Section 6-1 Antiderivatives and Indefinite Integrals

Transcription

1 Name Date Class Section 6- Antiderivatives and Indefinite Integrals Goal: To find antiderivatives and indefinite integrals of functions using the formulas and properties Theorem Antiderivatives If the derivative s of two functions are equal on an open interval (a, b), then the functions 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( ) + kfor some constant k. Formulas and Properties of Indefinite Integrals For C and k both a constant n+. n d= + C, n + n. ed= e + C. d = ln + C,. kf ( ) d = k f ( ) d. [ f ( ) ± g( )] d = f ( ) d ± g( ) d In Problems, find each indefinite integral and check by differentiating.. 6d y = 6d 6 y = + C y = + C Check: y = + C = 6 + d = 6 d 6-

2 . 9 d y = 9 d 9 y = + C 9 y = + C y = 6 + C Check: y = 6 + C 6 = + d = 9 d. 7e d y = 7e d y = 7e + C Check: y = 7e + C = 7e + d = 7e d. Find all the antiderivatives for = 7z + dz 7 z 7 z = (7z + ) dz = ( + ) dz = ( + ) dz y = 7ln z + z+ C dz = 7z +. 6-

3 In Problems 8, find each indefinite integral.. ( + 8) d ( + 8) d= ( + 8 ) d 6 8 = + + C 6 6 = + + C d ( 8 ) d= d 8 = + C 8 = + C d d = d + ( 8 ) = + d 8 = + + C = + C 6-

4 e 6 d 6 + e 6 d = + e d = ln + e + C In Problems 9, find the particular antiderivative of each derivative that satisfies the given conditions. 9. R'( ) = ; R () = First find the indefinite integral of the function. R( ) = ( ) d 6 R( ) = C R ( ) = C Using the condition given, find the specific value of C. R ( ) = C = () + () + () + C = C = 6 + C 6 = C Substituting the specific value of C yields the particular equation R ( ) =

5 . e t t ; = + y () = First find the indefinite integral of the function. t = (e + t ) t = (e + t ) t t t y = e + + C t t y = e + t+ C Using the condition given, find the specific value of C. t t y = e + t+ C () = e + () + C = C = + C = C Substituting the specific value of C yields the particular equation t t y = e + t+. 6-

6 . dd d 9 = ; D (9) = First find the indefinite integral of the function. 9 dd = d dd = ( 9 ) d dd = ( 9 ) d 9 D ( ) = + C 9 D ( ) = + + C Using the condition given, find the specific value of C. 9 D ( ) = + + C 9 = (9) + + C 9 = + + C = 6 + C = C Substituting the specific value of C yields the particular equation 9 D ( ) =

7 . h'( ) = ; h () = First find the indefinite integral of the function. h'( ) = h ( ) = ( + 7 ) d 7 h ( ) = 6ln + C 7 h ( ) = 6ln + + C Using the condition given, find the specific value of C. 7 h ( ) = 6ln + + C 7 = 6ln+ + C = 6() + 7+ C = 7+ C = C Substituting the specific value of C yields the particular equation 7 h ( ) = 6ln

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9 Name Date Class Section 6- Integration by Substitution Goal: To find the indefinite integrals using general indefinite integral formulas Formulas: General Indefinite Integral Formulas Definition: n+ n [ f( )] [ f( )] f '( ) d= + C, n + f ( ) f ( ) e f '( ) d= e + C '( ) ln ( ) ( ) f d = f f + C n+ n u u du = + C, n + u u edu= e + C du = ln u + C u Differentials n n If y = f( ) defines a differentiable function, then. The differential d of the independent variable is an arbitrary real number.. The differential of the dependent variable y is defined as the product of f '( ) and d: = f '( ) d Procedure: Integration by Substitution. Select a substitution that appears to simplify the integrand. In particular, try to select u so the du is a factor in the integrand.. Epress the integrand entirely in terms of u and du, completely eliminating the original variable and its differential.. Evaluate the new integral if possible.. Epress the antiderivative found in step in terms of the original variable. 6-9

10 In Problems 8, find each indefinite integral and check the result by differentiating.. (6 + ) ( + ) d Let u = 6 +, therefore du = ( + ) d. Rewrite the original integral in terms of the variable u and solve. y = (6 + ) ( + ) d y = u du u y = + C (6 ) y = + + C Check: y = (6 + ) + C = + d d + = (6 + ) ( + ) d d ()(6 ) (6 ). ( + )( + ) d Let u = +, therefore of the variable u and solve. du = ( + ) d. Rewrite the original integral in terms y = ( + )( + ) d y = ( + ) ( + ) d y = u du u C ( ) y = + y = + + C ( ) y = + + C 6-

11 Check: y = ( + ) + C y = ( + ) + C ( ) d ( = + + ) d d = (6 + ) ( + ) d = ( + )( + ) d. t + t + 6t Let u = t + 6t, therefore du = (8t + 6) du = 8( t + ). Rewrite the original integral in terms of the variable u and solve. 8( t + ) y = 8 t + 6t y = du 8 u y = ln u + C 8 ln t + 6t y = + C 8 Check: ln t + 6t y = + C 8 d = (t + 6t ) 8(t + 6t ) d = (8t + 6) 8(t + 6t ) t + = t + 6t 6-

12 . e. d Let u =., therefore du =. d. Rewrite the original integral in terms of the variable u and solve.. y = e d. y = e (. d). u y = e du. u y = e + C.. y = e + C Check:.. y = e + C. d = e (. ) d. = e (.) = e. ( + 7) 7 d Let u = + 7, therefore du = d. If we rewrite the original integral in terms of the variable u, the substitution would not be complete. We also need u 7 =. Now rewrite the original integral in terms of u and solve. 7 y = ( + 7) d 7 y = ( u 7) u du 8 7 y = ( u 7 u ) du 9 8 u 7u y = + C ( + 7) 7( + 7) y = + C

13 Check: 9 8 ( + 7) 7( + 7) y = + C d 7 7 d = (9)( + 7) ( + 7) (8)( + 7) ( + 7) 9 d 8 d 8 7 = ( + 7) 7( + 7) 7 = ( + 7) ( + 7 7) 7 = ( + 7) 6. (ln( )) d 6 Let u = ln( ), therefore du = d du d. = Rewrite the original integral in terms of the variable u and solve. ( ln ) (ln( )) y = y = u du u y = + C y = + C Check: ( ln ) d y = + C = d 6 = (ln ) (ln ) = d ()(ln ) (ln ) 6-

14 7. e d Let u =, therefore du = d du = d. Rewrite the original integral in terms of the variable u and solve. y = e d y = e d u y = e du u y = e + C y = e + C Check: y = e + C d = e ( ) d = e ( ) e = 6-

15 8. ( 7) ( 7) d d = + = + Let u = + 7, therefore variable u and solve. du = 6 d. Rewrite the original integral in terms of the 6 y = ( + 7) d y = ( + 7) 6 d y = u du u y = + C 6 ( 7) 6 y = + + C Check: 6 y = ( + 7) + C = + d + = ( + 7) (6 ) d (6)( 7) ( 7) = ( + 7) 9. The indefinite integral can be found in more than one way. Given the integral, ( + ) d, first use the substitution method to find the indefinite integral and then find it without using substitution. Using substitution, let u = +, therefore du = d. Rewrite the original integral in terms of the variable u and solve. y = ( + ) d y = u du u y = + C ( ) y = + + C 6-

16 Epanding the final function yields the following: ( ) y = + + C ( )( )( ) y = C ( ) y = C 6 y = C Note: The new value of C above is the addition of the constant 9 and the original constant C from the integration. Without using substitution to find the given indefinite integral, you first would multiply the integrand as follows ( + ) = ( + )( + ) = ( ) = Now find the indefinite integral using the above function as the integrand. y = ( ) d 6 8 y = C 6 6 y = C 6-6

17 Name Date Class Section 6- Differential Equations; Growth and Decay Goal: To solve differential equations that involve growth and decay. Theorem : Eponential Growth Law If dq rt = rq and Q() = Q, then Q= Q e, where Q = amount of Q at t = r = relative growth rate (epressed as a decimal) t = time Q = quantity at time t Table : Eponential Growth Description Model Solution Unlimited growth = ky kt y = ce Eponential decay Limited growth Logistic Growth kt>, y() = c = ky kt>, y() = c = km ( y) kt>, y () = = ky( M y) kt>, y() = M + c y = ce kt y = M( e kt ) M y = + ce kmt 6-7

18 In Problems, find the general or particular solution, as indicated, for each differential equation.. d = 6 The differential equation can be found by using the integration properties from Section 6.. y = 6 d 6 y = + C y = + C. d = e ; y () = The differential equation can be found by using the integration properties from Section 6.. Let u =, therefore du = d. Rewrite the original integral in terms of the variable u and solve. y = e d u u y = e du y = e + C y = e + C Using the condition given, find the specific value of C. y = e + C () = e + C = e + C = + C = C Substituting the specific value of C yields the particular equation y = e

19 . 6y d = The differential equation is in the form of eponential decay; therefore, using the eponential decay model in Table yield s the following solution: y = ce 6 d. ; = () = The differential equation is in the form of unlimited growth; therefore, using the unlimited growth model in Table yield s the following solution: = ce t Using the condition given, find the specific value of C. = ce = ce = ce = c t () Substituting the specific value of C yields the particular solution t = e.. Find the amount A in an account after t years if da =.7A and () 8 A = The eponential growth law (unlimited growth) applies to the situation. Since we are given the initial amount, the amount in an account after t years would be: A= 8e.7t 6-9

20 6. A single injection o a drug is administered to a patient. The amount Q in the bo decreases at a rate proportional to the amount present. For a particular drug, the rate dq is 6% per hour. Thus, =.6Q and Q() = Q where t is time in hours. a. If the initial injection is milliliters [ Q () = ], find Q= Q() t satisfying both conditions. b. How many milliliters (to two decimal places) are in the bo after 8 hours? c. How many hours (to two decimal places) will it take for half the drug to be left in the bo? a. By the eponential growth law (eponential decay) for the given conditions, we.6t have the equation Qt () = e. b. After 8 hours, t = 8, therefore Q(8) = e.6(8).8 Q(8) = e Q(8) = (.6878) Q(8).9 Therefore, after 8 hours, approimately.9 milliliters will remain in the bo. c. Half the initial injection is. milliliters. We need to find t such that Qt ( ) =.. Qt () = e. = e. = e.6t.6t.6t.6t ln. = ln e ln. =.6t. t Therefore, it will take approimately. hours for the amount of the drug to be half of the initial amount. 6-

21 7. A company is trying to epose a new product to as many people as possible through radio ads. Suppose that the rate of eposure to new people is proportional to the number of those who have not heard of the product out of L possible listeners. No one is aware of the product at the start of the campaign, and after days, % of L dn are aware of the product. Mathematically, = kl ( N ), N () =, and N() =. L. a. Solve the differential equation. b. What percent of L will have been eposed after 7 days of the campaign? c. How many days will it take to epose 7% of L? a. By the eponential growth law (limited growth) for the given conditions, we have the equation Nt () = L( e ) k k k k k. L= L( e ). = e. = e. = e ln. = ln e ln. = k.6 = k kt Therefore, the solution is.6t Nt () = L( e ). b. After 7 days, t = 7, therefore.6(7) N(7) = L( e ) N(7) = L(.7698) N(7) = L(.7) N(7).7L Therefore, after 7 days, approimately 7.% of L will have been eposed to the product. 6-

22 c. Find t such that Nt ( ) =.7 L..6t Nt () = L( e ).6t.7 L= L( e ).7 = e. = e. = e ln. = ln e.6t.6t.6t.6t ln. =.6t. t Therefore, it will take approimately days to epose 7% of L to the new product. 6-

23 Name Date Class Section 6- The Definite Integral Goal: To calculate the values of definite integrals using the properties. Theorem: Limits of Left and Right Sums If f( ) > and is either increasing or decreasing on [a, b], then its left and right sums approach the same real number as n. Theorem: Limit of Riemann Sums If f is a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n. Definition: Definite Integral Let f be a continuous function on [a, b]. The limit I of Riemann sums for f on [a, b] is called the definite integral of f from a to b and is denoted as a f ( ) d. b Properties: Definite Integrals a. a f ( ) d= b a a b b b akf d = k a f d b b b a ± = a ± a b c b a f d= a f d+ c f d. f( ) d= f( ) d. ( ) ( ), k a constant. [ f ( ) g( )] d f ( ) d g( ) d. ( ) ( ) ( ) 6-

24 In Problems and, calculate the indicated Riemann sum Sn for the function f( ) = 7.. Partition [, 9] into five subintervals of equal length, and for each subinterval [ k, k], let ck = ( k + k)/. 9 ( ) = = = c = = c = = c = = f () = 7 () f () = 7 f () = 7 () f () = 9 f () = 7 () f () = c = = 6 c = = 8 f (6) = 7 (6) f (6) = f (8) = 7 (8) f (8) = S = f( c ) + f( c ) + f( c ) + f( c ) + f( c ) S S S = (7)() + (9)() + ( )() + ( )() + ( )() = + 8 =. Partition [, 8] into four subintervals of equal length, and for each subinterval [ k, k], let ck = ( k + k) /. 8 ( ) = = = c ( ) + ( ) ( ) + = = c = = f ( ) = 7 ( ) f ( ) = f () = 7 () f () = 7 6-

25 c () + () + 8 = = c = = 6 f () = 7 () f () = f (6) = 7 (6) f (6) = S = f( c ) + f( c ) + f( c ) + f( c ) S S S = ( )() + (7)() + ( )() + ( )() = + 6 = In Problems 7, calculate the definite integral, given that d=. d= 7 d= 8. d d= = (.) d= 7. d. ( + ) d ( + ) = + d d d = (.) + = + ( + ) d= 6-

26 . 7 d 7 7 = 7 d d 7 ( d d) = + 8 = + = 686 d= 6. 7d 7d= 7 = 7(.) 7d= 87. d ( + ) d Since the upper and lower limits of integration are the same, the value of the integral 7 7 would be ( + ) d=. 6-6

27 Name Date Class Section 6- The Fundamental Theorem of Calculus Goal: To use the fundamental theorem of calculus to solve problems. Theorem : Fundamental Theorem of Calculus If f is a continuous function on [a, b], and F is any antiderivative of f, then b a f ( ) d= F ( b ) F ( a ) Definition: Average Value of a Continuous Function f over [a, b] b ( ) b a a f d In Problems 7, evaluate the integrals.. (6 + ) d 6 (6 + ) d= + (6 + ) d= 6 ( ) = + = (() + ()) (() + ()) = 6 6-7

28 . (7 e ) d (7 e ) d= 7e = 7e 7e = 7e 7 (7 e ) d 7.9. ( ) d Since the upper limit and lower limit of integration are the same value, the definite integral value will be ( ) d=.. + d d = ( + ) d + ( + ) = + = ( + ) = ( + ) (+ ) = () () d = 6-8

29 . 6 ( + 9) d We will solve the problem using the substitution method of integration. Let u = + 9, therefore variable u and solve. du = 6 d du = d. Rewrite the original integral in terms of the d u du 6 = 6 ( + 9) = = 6 u = 7 = = 7 ( + 9) = 7 7 (() + 9) (() + 9) = ( + 9) d 7, d + We will solve the problem using the substitution method of integration. Let u = +, therefore the variable u and solve. + = d = = du + u du = ( + ) d. Rewrite the original integral in terms of = ln u + d.7 + = = = ln + = ln () () + () ln () () + () = ln ln 6-9

30 d We will solve the problem using the substitution method of integration. Let u = 6, therefore du = d du = ( ) d. Rewrite the original integral in terms of the variable u and solve. d u du = 6 + = = = u = = (6 ) 6 = + = (6() + ) (6() + ) 6 6 = d 7.7 In Problems 8 and 9, find the average value of the function over the given interval. 8. f( ) = 8+ ; [,8] b 8 f ( ) d = ( 8 + ) d b aa 8 8 = ( + ) 6 = ((8) (8) + (8)) (() () + ()) 6 = ( 86 ) 6 = 6 ( ) 6-

31 9.. e g ( ) = [,] We will solve the problem using the substitution method of integration. Let u =., therefore du =. d. Rewrite the original integral in terms of the variable u and solve b. a f( ) d= ( e ) d b a = u = ( e ) du ()(.) u = = e =. e = =.6.().() ( e e ) ( e ). The total cost (in dollars) of manufacturing units of a product is C ( ) =, +. a. Find the average cost per unit if units are produced. [Hint: Recall that Cis ( ) the average cost per unit.] C ( ) C ( ) =, + C ( ) =, C ( ) = +, C() = + C() = 7 + C() = Therefore, the average cost for units is $ per unit. 6-

32 b. Find the average value of the cost over the interval [, ]. b 8 a f ( ) d = (, + ) b a d =, + (((, )() () ) ((, )() () )) = + + = (,, ) = 8, Therefore, the average value of the cost will be $8,.. A company manufactures a product and the research department produced the marginal cost function C'( ) = 8 where C'( ) is in dollars and is the number of units produced per month. Compute the increase in cost going from a production level of units per month to 8 units per month. Set up a definite integral and evaluate it. b a b a 8 C '( ) d = ( ) d 8 = 8 (8) () = (8) () 8 8 = 6,, C'( ) d= 6, Therefore, the increase in cost going from units per month to 8 units per month is $6,. 6-