Please read for extra test points: Thanks for reviewing the notes you are indeed a true scholar!

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1 Please read for extra test points: Thanks for reviewing the notes you are indeed a true scholar! See me any time B4 school tomorrow and mention to me that you have reviewed your integration notes and you will get extra points on the last test. Don't tell the other students about this as the more students who do this the less number of points you will earn. The most you will receive is 5 points, the least is 2 points. 1

2 Chapter 6 Integration There are two branches of calcoolus: 1. Differentiation The derivative is a measurement of how a function changes when the values of its input change. Applications of differential calcoolus include computations involving position, velocity, and acceleration, the slope of a curve, related rates, and optimization. 2. Integration The integral is an extension of the concept of a sum. The process of finding integrals is called integration. Used to find a measure of totality such as area, volume, mass, displacement, etc., when its rate of change with respect to some other quantity (position, time, etc.) is specified. Applications of integral calcoolus include finding areas bounded by curves, volumes of solids, lengths of curves, centers of gravity, etc. The area of a parabolic region can be approximated as the sum of the areas of rectangles. As you increase the number of rectangles, the approximation tends to become more accurate (this is called the Riemann method). Later, we will learn how the limit process can be used to find areas of a wide variety of regions. This process is called integration and is closely related to differentiation. There are two types of Integrals: 1. Indefinite (the antiderivative) 2. Definite * The Fundamental Theorem of Calculus connects the two types even though they are conceptually distinct. 2

3 6.1 Antiderivatives and Indefinite Integration Antiderivatives Find a function F whose derivative is f(x) = 3x 2. F(x) = x 3 b/c d/dx (x 3 ) = 3x 2 * the function F is an antiderivative of f Definition of an Antiderivative A function F is an antiderivative of f on an open interval I if F '(x) = f(x) for all x in I. * note that F is called an antiderivative of f, rather than the antiderivative of f F 1 (x) = x 3 F 2 (x) = x 3 4 F 3 (x) = x * all three functions are antiderivatives of f(x) = 3x 2 * for any constant C, F(x) = x 3 + C is an antiderivative of f Differential Calculus: Given f, find f ' Integral Calculus: Given f ', find f ex. G '(x) = 2x > differential equation (diffy q) G(x) = x 2 + C > general solution of the diffy q constant of integration 3

4 ex. Find the general solution of the diffy q y' = 2. y = 2x is an antiderivative y = 2x + C is the general solution sketch below: y = 2x + C for different C values: 5 y c=5 3 c=3 c=0 0 c= 4 x 4 4

5 Notation for Antiderivatives the operation of finding all solutions of a diffy q is called antidifferentiation or idefinite integration and is denoted by an integral sign the general solution is denoted by y = f(x) dx = F(x) + C integrand variable of constant of integration integration f(x) dx is read "the antiderivative of f with respect to x" * the differential dx identifies x as the variable of integration 5

6 Basic Integration Rules original integral > rewrite integrand > integrate > simplify 6

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8 Rewriting the Integrand: 8

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12 Initial Conditions and Particular Solutions y = f(x) dx has many solutions (each differing from the others by a constant) * the graphs are vertical translations of each other usually, enough information is given to determine a particular solution to do this, you only need to know the value of y = F(x) for one value of x this information is called an initial condition 12

13 ex. y = (3x 2 1) dx F(x) = x 3 x + C general solution say you know that F(2) = 4 F(2) = (2) 3 (2) + C 4 = C 2 = C therefore, F(x) = x 3 x 2 13

14 ex. Find the general solution of and find the particular solution that satisfies 14

15 ex. A ball is thrown up with an initial velocity of 64 feet per second from an initial height of 80 feet. a) Find the position function giving the height, s, as a function of the time, t. b) When does the ball hit the ground? Let: t = 0 represent the initial time s(0) = 80 ft. initial height s'(0) = 64 ft/sec initial velocity s''(t) = 32 ft/sec 2 accel. due to gravity This example shows how to use calcoolus to analyze vertical motion problems in which the acceleration is determined by a gravitational force. A similar strategy may be used to analyze other linear motion problems (vertical or horizontal) in which the acceleration (or deceleration) is the result of some other force. 15

16 Ex. (#64) With what initial velocity must an object be thrown upward (from a height of 2 meters) to reach a maximum height of 200 meters? 16

17 Ex. (#67) Consider a particle moving along the x axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''(t) is its acceleration. It's given that x(t) = t 3 6t 2 + 9t 2, 0 < t < 5. a) Find the velocity and acceleration of the particle. b) Find the open t intervals on which the particle is moving to the right (when the velocity is positive). c) Find the velocity of the particle when the acceleration is 0. a) Find the velocity and acceleration of the particle. b) Find the open t intervals on which the particle is moving to the right. c) Find the velocity of the particle when the acceleration is 0. 17

18 HW pg. 396 #60, 62 (use the equation in #61), 68, 70 18

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