Copyright © Cengage Learning. All rights reserved. 13 The Integral

Copyright © Cengage Learning. All rights reserved The Indefinite Integral

33 Antiderivative An antiderivative of a function f is a function F such that F = f. Quick Example An antiderivative of 4x 3 is x 4. Thus, If the derivative of A(x) is B(x), then an antiderivative of B(x) is A(x). We call the set of all antiderivatives of a function the indefinite integral of the function. Because the derivative of x 4 is 4x 3

44 The Indefinite Integral Indefinite Integral  f (x)dx is read “the indefinite integral of f (x) with respect to x” and stands for the set of all antiderivatives of f. Thus,  f (x)dx is a collection of functions; it is not a single function or a number. The function f that is being integrated is called the integrand, and the variable x is called the variable of integration.

55 The Indefinite Integral Quick Example  4x 3 dx = x 4 + C The constant of integration C reminds us that we can add any constant and get a different antiderivative. Every possible antiderivative of 4x 3 has the form x 4 + C.

66 Example 1 – Indefinite Integral Check that Solution: We check our answer by taking the derivative of the right-hand side: Because the derivative of the right-hand side is the integrand x, we can conclude that, as claimed.

77 The Indefinite Integral Power Rule for the Indefinite Integral, Part I In Words To find the integral of x n, add 1 to the exponent, and then divide by the new exponent. This rule works provided that n is not –1. Quick Example (if n ≠ –1)

88 The Indefinite Integral Power Rule for the Indefinite Integral, Part II Two other indefinite integrals that come from formulas for differentiation are the following: Indefinite Integral of e x and b x

99 The Indefinite Integral If b is any positive number other than 1, then Quick Example

10 The Indefinite Integral Sums, Differences, and Constant Multiples Sum and Difference Rules  [f(x)  g(x)]dx =  f(x)dx   g(x)dx In Words: The integral of a sum is the sum of the integrals, and the integral of a difference is the difference of the integrals.

11 The Indefinite Integral Constant Multiple Rule  kf(x)dx = k  f(x)dx(k constant) In Words: The integral of a constant times a function is the constant times the integral of the function.

12 The Indefinite Integral Quick Example Sum Rule: Constant Multiple Rule: f(x) = x 3 ; g(x) = 1 k = 5; f(x) = x 3

13 Applications

14 Example 5 – Finding Cost from Marginal Cost The marginal cost to produce baseball caps at a production level of x caps is 4 – 0.001x dollars per cap, and the cost of producing 100 caps is $500. Find the cost function. Solution: We are asked to find the cost function C(x), given that the marginal cost function is 4 – 0.001x. Recalling that the marginal cost function is the derivative of the cost function, we can write C (x) = 4 – 0.001x and must find C(x).

15 Example 5 – Solution Now C(x) must be an antiderivative of C (x), so C(x) =  (4 – 0.001x)dx Now, unless we have a value for K, we don’t really know what the cost function is. However, there is another piece of information we have ignored: The cost of producing 100 baseball caps is $500. In symbols C(100) = 500 cont’d K is the constant of integration.

16 Example 5 – Solution Substituting in our formula for C(x), we have C(100) = 4(100) – (100) 2 + K 500 = K K = 105. Now that we know what K is, we can write down the cost function: C(x) = 4x – x cont’d

17 Motion in a Straight Line

18 Motion in a Straight Line An important application of the indefinite integral is to the study of motion. Position, Velocity, and Acceleration: Derivative Form If s = s(t) is the position of an object at time t, then its velocity is given by the derivative In Words: Velocity is the derivative of position.

19 Motion in a Straight Line The acceleration of an object is given by the derivative In Words: Acceleration is the derivative of velocity.

20 Motion in a Straight Line Position, Velocity, and Acceleration: Integral Form Quick Example If the velocity of a particle moving in a straight line is given by v(t) = 4t + 1, then its position after t seconds is given by s(t) =  v(t)dt =  (4t + 1)dt = 2t 2 + t + C.

21 Example 7 – Motion in a Straight Line a.The velocity of a particle moving along in a straight line is given by v(t) = 4t + 1 m/s. Given that the particle is at position s = 2 meters at time t = 1, find an expression for s in terms of t. b.For a freely falling body experiencing no air resistance and zero initial velocity, find an expression for the velocity v in terms of t. [Note: On Earth, a freely falling body experiencing no air resistance accelerates downward at approximately 9.8 meters per second per second, or 9.8 m/s 2 (or 32 ft/s 2 ).]

22 Example 7(a) – Solution As we saw in the Quick Example, the position of the particle after t seconds is given by s(t) =  v(t)dt =  (4t + 1)dt = 2t 2 + t + C. But what is the value of C? Now, we are told that the particle is at position s = 2 at time t = 1.

23 Example 7(a) – Solution In other words, s(1) = 2. Substituting this into the expression for s(t) gives 2 = 2(1) C so C = –1. Hence the position after t seconds is given by s(t) = 2t 2 + t – 1 meters. cont’d

24 Example 7(b) – Solution Let’s measure heights above the ground as positive, so that a rising object has positive velocity and the acceleration due to gravity is negative. (It causes the upward velocity to decrease in value.) Thus, the acceleration of the stone is given by a(t) = –9.8 m/s 2. We wish to know the velocity, which is an antiderivative of acceleration, so we compute v(t) =  a(t)dt =  (–9.8)dt = –9.8t + C. cont’d

25 Example 7(b) – Solution To find the value of C, we use the given information that at time t = 0 the velocity is 0: v(0) = 0. Substituting this into the expression for v(t) gives 0 = –9.8(0) + C so C = 0. Hence, the velocity after t seconds is given by v(t) = –9.8t m/s. cont’d

26 Motion in a Straight Line Vertical Motion Under Gravity: Velocity and Position If one ignores air resistance, the vertical velocity and position of an object moving under gravity are given by British Units Metric Units Velocity: v(t) = –32t + v 0 ft/s v(t) = –9.8t + v 0 m/s Position: s(t) = –16t 2 + v 0 t + s 0 ft s(t) = –4.9t 2 + v 0 t + s 0 m v 0 = initial velocity = velocity at time 0 s 0 = initial position = position at time 0

27 Motion in a Straight Line Quick Example If a ball is thrown down at 2 ft/s from a height of 200 ft, then its velocity and position after t seconds are v(t) = –32t – 2 ft/s and s(t) = –16t 2 – 2t ft.