Antiderivatives and Indefinite Integrals

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

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.

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.

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

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.

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.

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).

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.

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

Example Find three different antiderivatives of f (x) = 4x 3. F(x) = x 4 G(x) = x 4 + 1 H(x) = x 4 + 5 Remark: other answers are possible.

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.

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.

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

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

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

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

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

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

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 + 1 3 x 3 + C

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

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

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

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. 15 10 5 3 2 1 1 2 3 x

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.

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) = 1 2 + 1 + C = C = 5 and the particular solution is F(x) = x 2 + x + 5.

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.

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 3 1 2 x 2 x + C 2. Find the particular solution that satisfies F(1) = 3.

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 2. 1. Find the position of the ball as a function of time. 2. Find the time the ball hits the ground. 100 80 60 40 20

Solution Position: s (t) = 32 s (t) = 32t + C s (0) = 32(0) + C = 80 s (t) = 32t + 80 (velocity) s(t) = 16t 2 + 80t + C s(0) = 16(0) 2 + 80(0) + C = 6 s(t) = 16t 2 + 80t + 6 (position) Impact: s(t) = 0 16t 2 + 80t + 6 = 0 t 5.07

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.

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 2 + 12x + C C(0) = 1 80 (0)2 + 12(0) + C = 5000 C(x) = 1 80 x 2 + 12x + 5000