is + ln + C. Since F () = +, we use F() = 4 +. By the Funamental Theorem, we have ( + ) = (4 + ) = (4 + ) (4 + ) = 4 = r = r r = + = 8/ sin θ θ = cos θ = cos If f() = /, then F() = ln (since ln = ). By the Funamental Theorem, we have = ln = ln ln = ln. 7. Since F () = =, we use F() = =. By the Funamental Theorem, we have 8. ( ) ( ) 4 + = + ( = ) = ( ) = + =. = = 6/ f() =, so F() = + C. F() = implies that + C =, so C =. Thus F() = is the only possibility.. f() =, so F() = + C. F() = implies that C =, so C =. Thus F() = /8 is the only possibility.

34 96 Chapter Seven /SOLUTIONS. We use the substitution w = +, w =. + = w / w = w / + C = + + C. ( + + C) = +.. We use the substitution w = +, w =. ( + ) 5 = w 5 w = w6 6 + C = 6 ( + ) 6 + C. ( 6 ( + ) 6 + C) = ( + ) 5.. We use the substitution w =, w =. e = e ( ) = e w w ( e + C) = ( )( e ) = e. 4. We use the substitution w = 4 +, w = 4. 4 = ew + C = e + C. 4 + = w w = ln w + C = ln(4 + ) + C. (ln(4 + ) + C) = We use the substitution w = +, w =. ( + ) = w w = w w = w + C = ( + ) + C. 6. We use the substitution w = + 4, w =. + 4 = w / w = w / / + C = C. 7. We use the substitution w = 5 7, w = 5. (5 7) = 5 w w = w 5 + C = 55 (5 7) + C. 8. We use the substitution w =.t, w =.t. e.t t = e w w = 5e w + C = 5e.t + C.. 9. We use the substitution w = +, w =. + = w / w = w/ + C = ( + ) / + C. ( ( + ) / + C) = ( + ) / = +.

35 SOLUTIONS to Review Problems For Chapter Seven We use the substitution w =, w =. sin( ) = sin w w = cos w + C = cos( ) + C. ( ) cos( ) + C = ( sin( )) = sin( ). 4. We use the substitution w = t, w = t t. tcos(t )t = t ( sin(t ) + C) = cos(t )(t) = t cos(t ). cos(w)w = sin(w) + C = sin(t ) + C. 4. To fin the area uner the graph of f() = e, we nee to evaluate the efinite integral e. This is one in Eample 4, Section 7., using the substitution w =, the result being e = (e4 ). 4. Since f() = /( + ) is positive on the interval = to =, we have Area = = ln( + ) + = ln ln = ln. The area is ln If f() =, the average value of f on the interval is efine to be + f() = +. We ll integrate by substitution. We let w = + an w =, an we have = w= + = w w = ln w = ln ln = ln. = Thus, the average value of f() on is ln.549. See Figure 7.. w=.549 f() = + Figure 7.

36 98 Chapter Seven /SOLUTIONS 45. (a) Since r gives the rate of energy use, between 5 an (where t = an t = 5), we have Total energy use = (b) The Funamental Theorem of Calculus states that b a 5 46e.9t t quarillion BTUs. f(t) t = F(b) F(a) provie that F (t) = f(t). To apply this theorem, we nee to fin F(t) such that F (t) = 46e.9t ; we take Thus, Total energy use = 5 F(t) = 46.9 e.9t = 4,6e.9t. 46e.9t t = F(5) F() = 4,6e.9t 5 = 4,6(e.95 e ) = 4 quarillion BTUs. Approimately 4 quarillion BTUs of energy were consume between 5 an. 46. Since P/t is negative for t < an positive for t >, we know that P is ecreasing for t < an increasing for t >. Between each two integer values, the magnitue of the change is equal to the area between the graph P/t an the t-ais. For eample, between t = an t =, we see that the change in P is. Since P = at t =, we must have P = at t =. The other values are foun similarly, an are shown in Table 7.4. Table 7.4 t 4 5 P / 47. See Figure 7.. Point of inflection f() 4 Figure See Figure 7.. Inflection point Local ma f() Local min Inflection point 4 Figure 7.

37 SOLUTIONS to Review Problems For Chapter Seven First notice that F will be ecreasing on the interval < < an on the interval < < 4 an will be increasing on the interval < <. The areas tell us how much the function increases or ecreases. By the Funamental Theorem, we have F() = F() + F() = F() + F(4) = F() + 4 F () = 5 + ( 6) =. F () = + 8 = 7. F () = 7 + ( ) = (a) If w = t/, then w = (/)t. When t =, w = ; when t = 4, w =. Thus, 4 g(t/)t = g(w)w = g(w)w = 5 =. (b) If w = t, then w = t. When t =, w = ; when t =, w =. Thus, g( t)t = g(w)( w) = + g(w)w = (a) We sketch f() = e ; see Figure 7.. The shae area to the right of the y-ais represents the integral e. f() = e Figure 7. (b) Using a calculator or computer, we obtain 5 e =.9596 e =.9995 (c) The answers to part (b) suggest that the integral converges to. 5. (a) A calculator or computer gives e = = 8 = 6. = 98. These values o not seem to be converging. (b) An antierivative of F () = is F() = ( since ( ) = ). So, by the Funamental Theorem, we have b = b = b = b. (c) The limit of b as b oes not eist, as b grows without boun. Therefore So the improper integral lim b b = lim ( b ) oes not eist. b oes not converge.

38 4 Chapter Seven /SOLUTIONS 5. The original ose equals the quantity of rug eliminate. The quantity of rug eliminate is the efinite integral of the rate. Thus, letting t, we have the improper integral Total quantity of rug eliminate = Using the fact that e kt t = k ekt + C, we have 5(e.t e.t ) t. ( Total quantity = lim 5 b. e.t + ) b. e.t = lim 5( e.b + 5e.b ( e + 5e )) b Since e.b an e.b as b, we have Total quantity = 5( 5) = 5 mg. 54. Integration by parts with u =, v = cos gives cos = sin sin + C = sin + cos + C. 55. We integrate by parts, with u = y, v = sin y. We have u =, v = cos y, an y sin y y = y cos y ( cos y)y = y cos y + sin y + C. ( y cos y + sin y + C) = cos y + y sin y + cos y = y sin y. y 56. Remember that ln( ) = ln. Therefore, ln( ) = ln = ln + C. (ln + C) = ln + = ln = ln( ). CHECK YOUR UNDERSTANDING. True. We see that the erivative of t / + 5 is t.. False. When we a one to the eponent, we get. The function is an antierivative of.. False. Antierivatives of e are of the form (/)e + C. 4. True. This is a correct integral statement. 5. True. We know 6. False. We know that e = e + C. z / z = z/ / + C = z + C. 7. False. The erivative of ln t is /t so the correct integral statement is (/t) t = ln t + C. 8. True, since the erivative of is (ln). 9. True.. True. We know that all antierivatives iffer only by a constant.. False, since w = (q + 6q ) q cannot be substitute.

39 . True, since w = (/). CHECK YOUR UNDERSTANDING 4. False. We have w =. Since the integral e oes not inclue an to be sustitute for w, this integral cannot be evaluate using this substitution. 4. True, since w =. 5. False. This is almost true, but is off by a minus sign since w = s. 6. True. Since w = t t, we have 7. True, since w = (e e ). t t + t = w = w 8. False. The substitution w = q + 5 woul give the integral (/)w w. 9. True, since w = cos α α.. False. This is almost true, but is off by a minus sign, since w = sin.. False. We nee to substitute the enpoints into an antierivative of /.. True, since ln is an antierivative of /.. True, since is an antierivative of. 4. False. We nee to first fin an antierivative of. w w. 5. False. For a efinite integral, we nee to substitute the enpoints into the antierivative. 6. True. An antierivative is e t an we substitute the limits of integration an subtract. 7. False. When we make the substitution w =, we must also substitute for the limits of integration. Since w = 5 = 5 when = 5 an w = when =, the result of the substitution is 5 e w w. 8. True, since w = (/) an w = when = e an w = when =. 9. False. The two efinite integrals represent two ifferent quantities.. True. The function y = e k is positive, so the integral represents the area uner the curve between = an = an so is positive.. True.. True. If a function is concave up, its secon erivative is positive which implies that its erivative is increasing.. True. This is the Funamental Theorem of Calculus. 4. False. The limits of integration on the integral nee to be from to to make this a true statement. 5. True. Since f is positive on the interval to 4, the function is increasing on that interval. 6. False. Since f is negative on the interval to, the function is ecreasing on that interval. 7. True. Since f is positive on the interval to, the function is increasing on that interval. 8. True. Since f is negative on the interval 5 to 6, the function is ecreasing on that interval. 9. False. Since f is negative on the interval to, the function is ecreasing on that interval. 4. True. The area below the curve of f between = an = is similar in size to the area above the curve between = an =. Between = an =, the function f increases approimately the same amount that it ecreases, so f() f(). 4. False. The integral has to be u v when u an v are substitute. In this case, we shoul have u = an v = e. 4. True. 4. False. We integrate v to fin v. We see v = e = (/)e. 44. False. We integrate v to fin v an / is not the antierivative of ln. It is the erivative of ln. In this case, the assignment of parts is wrong. We shoul try u = ln an v =. 45. False. This integral is more appropriately evaluate using the metho of substitution. 46. True. 47. True. 48. False. This integral is more appropriately evaluate using the metho of substitution. 49. False. This integral is more appropriately evaluate using the metho of substitution.

40 4 Chapter Seven /SOLUTIONS 5. True. We use u = ln an v =. PROJECTS FOR CHAPTER SEVEN. (a) Suppose Q(t) is the amount of water in the reservoir at time t. Then Q (t) = Rate at which water in reservoir is changing = Inflow rate Outflow rate Thus the amount of water in the reservoir is increasing when the inflow curve is above the outflow, an ecreasing when it is below. This means that Q(t) is a maimum where the curves cross in July 7 (as shown in Figure 7.4), an Q(t) is ecreasing fastest when the outflow is farthest above the inflow curve, which occurs about October 7 (see Figure 7.4). To estimate values of Q(t), we use the Funamental Theorem which says that the change in the total quantity of water in the reservoir is given by Q(t) Q(Jan 7) = or Q(t) = Q(Jan 7) + t Jan 7 t Jan 7 (inflow rate outflow rate) t (Inflow rate Outflow rate) t. rate of flow (millions of gallons/ay) Q(t) is ma Q(t) is min Outflow Inflow Jan (7) April July Oct Jan(8) Q(t) millions of gallons Q(t) is increasing most rapily Q(t) is ecreasing most rapily Jan (7) April July Oct Jan(8) Figure 7.4 (b) See Figure 7.4. Maimum in July 7. Minimum in Jan 8. (c) See Figure 7.4. Increasing fastest in May 7. Decreasing fastest in Oct 7.

41 SOLUTIONS TO PRACTICE PROBLEMS ON INTEGRATION 4 () In orer for the water to be the same as Jan 7 the total amount of water which has flowe into the reservoir minus the total amount of water which has flowe out of the reservoir must be. Referring to Figure 7.5, we have July 8 Jan 7 giving A + A = A + A 4 (Inflow Outflow)t = A + A A + A 4 = rate of flow (millions of gallons/ay) A A A 4 Inflow A Outflow Jan ( 7) April July Oct Jan ( 8) April July Figure 7.5 Solutions to Practice Problems on Integration q. + 5q + q + C. (u 4 + 5) u = u u + C. + + C. 4. Since (e t ) = e t, we have 5. 4 / + C 6. (a + b) = a + b + C C. e.5t t = (w 4 w + 6w ) w = w5 5 (4 + 5 ) e t t = e t + C. (.5 e.5t ) + C = e.5t + C w w w + C = w5 5 w4 + w w + C = 4 ln C = 4 ln 5 + C q / q = q/ / + C = q/ + C

42 44 Chapter Seven /SOLUTIONS sin θ θ = cos θ + C (p + 5 p )p = + 5ln p + C p ( ) P e kt t = P k ekt + C = P k ekt + C (q + 8q + 5) q = q q + 5q + C = q q + 5q + C ( ) e.75t t =.75 e.75t + C = e.75t + C (5sin + cos ) = 5cos + sin + C 8. ( + 5sin ) = 5cos + C 5 9. w = 5ln w + C w ( ) r. πr h r = πh + C = π hr + C. (q + q )q = q + q + C = q q + C ( ). 5p q 4 p p = 5 q 4 + C = 5p q 4 + C ( ) q. 5p q 4 q = 5p 5 + C = p q 5 + C 5 4. ( + 6e ) = + 6 e + C = + e + C 5. 5e q q = 5 eq + C =.5e q + C ( ) 6. p + p = p4 + ln p + C p 4 7. (A + B) = A4 4 + B + C 8. (6 / + 5) = 6 / / C = 4/ C 9. ( e ) = e + C. 5e.t + C t t + 5t + C. ( ( ) a + b ) = aln + b + C = a ln b + C (Aq + B)q = Aq + Bq + C (6 / + 8 / ) = 6 / 8/ / + / = + 6 / + C

43 (e t + 5)t = et + 5t + C sin() = cos() + C cos(4) = sin(4) + C SOLUTIONS TO PRACTICE PROBLEMS ON INTEGRATION We use the substitution w = y +, w = y: y + y = w = ln w + C = ln y + + C. w 9. We use the substitution w = y + 5, w = y y y(y + 5) 8 y = (y + 5) 8 (y y) = w 8 w = w C y ( 8 (y + 5) 9 + C) = 8 [9(y + 5) 8 (y)] = y(y + 5) 8. 4 sin(4) + C A sin(bt)t = A B cos(bt) + C = 8 (y + 5) 9 + C. 4. We use the substitution w = +, w = : + = w / w = w / / + C = 9 ( + )/ + C. 4. We use the substitution w = + e, w = e. e + e = w w = ln w + C = ln( + e ) + C. (We can rop the absolute value signs since + e for all.) [ln( + e ) + C] = + e e = e + e. 44. We use the substitution w = sin 5θ, w = 5 cos 5θ θ. sin 6 5θ cos 5θ θ = 5 w 6 w = 5 (w7 7 ) + C = 5 sin7 5θ + C. θ ( 5 sin7 5θ + C) = 5 [7 sin6 5θ](5 cos 5θ) = sin 6 5θ cos 5θ. Note that we coul also use Problem to solve this problem, substituting w = 5θ an w = 5 θ to get: sin 6 5θ cos 5θ θ = 5 sin 6 w cos w w = w 5 (sin7 ) + C = 7 5 sin7 5θ + C. 45. We use the substitution w = + sin, w = cos : cos = w / w = w/ + sin / + C = + sin + C.

44 46 Chapter Seven /SOLUTIONS 46. Integration by parts with u = ln, v = gives ln = ln = ln 4 + C. 47. e = e = e e + C, e (let = u, e = v, e = v) 48. where C is a constant. ze z z = [ ze z ] = e [e z ] = e e + = e +. e z z (let z = u, e z = v, e z = v)