introductory calculus understanding the integral pdf

Introductory Calculus Understanding the Integral Tunc Geveci www.EngineeringBooksPDF.com Introductory Calculus Understanding the Integral Tunc Geveci www.EngineeringBooksPDF.com Introductory Calculus: Understanding the Integral Copyright © Cognella Academic Publishing 2015 www.cognella.com All rights reserved No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means—electronic, mechanical, photocopy, recording, or any other except for brief quotations, not to exceed 250 words, without the prior permission of the publisher ISBN-13: 978-1-60650-856-5 (e-book) A Momentum Press publication www.momentumpress.net Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe Cover and interior design by S4Carlisle Publishing Services Private Ltd., Chennai, India www.EngineeringBooksPDF.com Contents Chapter Using the Integral in the Approximation of Area The Summation Notation The Area under the Graph of a Function Chapter Understanding the Concept of the Integral 15 The Riemann Integral and Signed Area 15 The Integrals of Piecewise Continuous Functions 27 The Precise Definition of the Integral 32 Chapter Introduction to the Fundamental Theorem of Calculus 35 The Fundamental Theorem of Calculus (Part 1) 35 The Proof of Theorem 35 The Indefinite Integrals of Basic Functions 44 A Short List of Antiderivatives 44 The Fundamental Theorem of Calculus and One-Dimensional Motion 52 Chapter The Antiderivative and the Fundamental Theorem of Calculus 57 Some Properties of the Integral 57 The Second Part of the Fundamental Theorem 65 Chapter The Indefinite and Definite Integrals of Linear Combinations of Functions 81 The Linearity of Indefinite and Definite Integrals 81 The Area of a Region between the Graphs of Functions 90 Chapter Using the Substitution Rule for Integrals 97 The Substitution Rule for Indefinite Integrals 97 The Substitution Rule for Definite Integrals 106 www.EngineeringBooksPDF.com iv CONTENTS Chapter The Fundamental Theorem of Calculus and the Differential Equation yƍ = fƍ 113 The Differential Equation y' = f and the Fundamental Theorem 114 Acceleration, Velocity and Position 119 Index 123 www.EngineeringBooksPDF.com CHAPTER Using the Integral in the Approximation of Area In this chapter, we will discuss the approximation of the area of a region between the graph of a positive-valued function and an interval The Summation Notation Let’s begin by introducing notation that will turn out to be convenient in expressing sums Given numbers a1, a2, , an, we can indicate the sum of the numbers as a1 + a2 + " + an We can also indicate the sum using the summation notation: n ¦a k k =1 (read “sigma ak as k runs from to n” ) The subscript k is the summation index, and is a “dummy index”, in the sense that it can be replaced by any convenient letter Thus, both n n j =1 l =1 ¦ a j and ¦ al denote the sum a1 + a2 + · · · + an Example The sum of the first n positive integers can be expressed succinctly: n ¦k =1+ + +" + n = k =1 n ( n + 1) www.EngineeringBooksPDF.com INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL Indeed, if we set Sn = ¦ k =1 k , we have n Sn = + + " + ( n − 1) + n Let’s add the terms in the opposite order: Sn = n + ( n − 1) + " + + Therefore, 2Sn = ( n + 1) + ( ( n − 1) + ) + " + ( + ( n − 1) ) + (1 + n ) = ( n + 1) + ( n + 1) + ! + ( n + 1) + ( n + 1) where the sum has n terms Thus, 2Sn = n ( n + 1) , so that Sn = n ( n + 1) Ƒ The following rules are natural and easy to confirm: n ¦(a k =1 k n n k =1 k =1 + bk ) = ¦ ak + ¦ bk , and n n k =1 k =1 ¦ cak = c ¦ ak (the constant rule for sums) Proof By the associativity of addition, n = ¦ ( ak + bk ) = ( a1 + b1 ) + " + ( an + bn ) k =1 = ( a1+ a2 + " + an ) + (b1 + b2 + " + bn ) n n k =1 k =1 = ¦ ak + ¦ bk www.EngineeringBooksPDF.com USING THE INTEGRAL IN THE APPROXIMATION OF AREA By the distributivity of multiplication with respect to sums, n n k =1 k =1 ¦ cak = ca1 + ca2 + " + can = c ( a1 + a2 + " + an ) = c ¦ ak Ƒ The Area under the Graph of a Function Assume that f is continuous on the interval [a, b] and f (x) ≥ for each x ∈ [ a , b ] Let G be the region in the xy-plane that is bounded by the graph of f, the interval [a, b] on the x-axis, the line x = a and the line x = b We will refer to G simply as the region between the graph of f and the interval [a, b] Our intuitive notion of the area of G is a measure of the size of G Even though we may not be able to compute the area of G exactly, we should be able to compute approximations We will devise a strategy that will be based on the approximation of G, in a geometric sense, by unions of rectangles y f G a b x Figure 1: The region between the graph of f and the interval [a, b] Definition The set of points P = {x0, x1, , xk−1, xk, , xn} is a partition of the interval [a, b] if a = x0 < x1 < x2 < " < x k −1 < x k < " < xn = b The interval [xk−1, xk] is the kth subinterval that is determined by the partition P We will denote the length of the kth subinterval by ǻxk, so that ǻxk = xk − xk−1 The maximum of the lengths of the subintervals determined by P is the norm of the partition P We will denote the norm of P by ŒPŒ, so that ŒPŒ is the maximum of ǻx1, ǻx2, , ǻxn we can abbreviate the expression “maximum of ǻx1, ǻx2, , ǻxn” as maxk=1,…,n ǻxk or maxk ǻxk Thus, www.EngineeringBooksPDF.com INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL P = max Δ xk k =1,!,n Let’s sample an arbitrary value of x in the kth subinterval [xk−1, xk] and denote it by x k* Thus, x k −1 ≤ x k* ≤ x k , but there is no other restriction on the choice of x k* Consider the rectangle that has as its base the interval [xk−1, xk] and has height equal to the value of f at x k* If ǻxk is small, it is reasonable to approximate the area of the slice of G between the lines x = xk−1 and x = xk by the area of such a rectangle y f xk xk xk xk x Figure 2: An approximating rectangle The area of the rectangle is f ( xk* ) ( xk − xk −1 ) = f ( xk* ) Δxk The sum of the areas of such rectangles should be a reasonable approximation to the area of G if the maximum of the lengths of the subintervals, i.e., ŒPŒ is small: n ¦ f ( x ) Δx * k k ≅ Area of G k =1 We would expect the approximation to be as accurate as desired if ŒPŒ = maxk ǻxk is sufficiently small a b Figure 3: Approximating rectangles www.EngineeringBooksPDF.com USING THE INTEGRAL IN THE APPROXIMATION OF AREA Example Let f (x) = x2 + 1, and let G be the region between the graph of f and the interval [0, 2] Figure shows G y G 1 x Figure Let P = {0, 0.5, 1, 1.2, 1.4, 1.6, 1.8, 2} , so that P is a partition of the interval [0, 2] With reference to the notation of Definition 1, we have x0 = 0, x1 = 0.5, x2 = 1, x3 = 1.2, x4 = 1.4, x5 = 1.6, x6 = 1.8 and x7 = The lengths of the subintervals determined by the partition P are Δ x1 = Δ x2 = 0.5 and Δ x3 = Δ x4 = " = Δ x7 = 0.2 Therefore, the norm of P is 0.5: P = 0.5 Let’s form the rectangle of height f (ck) on the kth subinterval [xk−1, xk], where ck is the midpoint of [xk−1, xk], k = 1, 2, , 7, and approximate the area of the region G by the sum of these rectangles Figure indicates the rectangles y 2 1 Figure www.EngineeringBooksPDF.com x USING THE SUBSTITUTION RULE FOR INTEGRALS 111 If f is odd, the graph of f is symmetric with respect to the origin With reference to Figure 6, the signed area of G− is (−1) × the area of G+ G a a x G Figure Thus, ³ a −a a −a f ( x ) dx = ³ f ( x ) dx + ³ f ( x ) dx = ( the signed area of G− ) + ( the area of G+ ) = The Proof of Proposition We will prove part a), and leave the similar proof of part b) as an exercise Thus, assume that f is even By the additivity of integrals with respect to intervals, ³ a −a a −a f ( x ) dx = ³ f ( x ) dx + ³ f ( x ) dx Since f is even, we have f (−x) = f (x) Therefore, ³ 0 f ( x ) dx = ³ f ( − x ) dx −a −a Let us apply the substitution rule to this integral by setting u = −x Then, du/dx = −1, so that ³ −a 0 −a −a f ( − x ) dx = − ³ f ( u )( −1) dx = − ³ f ( u ) = −³ u ( 0) u( − a ) du dx dx a a f ( u ) du = − ³ f ( u ) du = ³ f ( u ) du Thus, ³ −a a a 0 f ( − x ) dx = ³ f ( u ) du = ³ f ( x ) dx www.EngineeringBooksPDF.com 112 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL (the variable of integration is a dummy variable) Therefore, ³ a −a a a a −a 0 f ( x ) dx = ³ f ( x ) dx + ³ f ( x ) dx = ³ f ( x ) dx + ³ f ( x ) dx a = ³ f ( x )dx , as claimed Ŷ www.EngineeringBooksPDF.com CHAPTER The Fundamental Theorem of Calculus and the Differential Equation yƍ = fƍ In this chapter, we will take another look at the Fundamental Theorem of Calculus within the framework of differential equations and initialvalue problems In Section 4.6 we saw that the general solution of the differential equation yƍ (x) = ky(x), where k is a constant, is a constant multiple of ekx Thus, we were able to determine the unique solution of the initialvalue problem y ′ ( x ) = ky ( x ) , y ( x0 ) = y0 , where x0 and y0 are given numbers, as y ( x ) = y e k ( x − x0 ) Now we will consider differential equations of the form y′( x ) = f ( x ), where f is a given function, and initial-value problems of the form y ′ ( x ) = f ( x ) , y ( x ) = y0 , where x0 and y0 are given numbers www.EngineeringBooksPDF.com 114 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL The Differential Equation yƍ = f and the Fundamental Theorem We have yƍ (x) = f (x) for each x in an interval J if and only if y is an antiderivative of f on J Therefore, we can express y as the indefinite integral of f: y ( x ) = ³ f ( x ) dx We will refer to ³ f ( x ) dx as the general solution of the differential equation yƍ (x) = f (x) The indefinite integral involves an arbitrary constant The value of the constant is determined uniquely if an initial condition of the form y(x0) = y0 is specified, so that the solution of the initial-value problem, y ′ ( x ) = f ( x ) , y ( x ) = y0 is uniquely determined Example a) Determine the general solution of the differential equation y ′ ( x ) = 2x b) Determine the solution of the initial-value problem y ′ ( x ) = x and y ( ) = Solution a) We have yƍ (x) = 2x if and only if y ( x ) = ³ xdx = x + C , where C is an arbitrary constant Thus, y (x ) = x2 + C www.EngineeringBooksPDF.com THE FUNDAMENTAL THEOREM OF CALCULUS 115 is the general solution of the differential equation yƍ (x) = 2x Since C is an arbitrary constant, the general solution represents infinitely many functions that differ from x2 by the addition of a constant Figure displays the members of this family of functions corresponding to C = −4, 1, If (x, y) is on one of the solution curves, the slope of the line that is tangent to that particular solution curve at (x, y) is 2x Thus, the tangent lines to the solution curves corresponding to a given x are parallel to each other y 2,5 y x2 2 x Figure b) Since y(x) = x2 + C is the general solution of the given differential equation, we have y ( ) = ⇔ 22 + C = ⇔ C = Therefore, the required solution is y ( x ) = x + The graph of y = x2 + is the only member of the family of curves y = x2 + C that passes through the point (2, 5) … Example a) Determine the general solution of the differential equation y ′ ( x ) = sin ( x ) b) Determine the solution of the initial-value problems, y ′ ( x ) = sin ( x ) , y (π ) = 2, www.EngineeringBooksPDF.com 116 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL and y ′ ( x ) = sin ( x ) , y (π ) = −2 Sketch the graphs of the solutions Solution a) We have yƍ (x) = sin(4x) if and only if y ( x ) = ³ sin ( x ) dx = − cos ( x ) + C , where C is an arbitrary constant This is the general solution of the differential equation y ' ( x ) = sin ( x ) b) With reference to part a), 1 §π à y ă = cos ( ) + C = + C 4 â4ạ Therefore, y (π ) = ⇔ +C = ⇔ C = 4 Thus, the solution of the initial-value problem y ′ ( x ) = sin ( x ) , y (π ) = is F ( x ) = − cos ( x ) + 4 Similarly, the solution of the initial-value problem y ′ ( x ) = sin ( x ) , y (π ) = −2 is G ( x ) = − cos ( x ) − 4 www.EngineeringBooksPDF.com THE FUNDAMENTAL THEOREM OF CALCULUS 117 Figure displays the graphs of F and G Note that F ′ ( x ) = G ′ ( x ) = sin ( x ) , so that the tangent line to the graph of F at the point (x, F (x)) is parallel to the tangent line to the graph of G at (x, G(x)) … y π 4, 2 F π π π π 4 π 4, 2 x G Figure In the above examples, we were able to determine the relevant indefinite integral in terms of familiar functions This need not be the case Nevertheless, any continuous function has an antiderivative by the second part of the Fundamental Theorem of Calculus: We have d x f ( t ) dt = f ( x ) dx ³a for each x ∈ J if f is continuous on the interval J and a is a fixed point in J Therefore, we can express the general solution of the differential equation yƍ = f on the interval J as x y ( x ) = ³ f ( t ) dt + C , a where a is some point in J and C is a constant If we are given an initial condition of the form y (x0) = y0, it is convenient to set a = x0 In this case, x y ( x ) = C + ³ f ( t ) dt , x0 so that x0 y0 = y ( x0 ) = C + ³ f ( t ) dt = C x0 www.EngineeringBooksPDF.com 118 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL Therefore C = y0, and the unique solution of the initial-value problem y ' ( x ) = f ( x ) , y ( x ) = y0 can be represented as x y ( x ) = y0 + ³ f ( t ) dt x0 In the above expression, we may or may not be able to express the integral in terms of familiar functions In any case, the values of the solution can be approximated by approximating the integral Example a) Express the solution of the initial-value problem, y ′ ( x ) = sin ( x ) , y ( ) = 3, in terms of an integral b) Compute approximations to y (3) and y (4) with the help of the approximate integration facility of your computational utility c) Plot the graph of the solution of part a) on the interval [0, 4] with the help of your computational/graphing utility Solution a) We can express the solution as y ( x ) = + ³ sin ( t ) dt x b) We have y ( ) = + ³ sin ( t ) dt ≅ 2.968 79, and y ( ) = + ³ sin ( t ) dt ≅ 2.942 36 www.EngineeringBooksPDF.com THE FUNDAMENTAL THEOREM OF CALCULUS 119 c) Figure shows the graph of the solution on [0, 4] … y 2,3 1 x Figure Acceleration, Velocity and Position Let’s consider the relationships between acceleration, velocity and position within the framework of initial-value problems Assume that f (t) is the position, v(t) is the velocity and a(t) is the acceleration at time t of an object in one-dimensional motion Velocity is the rate of change of position, and acceleration is the rate of change of velocity: υ (t ) = df dυ and a ( t ) = dt dt When we introduced these concepts initially, we assumed that the position was given, and calculated velocity and acceleration by differentiation Now we are able to begin with a given acceleration function, and determine the velocity and position functions successively Thus assume that a(t) is given The velocity function v(t) is the solution of the differential equation dυ = a (t ) dt We have seen that such a differential equation does not have a unique solution On the other hand, if an initial condition is specified, the solution is uniquely determined Thus, assume that the velocity at a certain instant t0 is v0, so that v (t0) = v0 We can express the solution of the initial-value problem dυ = a ( t ) ,υ ( t ) = υ0 , dt www.EngineeringBooksPDF.com 120 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL as t υ ( t ) = υ0 + ³ a (τ ) dτ t0 The position function is uniquely determined if the position of the object is specified at some instant If f(t0) = f0, the position function is the solution of an initial-value problem df = υ (t ) , f (t0 ) = f dt The solution can be expressed as t f ( t ) = f + ³ υ (τ ) dτ t0 Example Assume that an object is falling under the influence of gravitational acceleration of 9.8 meters/second/second The effect of air resistance is neglected We model the motion as one-dimensional motion so that the number line is vertical, points downward, and the origin coincides with the point at which the object is released We assume that the object is released from rest Thus, with the above notation, a(t) = 9.8, v(0) = and f (0) = Determine v(t) and f (t) at any instant t before the object hits the ground Solution We have dυ = a ( t ) = 9.8, υ ( ) = dt Therefore, t t 0 υ ( t ) = ³ a (τ ) dτ = ³ 9.8dτ = 9.8t ( meters/sec.) We have df = υ ( t ) = 9.8t and f ( ) = dt www.EngineeringBooksPDF.com THE FUNDAMENTAL THEOREM OF CALCULUS 121 Therefore, t t 0 f ( t ) = ³ υ (τ ) d τ = ³ 9.8τ d τ = t 9.8 τ = 4.9t ( meters ) … Example Assume that the acceleration of an object in simple harmonic motion is 20 cos (6t) at the instant t Determine the velocity and the position of the object at the instant t if v(π/6) = and f (π/6) = (with the notation preceding Example 4) Solution We have dυ = a ( t ) = 20 cos ( 6t ) and υ (π ) = dt Therefore, t υ ( t ) = ³ 20cos ( 6τ ) dτ = π t 10 sin ( 6τ ) π 10 10 sin ( 6t ) − sin (π ) 3 10 = sin ( 6t ) = We have df 10 §π · = υ ( t ) = sin ( 6t ) and f ă = dt â6ạ Therefore, f (t ) = + ³ t π 10 sin ( 6τ ) d τ t § 10 · = + ă cos ( ) ă 18 6ạ â 10 Đ 10 à = + ă cos ( 6t ) + cos ( ) 18 â 18 10 Ã Đ 10 = + ă cos ( 6t ) 18 18 â 13 = cos ( 6t ) 9 www.EngineeringBooksPDF.com 122 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL Figure shows the graph of f Note that the motion is periodic with period π/3 … y π 6, 2 1.5 0.5 π π π 2π Figure www.EngineeringBooksPDF.com t Index Acceleration, 119 Additivity of integral with respect to intervals, 23–24, 64 Antiderivatives of function, 40 and fundamental theorem of Calculus, 57–79 short list of, 44–45 Antidifferentiation, 44 Calculus Corollary to Fundamental Theorem of, 39 and differential equation yƍ=fƍ, 113–122 fundamental theorem of, 35–44 indefinite integrals of basic functions, 44–52 and one-dimensional motion, 52–55 Corollary to Fundamental Theorem of Calculus, 39 Definite integral, 41–42 constant multiple rule for, 87 linearity of, 88–95 substitution rule for, 106–112 sum rule for, 88 Differential equation yƍ=fƍ, and fundamental theorem, 14–122 Fundamental theorem parts of, 79 second part of, 65–79 Indefinite integrals of basic functions, 44–52 constant multiple rule, 81 linearity of, 84–87 substitution rule for, 97–106 sum rule for, 81–84 Initial value problem, unique solution of, 113–122 Integral in approximation of area under graph of function, 3–13 summation notation, 1–3 basic functions, indefinite, 44–52 Mean Value Theorem for, 61–65 natural logarithm defined as, 74–75 of piecewise continuous functions, 27–31 precise definition of, 33–33 properties of, 57–60 Riemann integral and signed area, 15–27 substitution rule for definite integrals, 106–112 indefinite integrals, 97–106 Integrand, 42 Left-endpoint sum, 6–7 Error function erf, 73 Error tolerance, 32 Functions area of region between the graphs of, 90–95 indefinite integrals of basic, 44–52 integrals of piecewise continuous, 27–31 Mean Value Theorem, 36 for integrals, 61–65 Midpoint sum, 7, 12 Numerical integration schemes, 27 One-dimensional motion, Calculus and, 52–55 www.EngineeringBooksPDF.com 124 INDEX Piecewise continuous functions, integrals of, 27–31 Plausibility argument, 70–72 Polynomial, 86 Power rule, 37 Reverse power rule, 45 Riemann integral and signed area, 15–27 Riemann integral of f, 16–17, 32 Riemann sum for f, 16 Right-endpoint sum, Sine integral function, 75–77 Substitution rule for definite integrals, 106–112 for indefinite integrals, 97–106 Summation index, Summation notation, 1–3 Triangle inequality for integrals, 58–60 Trigonometric polynomial, 87 Velocity, 119 www.EngineeringBooksPDF.com EBOOKS FOR THE COLLEGE MATH LIBRARY Introductory Calculus Understanding the Integral Tunc Geveci Create your own Customized Content Bundle—the more books you buy, the greater your discount! THE CONTENT • Calculus • Algebra (all levels) THE TERMS • Perpetual access for a one time fee • No subscriptions or access fees • Unlimited concurrent usage • Downloadable PDFs • Free MARC records Professor Geveci has published research papers on the stability and accuracy of approximation schemes for partial differential equations In recent years his emphasis has been on the improvement of the teaching and exposition of calculus He has taught calculus, advanced calculus and complex analysis courses for many years at San Diego State University For further information, a free trial, or to order, contact:  sales@momentumpress.net www.EngineeringBooksPDF.com ... www.EngineeringBooksPDF.com 18 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL In the notation, b ³ f ( x ) dx , a for the integral of f on [a, b], the number a is referred to as the lower limit of the integral, ... n 3 Therefore, the area of the region G between the graph of f and the interval [1, 3] is 26/3 www.EngineeringBooksPDF.com 12 INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL Figure shows the. .. approximate the area of the region G by the sum of these rectangles Figure indicates the rectangles y 2 1 Figure www.EngineeringBooksPDF.com x INTRODUCTORY CALCULUS: UNDERSTANDING THE INTEGRAL The approximation