Name Date Class Section 6-1 Antiderivatives and Indefinite Integrals Goal: To find antiderivatives and indefinite integrals of functions using the formulas and properties Theorem 1 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'( x) = G'( x) for all x in (a, b), then F( x) = G( x) + kfor some constant k. Formulas and Properties of Indefinite Integrals For C and k both a constant n+ 1 1. n x x= + C, n + 1 n 1 2. x x e= e + C 3. 1 = ln x + C, x x 0 4. kf ( x) = k f ( x) 5. [ f ( x) ± g( x)] = f ( x) ± g( x) In Problems 1 3, find each indefinite integral and check by differentiating. 1. 6x 6-1
2. 9x 1 2 3. 7e x 4. Find all the antiderivatives for dy 1 = 7z + 4. dz 6-2
In Problems 5 8, find each indefinite integral. 5. 3 2 x ( x + 2x 8) 6. 8 4 x x 3 7. 8x + 5 4 x 6-3
8. 5 6 5x + 3x e 6 x x In Problems 9 12, find the particular antiderivative of each derivative that satisfies the given conditions. 9. 3 2 R'( x) = 4x + 6x + 3; R (1) = 12 6-4
dy 10. 3e t 3t 5; dt = + y (0) = 5 6-5
11. dd 2 5x 9 = 2 ; D (9) = 50 x 6-6
12. 1 2 h'( x) = 6x + 7 x ; h (1) = 3 6-7
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Name Date Class Section 6-2 Integration by Substitution Goal: To find the indefinite integrals using general indefinite integral formulas Formulas: General Indefinite Integral Formulas 1. 2. 3. 4. 5. 6. Definition: n+ 1 n [ f( x)] [ f( x)] f '( x) = + C, n + 1 f ( x) f ( x) e f '( x) = e + C 1 '( ) ln ( ) ( ) f x = f x f x + C n+ 1 n u u du = + C, n + 1 u u edu= e + C 1 du = ln u + C u Differentials n 1 n 1 If y = f( x) defines a differentiable function, then 1. The differential of the independent variable x is an arbitrary real number. 2. The differential dy of the dependent variable y is defined as the product of f '( x) and : dy = f '( x) Procedure: Integration by Substitution 1. Select a substitution that appears to simplify the integrand. In particular, try to select u so the du is a factor in the integrand. 2. Express the integrand entirely in terms of u and du, completely eliminating the original variable and its differential. 3. Evaluate the new integral if possible. 4. Express the antiderivative found in step 3 in terms of the original variable. 6-9
In Problems 1 8, find each indefinite integral and check the result by differentiating. 1. 2 3 (6x + 3x 5) (12 x+ 3) 2. 3 2 3 ( x + 3 x 5)(3 x + 3) 6-10
3. 3 t + 2 dt 4 2t + 16t 1 4. e 0.09x 6-11
5. xx ( + 7) 7 6. 2 4 2(ln(3 x )) x 6-12
7. x 1 1 x 2 e 3 8. dy 12 x (2x 7) = + 2 3 5 6-13
9. The indefinite integral can be found in more than one way. Given the integral, 2 2 2 xx ( + 3), first use the substitution method to find the indefinite integral and then find it without using substitution. 6-14
Name Date Class Section 6-3 Differential Equations; Growth and Decay Goal: To solve differential equations that involve growth and decay. Theorem 1: Exponential Growth Law If dq rt = rq and Q(0) = Q0, then Q= Q0 e, dt where Q 0 = amount of Q at t = 0 r = relative growth rate (expressed as a decimal) t = time Q = quantity at time t Table 1: Exponential Growth Description Model Solution Unlimited growth dy kt = ky y = ce dt kt>, 0 y(0) = c Exponential decay = ky y = ce kt Limited growth Logistic Growth dy dt kt>, 0 y(0) = c dy dt = km ( y) kt>, 0 y (0) = 0 dy dt = ky( M y) kt>, 0 y(0) = M 1+ c y = M(1 e kt ) M y = 1 + ce kmt 6-15
In Problems 1 4, find the general or particular solution, as indicated, for each differential equation. 1. dy = 6x 3 2. dy 3 2 x = 3 xe ; y (0) = 4 6-16
dy 3. 6y = 4. 4 x; dt = x (0) = 2 5. Find the amount A in an account after t years if da = 0.07A and (0) 8000 dt A = 6-17
6. A single injection o a drug is administered to a patient. The amount Q in the body decreases at a rate proportional to the amount present. For a particular drug, the rate dq is 6% per hour. Thus, = 0.06Q and Q(0) = Q0 where t is time in hours. dt a. If the initial injection is 5 milliliters [ Q (0) = 5], find Q= Q() t satisfying both conditions. b. How many milliliters (to two decimal places) are in the body after 8 hours? c. How many hours (to two decimal places) will it take for half the drug to be left in the body? 6-18
7. A company is trying to expose a new product to as many people as possible through radio ads. Suppose that the rate of exposure 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 15 days, 50% of L dn are aware of the product. Mathematically, = kl ( N ), N (0) = 0, and dt N(15) = 0.5 L. a. Solve the differential equation. b. What percent of L will have been exposed after 7 days of the campaign? c. How many days will it take to expose 75% of L? 6-19
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Name Date Class Section 6-4 The Definite Integral Goal: To calculate the values of definite integrals using the properties. Theorem: Limits of Left and Right Sums If f( x ) > 0and 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 ( x ). b Properties: Definite Integrals a 1. a f ( x ) = 0 b a a b b b akf x = k a f x b b b a ± = a ± a b c b a f x = a f x + c f x 2. f( x) = f( x) 3. ( ) ( ), k a constant 4. [ f ( x) g( x)] f ( x) g( x) 5. ( ) ( ) ( ) 6-21
In Problems 1 and 2, calculate the indicated Riemann sum Sn for the function 2 f( x) = 17 2 x. 1. Partition [ 1, 9] into five subintervals of equal length, and for each subinterval [ xk 1, xk], let ck = ( xk 1 + xk)/2. 2. Partition [ 4, 8] into four subintervals of equal length, and for each subinterval [ xk 1, xk], let ck = (2 xk 1 + xk) / 3. 6-22
In Problems 3 7, calculate the definite integral, given that 5 0 x= 12.5 5 2 0 x = 125 3 7 2 5 x = 118 3 3. 5 0 3x 4. 5 2 0 (2 x+ x ) 6-23
5. 7 2 0 2x 6. 0 5 7x 7. 7 2 7 ( x + 2x 15) 6-24
Name Date Class Section 6-5 The Fundamental Theorem of Calculus Goal: To use the fundamental theorem of calculus to solve problems. Theorem 1: Fundamental Theorem of Calculus If f is a continuous function on [a, b], and F is any antiderivative of f, then b a f ( x ) = F ( b ) F ( a ) Definition: Average Value of a Continuous Function f over [a, b] 1 b ( ) b a a f x In Problems 1 7, evaluate the integrals. 1. 5 2 1 x (6 + 2) 6-25
2. 4 x 0 (7 e ) 3. 3 3 2 3 ( x + 2x 8x+ 7) 4. 22 13 5 x + 3 6-26
5. 2 2 3 6 1 2 (2 + 9) x x 6. 2 2 x 1 3 2 3 2x+ 5 x x + 5x 3 6-27
7. 2 2 1 3x 6x + 1 In Problems 8 and 9, find the average value of the function over the given interval. 8. 3 f( x) = 4x 8x+ 2; [2,8] 6-28
9. 0.3x e gx ( ) = 2 [0,10] 10. The total cost (in dollars) of manufacturing x units of a product is Cx ( ) = 30,000 + 250 x. a. Find the average cost per unit if 400 units are produced. [Hint: Recall that Cxis ( ) the average cost per unit.] b. Find the average value of the cost over the interval [0, 400]. 6-29
11. A company manufactures a product and the research department produced the marginal cost function x C'( x ) = 300 0 x 800 4 where '( ) C x is in dollars and x is the number of units produced per month. Compute the increase in cost going from a production level of 400 units per month to 800 units per month. Set up a definite integral and evaluate it. 6-30