Antiderivatives and Indefinite Integrals

Transcription

1 Antiderivatives and Indefinite Integrals MATH 151 Calculus for Management J. Robert Buchanan Department of Mathematics Fall 2018

2 Objectives After completing this lesson we will be able to use the definition of the antiderivative, and use the rules of integration to find integrals.

3 Motivation Many times a mathematical description of a situation specifies the rates of change of quantities. Example An object initially at altitude y 0 and traveling at velocity v 0 is allowed to move under the influence of gravitational acceleration g. Find the position of the object at any time t.

4 Motivation Many times a mathematical description of a situation specifies the rates of change of quantities. Example An object initially at altitude y 0 and traveling at velocity v 0 is allowed to move under the influence of gravitational acceleration g. Find the position of the object at any time t. y (t) = g

5 Motivation Many times a mathematical description of a situation specifies the rates of change of quantities. Example An object initially at altitude y 0 and traveling at velocity v 0 is allowed to move under the influence of gravitational acceleration g. Find the position of the object at any time t. y (t) = g y (t) = gt + v 0

6 Motivation Many times a mathematical description of a situation specifies the rates of change of quantities. Example An object initially at altitude y 0 and traveling at velocity v 0 is allowed to move under the influence of gravitational acceleration g. Find the position of the object at any time t. y (t) = g y (t) = gt + v 0 y(t) = 1 2 gt2 + v 0 t + y 0

7 Analysis y (t) = g y (t) = gt + v 0 y(t) = 1 2 gt2 + v 0 t + y 0 Note: y(0) = y 0.

8 Analysis y (t) = g y (t) = gt + v 0 y(t) = 1 2 gt2 + v 0 t + y 0 Note: y(0) = y 0. y (t) = d [ 1 ] dt 2 gt2 + v 0 t + y 0 = gt + v0 and y (0) = v 0.

9 Analysis y (t) = g y (t) = gt + v 0 y(t) = 1 2 gt2 + v 0 t + y 0 Note: y(0) = y 0. y (t) = d [ 1 ] dt 2 gt2 + v 0 t + y 0 = gt + v0 and y (0) = v 0. y (t) = d dt [gt + v 0] = g.

10 Antiderivatives Definition A function F(x) is an antiderivative of f (x) if for every x in the domain of f we have F (x) = f (x).

11 Antiderivatives Definition A function F(x) is an antiderivative of f (x) if for every x in the domain of f we have F (x) = f (x). Antidifferentiation can be thought of as the reverse process of differentiation. Antidifferentiation is the second major theme of calculus. Antidifferentiation can be thought of as a process of accumulation or summing.

12 Example Find three different antiderivatives of f (x) = 4x 3.

13 Example Find three different antiderivatives of f (x) = 4x 3. F(x) = x 4 G(x) = x H(x) = x Remark: other answers are possible.

14 Constants of Integration Theorem Suppose that F and G are both antiderivatives of f on interval I. Then F (x) = G(x) + C for some constant C. C is called a constant of integration.

15 Indefinite Integrals Definition Let F be any antiderivative of f. The indefinite integral of f (x) with respect to x is defined by f (x) dx = F(x) + C where C is called the constant of integration. The function f (x) is called the integrand and dx is the differential.

16 Anatomy of an Indefinite Integral f (x) dx = F(x) + C integral sign f (x) integrand dx differential F(x) antiderivative C constant of integration f (x) dx indefinite integral

17 Formulas for Integration k dx = kx + C x r 1 dx = r + 1 x r+1 + C 1 (if r 1) (Power Rule) dx = ln x + C x e x dx = e x + C k f (x) dx = k f (x) dx (Constant Multiple Rule) [f (x) ± g(x)] dx = f (x) dx ± g(x) dx (Sum/Difference Ru

18 Examples Find the indefinite integrals of the following functions. f (x) = 3 g(x) = 2x h(x) = 9x 2 k(x) = 1 + x 2

19 Examples Find the indefinite integrals of the following functions. f (x) = 3 F(x) = 3x + C g(x) = 2x h(x) = 9x 2 k(x) = 1 + x 2

20 Examples Find the indefinite integrals of the following functions. f (x) = 3 F(x) = 3x + C g(x) = 2x G(x) = x 2 + C h(x) = 9x 2 k(x) = 1 + x 2

21 Examples Find the indefinite integrals of the following functions. f (x) = 3 F(x) = 3x + C g(x) = 2x G(x) = x 2 + C h(x) = 9x 2 H(x) = 3x 3 + C k(x) = 1 + x 2

22 Examples Find the indefinite integrals of the following functions. f (x) = 3 F(x) = 3x + C g(x) = 2x G(x) = x 2 + C h(x) = 9x 2 H(x) = 3x 3 + C k(x) = 1 + x 2 K (x) = x x 3 + C

23 Relationship Between Differentiation and Antidifferentiation Differentiation and antidifferentiation are inverse operations of each other.

24 Relationship Between Differentiation and Antidifferentiation Differentiation and antidifferentiation are inverse operations of each other. d f (x) dx = f (x) dx

25 Relationship Between Differentiation and Antidifferentiation Differentiation and antidifferentiation are inverse operations of each other. d f (x) dx = f (x) dx f (x) dx = f (x) + C

26 Particular Solutions (1 of 2) An equation such as y = (2x + 1) dx = x 2 + x + C has many solutions which differ from one another by a constant x

27 Particular Solutions (2 of 2) We will call F(x) = x 2 + x + C the general solution to the differential equation dy dx = 2x + 1.

28 Particular Solutions (2 of 2) We will call F(x) = x 2 + x + C the general solution to the differential equation dy dx = 2x + 1. If we are given the coordinates of a point the solution must pass through, then we can find a particular solution. Suppose F(1) = 7 (so that the graph of the solution must pass through point (1, 7)), then 7 = F(1) = C = C = 5 and the particular solution is F(x) = x 2 + x + 5.

29 Example Consider the differential equation dy dx = 3 2 x 2 x 1 1. Find the general solution F(x). 2. Find the particular solution that satisfies F(1) = 3.

30 Example Consider the differential equation dy dx = 3 2 x 2 x 1 1. Find the general solution F(x). F(x) = [ ] 3 2 x 2 x 1 dx = 1 2 x x 2 x + C 2. Find the particular solution that satisfies F(1) = 3.

31 Example Consider the differential equation dy dx = 3 2 x 2 x 1 1. Find the general solution F(x). F(x) = [ ] 3 2 x 2 x 1 dx = 1 2 x x 2 x + C 2. Find the particular solution that satisfies F(1) = 3. F(1) = 1 2 (1)3 1 2 (1)2 1 + C 3 = 1 + C C = 4 F(x) = 1 2 x x 2 x + 4

32 Position, Velocity, and Acceleration Suppose a ball is thrown straight up from shoulder height of 6 feet at speed of 80 feet per second. Gravitational acceleration on the ball is 32 ft/s Find the position of the ball as a function of time. 2. Find the time the ball hits the ground

33 Solution Position: s (t) = 32 s (t) = 32t + C s (0) = 32(0) + C = 80 s (t) = 32t + 80 (velocity) s(t) = 16t t + C s(0) = 16(0) (0) + C = 6 s(t) = 16t t + 6 (position) Impact: s(t) = 0 16t t + 6 = 0 t 5.07

34 Finding a Cost Function Suppose the marginal cost of producing x units is C (x) = 1 40 x + 12 and the fixed cost (cost to produce zero units) is $5,000. Find the cost function.

35 Finding a Cost Function Suppose the marginal cost of producing x units is C (x) = 1 40 x + 12 and the fixed cost (cost to produce zero units) is $5,000. Find the cost function. C(x) = [ ] 1 40 x + 12 dx C(x) = 1 80 x x + C C(0) = 1 80 (0)2 + 12(0) + C = 5000 C(x) = 1 80 x x