Foundations of Calculus. November 18, 2014

Foundations of Calculus November 18, 2014

Contents 1 Conic Sections 3 11 A review of the coordinate system 3 12 Conic Sections 4 121 Circle 4 122 Parabola 5 123 Ellipse 5 124 Hyperbola 6 2 Review of Functions and Their Graphs 8 21 Functions 8 22 Operations on Functions 9 23 Types of Functions 10 3 Limits 12 31 Limits 13 32 One Sided Limits 16 33 Limits at Infinity 18 4 Differentiation 20 41 The Tangent Line and the Derivative 20 42 Differentiability and Continuity 23 43 Theorems on Differentiation and Higher-Order Derivatives 24 44 Derivatives of Trigonometric Functions 24 45 Derivative of a Composite Function and the Chain Rule 24 46 Derivative of the Power Function for Rational Exponents and Implicit Differentiation 24 47 Rectilinear Motion and Rates of Change 24 48 Related Rates 24 49 Rolles Theorem and Mean Value Theorem 24 410 Maximum and Minimum Function Values 24 411 Applications Involving an Absolute Extremum on a Closed Interval 24 412 Increasing and Decreasing Functions and the First Derivative Test 24 413 Concavity and Points of Inflection and the Second Derivative Test 24 414 Derivatives of Transcendental Functions 24 1

5 Integration and Its Applications 27 51 Differentials 27 52 Antiderivatives 27 53 Some Techniques of Antidifferentiation 29 54 The Definite Integral and Area 31 55 Mean Value Theorem for Integrals 37 56 Fundamental Theorem of Calculus 39 57 Area between two curves 40 58 Volumes of Solids 44 581 Disk Method 44 582 Washer Method 47 583 Shell Method 49 59 Integrals of Transcendental Functions 52 2

Chapter 1 Conic Sections 11 A review of the coordinate system To identify points in the plane, we set up a coordinate system We start by setting up a pair of perpendicular lines The vertical line is called the y-axis, while the horizontal line is called the x-axis The point of intersection of these two axes is called the origin, and this point is assigned the pair of coordinates (0, 0) If P is any other point on the plane, then P is assigned a pair of coordinates (x, y) that will give its address or location in the plane The first coordinate x gives the directed distance of P from the y-axis, which means that x is positive when P is to the right of the y-axis, and x is negative if P lies to the left of the y-axis Similarly, the y coordinate of P gives its directed distance from the x-axis Figure 11: The quadrants of the cartesian coordinate system Second Quadrant First Quadrant Third Quadrant Fourth Quadrant 3

12 Conic Sections The graphs of some special second degree equations in two variables are called conic sections and are illustrated below 121 Circle A circle is the set of all points P (x, y) in plane equidistant from a fixed point The fixed point is called the center of the circle, and the constant distance is called the radius The standard equation of a circle with center at (h, k) and radius r is given by: (x h) 2 + (y k) 2 = r 2 Example 1 The figure shows the graph of a circle centered at (1,1) and radius 2cm 2 cm 4

122 Parabola The set of all points P (x, y) on the plane equidistant from a fixed point F and a fixed line l forms a continuous curve called a parabola The fixed point is called the focus while the line l is called the directrix of the parabola Example 2 The figure shows the graph of a parabola with vertex at (0, 0) and focus at (1, 1) focus (0,2) The following are the standard forms of a parabola (x h) 2 = 4p(y k) where the axis of symmetry is vertical and the parabola open upward if p > 0 or downward if p < 0 (y k) 2 = 4p(x h) where the axis of symmetry is vertical and the parabola open to the right if p > 0 or to the left if p < 0 In the above standard forms of a parabola, the vertex is at (h, k) and the distance between the vertex and the foci is given by p and the length of the focal chord is 4 p 123 Ellipse If F 1 and F 2 are two fixed points on the plane, the set of all points P (x, y) for which the sum P F 1 + P F 2 of the distances from P to each of the two fixed points is a constant forms a continuous closed curve called an ellipse Each fixed point is called a focus (plural: foci) of the ellipse The following are the standard forms of an ellipse 5

(x h)2 (y k)2 + = 1, a > b where the center is at (h, k) and the orientation a 2 b 2 of the ellipse is horizontal, that is, the major axis is horizontal (y k)2 (x h)2 + = 1, a > b where the center is at (h, k) and the orientation a 2 b 2 of the ellipse is vertical, that is, the major axis is vertical The following are some properties of an ellipse The line passing through the two foci of an ellipse is called the principal axis of the ellipse The points of intersection of the ellipse and the principal axis are called the vertices (singular: vertex) of the ellipse The line segment of the principal axis whose endpoints are the vertices is called the major axis of the ellipse The length of the major axis is denoted by 2a The midpoint of the major axis is called the center of the ellipse, usually denoted by C(h, k) The distance from the center to each of the two foci is denoted by c, while the distance from the center to each vertex is a The line segment passing through the center, perpendicular to the major axis and whose endpoints are on the ellipse is called the minor axis, and its length is denoted by 2b 124 Hyperbola If F 1 and F 2 are two fixed points on the plane, then the set of all points P (x, y) for which the absolute value of the difference, that is, P F 1 P F 2, of the distances from P to each of the two fixed points is a constant, forms a pair of continuous curves called the branches of a hyperbola Each fixed point is called a focus (plural: foci) of the hyperbola The following are some properties of a hyperbola The midpoint of the line segment joining the two foci is called the center of the hyperbola, usually denoted by C(h, k) The distance from the center to each of the two foci is denoted by c The line through the two foci intersects the hyperbola at two distinct points called the vertices of the hyperbola The distance from the center to each vertex is denoted by a The line segment joining the two vertices of the hyperbola is called the transverse axis 6

Let b = c 2 a 2 The line segment of length 2b whose midpoint is the center of the hyperbola and perpendicular to the transverse axis is called the conjugate axis of the hyperbola The rectangle formed by drawing vertical/horizontal lines through the endpoints of the transverse axis and the conjugate axis is called the auxiliary rectangle of the hyperbola The two lines joining opposite vertices of the auxiliary rectangle are called the asymptotes of the hyperbola 7

Chapter 2 Review of Functions and Their Graphs 21 Functions We give one definition of a function and an alternative definition If there is an association or a correspondence between each element x of a set X to exactly one element y of a set Y then the association is a function from X to Y In symbols we have : f : X Y and is read as a function f from a set X to a set Y Remarks We use the following notation The first set is called the domain, while the second set is called the codomain The set of all values y Y that corresponds to a value in the domain is called the range If an element y of a set Y corresponds to the element x of a set X, we call y the value of the function at x or the image of x under the function, and we write is as follows: f(x) = y, read as f of x is equal to y The following statement is an alternative definition of a function A function, f, from X to Y is a set of ordered pairs of real numbers,(x, y), in which not two distinct ordered pairs have the same first number Remarks 1 An element of the range may correspond to more than one element of the domain 2 Some elements in the set Y (co-domain) may not be in the range 8

3 An element x of the domain NEVER corresponds to more than one element in the co-domain If f is a function, then the graph of f is the set of all points (x, y) in the plane R 2 for which (x, y) is an ordered pair in f 22 Operations on Functions Given two functions f and g, 1 the sum of f and g denoted by f + g is the function defined by (f + g)(x) = f(x) + g(x) 2 the difference of f and g denoted by f + g is the function defined by (f g)(x) = f(x) g(x) 3 the product of f and g denoted by f + g is the function defined by (fg)(x) = f(x)g(x) 4 the quotient of f and g denoted by f + g is the function defined by ( ) f (x) = f(x) g g(x) where g(x) 0 5 the composite function of f by g denoted by f g is defined by (f g) (x) = f (g(x)) Example 3 Given the functions f(x) = 2x + 1 and g(x) = x 2 1 then (f + g)(x) = f(x) + g(x) = (2x + 1) + (x 2 1) = x 2 + 2x (fg)(0) = f(x) g(x) = (2(0) + 1) ((0) 2 1) = (1) ( 1) = 1 (f g)(x) = f (g(x)) = f (x 2 1) = 2(x 2 1) + 1 = 2x 2 2 + 1 = 2x 2 1 ( ) f (2y) = f(2y) g g(2y) = 2(2y) + 1 (2y) 2 1 = 4y + 1 4y 2 1 9

23 Types of Functions In this section we study some special types of functions A polynomial function is a function defined by an equation of the form f(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0, where n Z + Example 4 The following are examples of polynomial functions: 1 f(x) = 2 2 f(x) = 1 x x 2 + 2x 5 3 f(x) = 1 4 x + 5 4 f(x) = x3 3 2x 5 The greatest integer function x or the floor function is defined by f(x) = x where x = n if n x < n + 1 The domain of this function is R and its range is the set of integers Z Example 5 The following are examples of the greatest integer of a specified number 1 03 = 0 2 19 = 1 3 21 = 3 4 002 = 1 5 5 = 5 6 009 = 0 Figure 21: The graph of the function f(x) = [x] is shown below 10

The signum function is a piecewise-defined function defined by 1 if x < 0 Sgn(x) = 0 if x = 0 1 if x > 0 We note that the domain of the signum function is the set of all real numbers, R and its range is the set { 1, 0, 1} A function f defined by f(x) = { x if x < 0 x if x 0 is called an absolute value function We note that the domain of the absolute value function is the set of all real numbers, R and its range is the set [0, + ) 11

Chapter 3 Limits In this chapter we will study the concept of the limit of a function Intuitively, we can think if the limit of a function as a value that the function approaches to as the values of the independent variable approach a certain real number Let us first consider the following example Example 6 Given the function f(x) = x 2 We obtain the values of the function f given some values arbitrarily close to 2 as seen in the table below x f(x) x f(x) 10000 10000 30000 90000 15000 22500 25000 62500 17000 28900 23000 52900 18000 32400 22000 48400 19000 36100 21000 44100 19900 39601 20100 40401 19990 39960 20010 40040 19999 39996 20001 40004 Notice that the value of f(x) becomes close to 4 as we let the value of x approach to 2, in symbols, lim x 2 x 2 = 4 read as, the limit of f(x) as x approaches to 2 is 4 Example 7 Consider the function defined by f(x) = x2 + x 6 which is defined everywhere except at 2 When x 2, we may write f(x) = x + 3, since x 2 x 2 + x 6 (x 2)(x + 3) = = x + 3 Consider the following table of function values x 2 (x 2) for f(x) at numbers close (but not equal) to a = 2 12

x f(x) x f(x) 2500000 5500000 1500000 4500000 2100000 5100000 1900000 4900000 2003000 5003000 1997000 4997000 2000040 5000040 1999900 4999900 2000030 5000030 1999997 4999997 Notice that the value of f(x) becomes close to 5 as we let the value of x be close to 2 This tells us that the limit of the function is 5 as the value of x approaches to x 2 + x 6 2, in symbols we have, lim = 5 x a x 2 This tells us that even if the function is undefined at a number a, the limit of the function at a may exist 31 Limits Let f be a function defined at every number in some open interval containing a, except possibly at the number a itself The limit of f(x) as x approaches a is L written as lim f(x) = L x a if ε > 0, δ > 0 such that f(x) L < ε whenever 0 < x a < δ Example 8 Prove that lim(2x + 1) = 7 x 3 Preliminary to the proof: We choose ε = 1 We need to find a δ > 0 such that f(x) L < ε whenever 2 x a < δ We take note of the following: f(x) = 2x + 1, L = 7,and a = 3 Thus, we are to find δ > 0 such that 2x + 1 7 < 1 whenever 0 < x 3 < δ Now 2 consider Proof: Let ε > 0 be given such that (2x + 1) 7 < 1 2 2x 6 < 1 2 2(x 3) < 1 2 2 x 3 < 1 2 x 3 < 1 4 (2x + 1) 7 < ε 2x 6 < ε 2(x 3) < ε 2 x 3 < ε x 3 < ε 2 13

We take δ = ε 2 Hence: 0 < x 3 < δ x 3 < ε 2 2 x 3 < ε 2(x 3) < ε 2x 6 < ε (2x + 1) 7 < ε Some Limits Theorems We present here some of the limits theorems Theorem (1) The limit of a constant at any number a is also the constant c = c, c R lim x a (2) lim x a x = a (3) lim x a mx + b = ma + b (4) lim x a cf(x) = c [ ] lim f(x), c R x a (5) lim [f(x) + g(x)] = lim f(x) + lim g(x) x a x a x a (6) lim x a [f(x)g(x)] = (7) lim x a [f(x)] n = (8) lim x a [ ] f(x) = g(x) ( ) ( ) lim f(x) lim g(x) x a x a ( ) n lim f(x), n Q x a lim f(x) x a lim x a 1 (9) lim x a x = 1 a, a 0 g(x), where lim x a g(x) 0 (10) lim x a n x = n a if n is odd or if n is even and a > 0 (11) If P (x) is a polynomial function then lim x a P (x) = P (a) (12) If R(x) = N(x) D(x) is a rational function then lim R(x) = R(a) x a provided that a is in the domain of R(x) 14

Example 9 Consider the following functions f(x) = 3x 2 and g(x) = 2x 7 lim x 1 f(x) = 3(1) 2 = 1 lim g(x) = 2( 3) 7 = 1 x 3 Example 10 Evaluate the following limits: (a) lim x 1 (3x 2 2x + 1) x 2 (c) lim x 4 x 4 (b) lim x 0 7x 1 2x 1 (d) lim x 1 x + 5 2 x + 1 Items (a) and (b) can be evaluated using direct substitution but direct substitution fails in items (c) and (d) For (c) and (d), using direct substitution leads to a form 0 but by using some algebraic manipulation we can avoid this and rewrite the expression in a form where we 0 can use direct substitution For (c), consider the following: x 2 x 2 x + 2 x 4 = x 4 = ( x + 2)( x 2) x + 2 (x 4)( x 4 = x + 2) (x 4)( x + 2) = 1 x + 2 Thus, For (d), consider the following: x + 5 2 x + 1 Thus, x 2 lim x 4 x 4 x + 5 + 2 x + 5 + 2 = lim x 1 x + 5 2 x + 1 Exercise 1 Evaluate the following limits = lim 1 = 1 x 4 x + 2 4 (x + 5) 4 (x + 1)( x + 5 + 2) = 1 x + 5 + 2 1 = lim = 1 x 1 x + 5 + 2 4 3 1 lim x 1 x 2 lim x 1 2x2 3 3 lim x 2 x 2 + 2x 1 4 lim x 2 x3 + 8 5 lim x 0 2x 3 3x 1 + x + 1 6 lim x 2 (4x + 3) 15

7 lim x 0 5x + 7 3x 2 + 1 8 lim x 2 2x + 1 x + 2 (x 1) 2 9 lim x 3 (2x 5) 3 10 lim x 1 (2x + 1) (3x 2) 11 lim x 3 3 5 + 2x 5 x 8x + 1 12 lim x 1 x + 3 x 13 lim x 1 x Remarks The following should be clear given the functionf(x) = x If a is not an integer then lim x a If a is an integer then x = lim x = lim x = a + x a lim x = a 1 and lim x a x a x = a x a + 32 One Sided Limits Let f be a given function, and let a be a fixed number (a) The limit of f(x) as x approaches a from the left is equal to L 1 and we denote this by lim x a f(x) = L 1 if the values f(x) becomes closer and closer to L 1 as x becomes closer to a with x < a L 1 is called the left-hand limit of f(x) at a (b) The limit of f(x) as x approaches a from the right is equal to L 2 and we denote this by lim x a + f(x) = L 2 if the values f(x) becomes closer and closer to L 2 as x becomes closer to a with x > a L 2 is called the right-hand limit of f(x) at a Exercise 2 Evaluate the following limits 16

1 Let f(x) = x x (a) (b) lim f(x) x 0 lim f(x) x 0 + (c) lim x 0 f(x) 2 Let f(x) = (a) (b) lim f(x) x 0 lim f(x) x 0 + (c) lim x 0 f(x) (d) (e) lim f(x) x 1 lim f(x) x 1 + (f) lim x 1 f(x) { x 2 1, x 1 1 x, x > 1 x + 5, x < 3 3 Let f(x) = 9 x 2, 3 x 3 3 x, x > 3 (a) (b) (c) (d) lim f(x) x 0 + lim f(x) x 3 lim f(x) x 3 + lim f(x) x 3 (e) lim f(x) x 3 (f) lim f(x) x 3 + (g) lim f(x) x 3 4 Let f(x) = (a) lim f(x) x 3 (b) lim x 0 f(x) { x 2 9, x 3 4, x = 3 Remarks The following are some facts regarding the limit of a function The limit of a function f(x) as x approaches a may exists even when f(a) is not a real number Inversely, the limit as x approaches a may fail to exists even when the function is defined at the number a, that is, even if f(a) is a real number A function f can only have on limit at a number a The limit of a function f(x) as x approaches a fails to exists when f(x)is unbounded at a 17

I Evaluate the following limits 1 lim x 2 (x 2 4x) 2 lim x 1 x3 2x 2 + x 1 ( ) x + 1 3 lim x 0 x 2 + 6x + 9 ( ) 3 x 3 x 4 lim x 0 3 x + 3 x ( ) x 2 4 5 lim x 2 x 2 5x + 6 ( ) x 2 6 lim x 2 x2 4 ( 1 7 lim 1 ) x 2 x 3 x 3 x 1 8 lim x 1 x2 + 3 2 x + 4 2 9 lim x 0 x 3 10 lim (x + 1) 6 x 4 PRACTICE EXERCISES 7 x 11 lim x 7 x 7 12 lim x + 1 x 1 13 lim x + 3 + 1 x 2 14 lim x 3 1 x 15 lim 3x 2 x 5 16 lim x 1 1 x (1 x) 2 1 x 2 17 lim x 1 + 1 x 18 lim x 4 + 16 x 2 16 x 2 1 + x 1 x 19 lim x 0 x x 20 lim x 0 x x II Let f(x) = x be the greatest integer function For what values of a does the lim x a f(x) exists? III Find the limits lim f(x) and lim f(x) for each integer n given the following x n x n + functions: 1 f(x) = ( 1) x x 2 f(x) = 2 IV Determine the vertical asymptotes of the following functions Describe the behavior of the graph of f(x) as the values of x becomes close the asymptote/s: 1 f(x) = 3x2 x + 5 x 2 4 2 f(x) = x2 4 x + 1 + x x 2 3x + 2 33 Limits at Infinity 3 f(x) = 2 3x2 6x 2 2x 5 4 f(x) = x 2 4 x 2 In the preceding sections of this chapter, we focused on the behavior of a function as the values of the independent variable x approach a fixed number a We now look at 18

how the function behaves when we allow the values of x to either increase or decrease without bound Let f be a function If we allow the values of x to increase or decrease indefinitely, and the corresponding values of f(x) becomes very close to a unique number L, then we say that the limit of f(x) as x approaches ± is equal to L, and we denote this by lim x + f(x) = L or by Theorem Let r be any positive integer Then 1 1 lim x + x = 0 r 1 2 lim x x = 0 r lim f(x) = L x Example 11 Consider the function f(x) = 1 We see here that lim x 1 lim x + x = 0 1 x x = 0 and Figure 31: The graph of the function f(x) = 1 x 19

Chapter 4 Differentiation 41 The Tangent Line and the Derivative Euclid s notion of a tangent as a line touching a curve at just one point is all right for circles The idea of a tangent to a curve at P as that line which best approximates the curve near P is better, but is still too vague for mathematical precision The concept of limit provides a way of getting the best description Let P be a fixed point on a curve and let Q be a nearby movable point on that curve Consider the line through P and Q, called a secant line The tangent line at P is the limiting position (if it exists) of the secant line as Q moves toward P along the curve 20

Let P (x 1, f(x 1 )) and Q(x 2, f(x 2 )) and put x = x 2 x 1 This change in abscissa maybe + or and this is called an increment of x In the given figure, the secant line P Q has slope m P Q = f(x 2) f(x 1 ) x Think of P as a fixed point, and move Q along the curve toward P ; that is, Q approaches P This is equivalent to letting x approaches 0 As this occurs, the secant line turns about the fixed point P If this secant line has a limiting position, it is this limiting position that we wish to be the tangent line to the curve at P Suppose that the f(x) is continuous at x 1 The tangent line to the graph of f at the point P (x 1, f(x 1 )) is (i) the line through P having the slope m(x 1 ) given by if this limit exists; (ii) the line x = x 1 if f(x 1 + x) f(x 1 ) m x1 = lim x 0 x f(x 1 + x) f(x 1 ) lim x 0 + x is + or and f(x 1 + x) f(x 1 ) lim x 0 x is + or 21

The normal line to a graph at a given point is the line perpendicular to the tangent line at that point The derivative of the function f is the function f such that its value at a number x in the domain of f is given by if this limit exists f f(x + x) f(x) (x) = lim x 0 x If x 1 is a particular number in the domain of f, then f f(x 1 + x) f(x 1 ) (x 1 ) = lim x 0 x In this formula, if we let x = x 1 + x, then the statement x approaches 0 is equivalent to x approaches x 1 We see that f (x 1 ) = lim x x1 f(x) f(x 1 ) x x 1 Example 12 If f(x) = 3/x, find f (2) Other notations for the derivative of y = f(x) at x are dy dx, y, D x y, and D x [f(x)] If we use y = f(x + x) f(x), then we have dy dx = lim y x 0 x 22

42 Differentiability and Continuity The process of computing the derivative at x = x 1 is called differentiation If a function has a derivative at x 1, then we say that f is differentiable at x 1 A function is said to be differentiable on an open interval if is differentiable at every number in the open interval If a function is differentiable at every number in its domain, then it is called a differentiable function Theorem If a function is differentiable at x 1, then f is continuous at x 1 Proof (i) By hypothesis f (x 1 ) exists Now, we define f (x 1 ) as f (x 1 ) = lim x x1 f(x) f(x 1 ) x x 1 and so it is necessary for f(x 1 ) to exist (ii) Consider lim [f(x) f(x 1 )] x x 1 = lim (x x 1 ) f(x) f(x 1) x x1 x x 1 = f(x) f(x 1 ) lim (x x 1 ) lim x x1 x x1 x x 1 = 0 f (x 1 ) = 0 This is equivalent to saying that lim x x1 f(x) = f(x 1 ) 23

43 Theorems on Differentiation and Higher-Order Derivatives 44 Derivatives of Trigonometric Functions 45 Derivative of a Composite Function and the Chain Rule 46 Derivative of the Power Function for Rational Exponents and Implicit Differentiation 47 Rectilinear Motion and Rates of Change 48 Related Rates 49 Rolles Theorem and Mean Value Theorem 410 Maximum and Minimum Function Values 411 Applications Involving an Absolute Extremum on a Closed Interval 412 Increasing and Decreasing Functions and the First Derivative Test 413 Concavity and Points of Inflection and the Second Derivative Test 414 Derivatives of Transcendental Functions If a is any positive number and x is any real number, then the function f defined by f(x) = a x is called the exponential function to the base a Theorem 24

If a is any positive real number and u is a differentiable function of x, then D x (a u ) = a u ln ad x u Example 13 Consider the function f(x) = 3 x2 We note that u(x) = x 2 is a differentiable function and D x (x 2 ) = 2x thus, f (x) = 3 x2 ln 3 (2x) = 2(ln 3)x3 x2 Remarks We recall the following properties of exponential functions functions The graph of f(x) = a x lies above the x-axis The graph is asymptotic to the x-axis from right to left if a > 1 and asymptotic to the x-axis from left to right if 0 < a < 1 The graph passess through the points (0, 1), (1, a) and ( 1, 1 a ) If a > 1, then the entire graph is increasing If 0 < a < 1, then the entire graph is decreasing If a = 1, then the function is a constant function equal to f(x) = 1 If a is any positive real number except 1, the logarithmic function to the base a is the inverse of the exponential function to the base a, that is y = log a x if and only if a y = x In a logarithmic function, if a = 10 we call the function the common logarithm and denote log a x simply by log x If a = e, the natural number e 2718 then the logarithmic function is called the natural logarithmic function and is denoted by ln x instead of log e x Remarks Logarithms which have a common base a satisfiy the following properties: log a MN = log a M log a N log a M N = log a M log a N log a M n = n log a M log a a = 1 log a x = ln x ln a 25

Theorem If u is a differentiable function of x then ( loga e D x (log a u) = u ( loge e If a = e, then D x (ln u) = u ) D x u = D xu u ln a ) D x u = D xu u ln e = D xu u Example 14 Consider y = f(x) = log x + 1 dy To find, we consider the property x 2 + 1 dx of logarithms and re-write f(x) as f(x) = log x + 1 + log x 2 + 1 Thus, dy dx = log 10e x + 1 + 2xlog 10e x 2 + 1 = log 10 e(1 2x x 2 ) (x + 1)(x 2 + 1) Exercise 3 Find the derivative of the following functions (TC7 Exer 55 pp476) 1 f(x) = 4 3x2 sin 2x 2 f(x) = 4 3 f(x) = 2 5x 3 4x2 4 f(x) = log x x 5 f(x) = tan 2 3x 6 f(x) = log 2 x 26

Chapter 5 Integration and Its Applications 51 Differentials Let y = f(x) be differentiable at x and suppose that dx, the differential of the independent variable x, denotes an arbitrary increment of x The corresponding differential dy of the dependent variable y is defined by dy = f (x)dx Because, y = f(x) and dy = f (x) x, we have a linear approximation to the function f(x) near the point a given by f(x) f(a) + f (a)(x a) 52 Antiderivatives We call F an antiderivative of f on the interval I if F (x) = f(x) for all x in I (If x is an endpoint of I, F (x) need be only a one-sided derivative) If F is an antiderivative of f, then the family of functions F (x) + C (C any real number), is the general antiderivative of f over the interval I The constant C is called the arbitrary constant 27

Antidifferentiation is the process of finding the set of all antiderivatives of a given function The symbol denotes the operation of antidifferentiation, and we write f(x)dx = F (x) + C where F (x) = f(x) and d(f (x)) = f(x)dx The expression F (x) + C is the general antiderivative of f Theorem If f and g are two functions defined on an interval I, such that f (x) = g (x) for all x I then there is a constant K such that f(x) = g(x) + K for all x I Theorem If F is a particular antiderivative of f on an interval I, then every antiderivative of f on I is given by F (x) + C (51) where C is an arbitrary constant, and all antiderivatives of f on I can be obtained from (51) by assigning particular values to C The following are some common formulas for antiderivatives 1 dx = x + C 2 af(x)dx = a f(x)dx 3 (f(x) + g(x))dx = f(x)dx + g(x)dx 4 (c 1 f 1 (x) + + c n f n (x))dx = c 1 f1 (x)dx + + c n fn (x)dx 5 If n is a rational number not equal to, x n dx = 1 n+1 xn+1 + C n 1 6 sin xdx = cos x + C 7 cos xdx = sin x + C 8 sec 2 xdx = tan x + C 9 csc 2 xdx = cot x + C 10 sec x tan xdx = sec x + C 11 csc x cot xdx = csc x + C 28

53 Some Techniques of Antidifferentiation Theorem The Chain Rule for Antidifferentiation Let g be a differentiable function, and let the range of g be an interval I Suppose that f is a function defined on I and that F is an antiderivative of f on I Then f(g(x))[g (x)]dx = F (g(x)) + C Proof It is known that F (g(x)) = f(g(x)) Hence, D x [F (g(x))] = F (g(x))[g (x)] = f(g(x))g (x) Therefore, it follows that f(g(x))[g (x)]dx = F (g(x)) + C Theorem If g be a differentiable function and n is a rational number, [g(x)] n [g (x)dx] = [g(x)]n+1 + C n 1 n + 1 Exercises (TC7, pages 334-335) 1 1 4y dy 2 x 3 x 2 9 dx 3 x 2 (x 3 1) 10 dx y 3 4 (1 2y 4 ) dy 5 5 x x + 2 dx 6 2r (1 r) 7 dr 7 3 2xx 2 dx 8 cos 4θ dθ 9 6x 2 sin x 3 dx 10 y csc 3y 2 cot 3y 2 dy 29

EXERCISES Evaluate the following integrals 1 3x 4 dx 19 (x 2 sec 2 x)dx 2 2x 7 dx 3 dx x 3 4 x 4 (5 x 2 )dx 5 x(x + 1)dx 6 3 sin x 2 cos xdx 7 ds (2x 3 5x 2 + 6x 10)dx 8 (5x 2 + 3x 1)dx 9 xdx 10 (x 2 + 1) xdx 11 tan 2 xdx 1 12 x dx 3 ( ) 7 1 13 2 x2 + 5 xdx 14 2x 5dx 15 x 2 cos x 3 dx (x 2 + 2x)dx 16 x3 + 3x 2 + 1 17 x 2 x + 1dx 18 ( 3 x 5x 2 )dx 20 21 22 23 24 25 26 27 28 29 30 31 sin x dx x xdx x + 3 (x + 1) 2 x + 3dx (x + 3)dx 3 (x + 3) 2 xdx x + 9 ( sin 1 ) ( x cos 1 x) x 2 dx 2 cot x 3 sin 2 x dx sin x sin x cos 2 x dx sin x 1 cos xdx tan x sec 2 xdx ( 1 t 2 ) 1 t adt sin x sin (cos x)dx 30

54 The Definite Integral and Area n F (i) = F (m) + F (m + 1) + + F (n) i=m where m and n are integers and m n The following are some properties of the summation notation 1 2 3 n c = cn, where c is any constant i=1 n n cf (i) = c F (i) i=1 i=1 n n n [F (i) + G(i)] = F (i) + G(i) i=1 i=1 i=1 4 5 b F (i) = i=a b+c i=a+c F (i c) and b F (i) = i=a n [F (i) F (i 1)] = F (n) F (0) i=1 6 If n is a positive integer, then (a) (b) (c) (d) n i = i=1 n i 2 = i=1 n i=1 n i=1 n(n + 1) 2 n(n + 1)(2n + 1) 6 i 3 = n2 (n + 1) 2 4 b c i=a c i 4 = n(n + 1)(2n + 1)(3n2 + 3n 1) 30 F (i + c) 31

Example 15 Consider the function f(x) = x 2 + 5 If we are to approximate the area under the curve of f(x), on the interval [0, 2] we can do the following partitions of the interval [0, 2]: Figure 51: f area : The approximated area where the upper limit of the intervals are used to get the heights of the rectangles Figure 52: F area : The approximated area where the lower limit of the intervals are used to get the heights of the rectangles We note that the right endpoints of the five intervals are 2i 5 where i = 1, 2, 3, 4, 5 32

The width of each rectangle is 2 and the height of each rectangle can be obtained by 5 evaluating at the right endpoint of each interval [0, 2], [ 2, 4], [ 4, 6], [ 6, 8], and, [ 8, 10] 5 5 5 5 5 5 5 5 5 Thus, the sum of the areas fo the rectangles is 5 f i=1 ( 2i 5 ) ( ) 2 = 162 5 255 On the other hand, the left endpoints of the five intervals are 2 (i 1) where 5 i = 1, 2, 3, 4, 5 The width of each rectangle is 2 and the height of each rectangle can 5 be obtained by evaluating at the left endpoint of each interval So, the sum is 5 ( ) ( ) 2i 2 2 f = 202 5 5 255 i=1 We note that the actual area under the curve satisfy f area < AreaofRegion < F area Area of a Plane Region Suppose that the function f is continuous in the closed interval [a, b], with f(x) 0 for all x [a, b], and that R is the region bounded by the curve y = f(x), the x axis, and the lines x = a and x = b Divide the interval into n subintervals, each of length x = (b a)/n, and denote the ith subinterval by [x i 1, x i ] Then if f(c i ) is the absolute minimum function value on the ith subinterval, the measure of the area of region R is given by A = lim n n f(c i ) x i=1 This equation means that for any ɛ > 0 there is a number N > 0 such that if n is a positive integer and n if n > N then f(c i ) x A < ɛ i=1 33

Example 16 Find the area of the region bounded by the graph f(x) = x 3, the x-axis, and the lines x = 0, x = 1 Note that the function f(x) is continuous on on Figure 53: The area under the curve of f(x) = x 3, bounded by x-axis and the lines x = 0, x = 1 the interval [0, 1] We partition the interval [0, 1] into n sub-intervals each with width x = 1 Choosing the right endpoint of each interval as the representative of each n interval (for convenience), we have Area = lim n n i=1 f(c i ) x = lim n n ( i n n i=1 1 = lim n n 4 i=1 ) 3 ( ) 1 n i 3 1 = lim n 4 + 1 2n + 1 4n 2 = 1 4 34

Function Integrable on a Closed Interval Let f be a function whose domain includes the closed interval [a, b] Then f is said to be integrable on [a, b] if there is a number L satisfying the condition that, for any ɛ > 0, there exists a δ > 0 such that for every partition for which < δ, and for any w i [x i 1, x i ], i = 1,, n, then n f(w i ) i x L < ɛ i=1 For such we write lim 0 n f(w i ) i x = L i=1 Definite Integral If f is a function defined on the closed interval [a, b], then the definite integral of f from a to b, denoted by b f(x)dx, is given by a b a f(x)dx = lim 0 n f(w i ) i x Theorem If a function is continuous on the closed interval [a, b], then it is integrable on [a, b] Area of a Plane Region Let the function f be continuous on [a, b] and f(x) 0 for all x [a, b] Let R be the region bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b Then the measure A of the area of the region R is given by If a > b and a f(x)dx exists, then b A = lim 0 b a = b a i=1 n f(w i ) i x i=1 f(x)dx f(x)dx = a b f(x)dx 35

If f(a) exists, then a a f(x)dx = 0 Theorem If is any partition of the closed interval [a, b], then lim 0 n i x = b a i=1 Theorem If is any partition of the closed interval [a, b], then lim 0 n f(w i ) i x exists, where is any partition of [a, b], then if k is any constant, lim 0 n i=1 i=1 kf(w i ) i x = k lim 0 n f(w i ) i x i=1 Theorem If the function f is integrable on the closed interval [a, b], and if k is any constant, then b a kf(x)dx = k b a f(x)dx Theorem If the functions f and g are integrable on the closed interval [a, b], then f + g is integrable on [a, b] and b a [f(x) + g(x)]dx = b a f(x)dx + b a g(x)dx Theorem If the function f is integrable on the closed intervals [a, b], [a, c], and [c, b], then where a < c < b b f(x)dx = c f(x)dx + b a a c f(x)dx 36

Theorem If the function f is integrable on a closed interval containing the numbers a, b and c, then b a f(x)dx = regardless of the order of a, b and c c a f(x)dx + b c f(x)dx Exercise 4 Evaluate the following integrals (TC7 EXER 45 pp368) 1 4 2 xdx 4 2 0 2x x2 dx 2 3 0 (4x 2 2x + 5)dx 5 5π 2 π 2 sin xdx 3 2 1 (5 2x)dx 6 π 0 3 cos 2 xdx 55 Mean Value Theorem for Integrals Theorem If the functions f and g are integrable on the closed interval [a, b] and if f(x) g(x) for all x in [a, b], then b a f(x)dx b a g(x)dx Theorem Suppose that the function f is continuous on the closed interval [a, b] If m and M are, respectively, the absolute minimum and absolute maximum function values of f on [a, b] so that m f(x) M for a x b then m(b a) b a f(x)dx M(b a) Theorem The Mean Value Theorem for Integrals If the function f is continuous on the closed interval [a, b], there exists a number c in [a, b] such that b a f(x)dx = f(c)(b a) 37

If the function f is integrable on the closed interval [a, b], then the average value of f on [a, b] is b a f(x)dx b a Example 17 If f(x) = x 2, find the average value of f on the interval [1, 3] Solution: Consider the graph of f(x) = x 2 and 3 1 x2 dx = 8 2 From the theorem 3 26 3 13 above we note that the average value of f(x) on [1, 3] is which is equal to 3 1 3 Note that f(208) 13 We see from the figure that when we use x = 208 the height 3 of the rectangle is f(208) 13 is equal to the area under the curve of f(x) from 3 [1, 3] Figure 54: Area of Rectangle ABCD Figure 55: Area of f(x) = x 2 from [1, 3] 38

56 Fundamental Theorem of Calculus Theorem The First Fundamental Theorem of the Calculus Let f be continuous on the closed interval [a, b] and let x be any number in [a, b] If F is the function defined by then d dx F (x) = x a x a f(t)dt F (x) = f(x) (52) f(t)dt = f(x) (53) (If x = a, the derivative in (2) may be a derivative from the right, and if x = b, it may be a derivative from the left) Theorem The Second Fundamental Theorem of the Calculus Let f be continuous on the closed interval [a, b] and let g be a function such that g (x) = f(x) for all x in [a, b] Then b a f(t)dt = g(b) g(a) Example 18 Evaluate 4 3 x + 2 dx Solution: Recall that x + 2 = x 2 when x 2 and x + 2 = x + 2 when x > 2 Thus we have, 4 3 2 4 x + 2 dx = ( x 2)dx + (x + 2)dx 3 ] 2 2 [ ] = [ x2 x 2 4 2 2x + 3 2 + 2x 2 = [( 2 + 4) ( 9 + 6)] + [(8 + 8) (2 4)] 2 = 1 2 + 18 = 37 2 39

Exercise 5 Evaluate the following definite integrals 1 3 0 (3x 2 4x + 1)dx 4 π 2 0 sin 2x)dx 2 6 3 (x 2 2x)dx 5 3 1 1 (x + 2) 3 )dx 3 1 0 z ( (z 2 + 1) )dx 3 6 4 2 ( x4 x )dx x 3 57 Area between two curves In the previous sections, we considered only area of a plane region for which the function values are non-negative on [a, b] Suppose now that f(x) < 0 for all x [a, b] Then each f(c i ) is a negative number for any c i [a, b] Thus the area represented by this particular rectangle is f(c i ) i x where i x represents the width of the i th rectangle Figure 56: Area of f(x) = 2x from [ 2, 1] 40

Consider the equation f(x) = 2x and the area of the region bounded by f(x), x = 1 and x = 2 We see that in the interval [ 2, 0] the area should be computed as 1 0 2xdx 0 2 2xdx = 5 It can be easily computed using the formula for area of a triangle that the area bounded by f(x), x = 1 and x = 2 is 5 square units Example 19 Find the area of the region bounded by the curve f(x) = x 3 2x 2 5x + 6, the x-axis, and the lines x = 1 and x = 2 Consider the graph of f(x) Figure 57: The region bounded by the curve f(x) = x 3 2x 2 5x + 6, the x-axis, and the lines x = 1 and x = 2 ANS: 157 12 square units In general, we have the following idea to get the area between two curves as where (x 1, y 1 ) and (x 2, y 2 ) are either adjacent points of intersection of the two curves involved or points on the specified boundary lines 41

Example 20 Find the area of the region bounded by the parabola y 2 = 2x 2 and the line y = x 5 Figure 58: The region bounded by the curve y 2 = 2x 2 and and the line y = x 5 ANS: 18 square units Example 21 Find the area of the region bounded by the function f(x) = sin x and the lines x = π and x = 2π Consider the graph of f(x) 3 3 Figure 59: The region bounded by the function f(x) = sin x and the lines x = π 3 and x = 2π 3 ANS: 1 square units 42

Example 22 Find the area of the region bounded by the function f(x) = sin x and the parabola y = x 2 Consider the graph of f(x) Figure 510: The region bounded by the function f(x) = sin x and the parabola y = x 2 ANS: 01357 square units Example 23 Find the area of the region bounded by the function f(x) = 3x 3 x 2 10x and the parabola g(x) = x 2 + 2x Consider the graph of f(x) and g(x) Figure 511: The region bounded by the the functions f(x) = 3x 3 x 2 10x and g(x) = x 2 + 2x ANS: 01357 square units 43

Example 24 Find the area of the region bounded by the graphs x = 3 y 2 and x = y + 1 Consider the graph of f(x) and g(x) Figure 512: The region bounded by the graphs x = 3 y 2 and x = y + 1 ANS: 9 2 square units 58 Volumes of Solids 581 Disk Method If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution The simplest such solid is a right circular cylinder or disk, which is formed by revolving a rectangle about an axis adjacent to one side of the rectangle, refer to the figure below Recall that the volume of a right circular cylinder with circular base of radius R and a height of w is given by V olume = πr 2 w 44

Schematically, we have the following idea to determine the volume of a solid of revolution using the disk method To find the volume of a solid of revolution we use one of the following formulas: Horizontal Axis of Revolution: V = π Vertical Axis of Revolution: V = π where d c b a [R(x)] 2 dx [R(y)] 2 dy Example 25 Find the volume of the solid generated by revolving about the line x = 1 the region bounded by the curve (x 1) 2 = 20 4y and the lines x = 1, y = 1 45

Figure 513: The region bounded by the graphs(x 1) 2 x = 1, y = 1 and y = 3 = 20 4y and the lines and y = 3 as shown in the figure below Consider the graph of (x 1) 2 = 20 4y ANS: 24π cubic units Example 26 Find the volume of the solid formed by revolving the region bounded by the graph of f(x) = sin x and the x-axis (0 x π) about the x-axis ANS: 2π cubic units 46

582 Washer Method The disk method can be extended to cover solids of revolution with holes by replacing the representative disk with a representative washer The washer is formed by revolving a rectangle about an axis, as shown below If r and R are the inner and outer radii of the washer and w is the width of the washer, the volume is given by In terms of our idea of integrals we have, V = π V olume = π(r 2 r 2 )w b a ( [R(x)] 2 r(x)] 2) dx Example 27 Find the volume of the solid formed by revolving the region bounded by the graphs of y = x and y = x 2 ANS: 3π 10 cubic units 47

Example 28 Find the volume of the solid formed by revolving the region bounded by the graphs of y = x 2 + 1, y = 0, x = 0 and x = 1 about the y-axis ANS: 3π 2 cubic units Example 29 A manufacturer drills a hole through the center of a metal sphere of radius 5 inches The hole has a radius of 3 inches What is the volume of the resulting metal ring? See the illustration below ANS: 256π 3 cubic units 48

583 Shell Method An alternative method for finding the volume of a solid of revolution is called the shell method because it uses cylindrical shells The disk and shell methods can be distinguished as follows For the disk method, the representative rectangle is always perpendicular to the axis of revolution, whereas for the shell method, the representative rectangle is always parallel to the axis of revolution, as shown below: 49

Example 30 Consider the region in Example 28 Use the Shell Method to determine the volume of the solid of revolution Using the shell method we have the following V = 2π b a p(x)h(x)dx = 2π 1 0 [ ] x x(x 2 4 1 + 1)dx = 2π 4 + x2 = 3π 2 0 2 50

Exercise 6 Do as indicated A Set up and evaluate the integral that gives the volume of the solid formed by revolving the region about the x-axis B Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines 1 y = x, y = 0, x = 3 about the x-axis 2 y = x, y = 0, x = 3 about the line x = 3 3 y = x, y = 0, x = 3 about the line x = 6 4 y = 2x 2, y = 0, x = 2 about the y-axis 5 y = 2x 2, y = 0, x = 2 about the line y = 8 6 y = 2x 2, y = 0, x = 2 about the line x = 2 51

59 Integrals of Transcendental Functions Theorem Let u be a differentiable function of x dx 1 = ln x + C x du 2 = ln u + C u 3 e u du = e u + C 4 5 a u du = au ln a + C e u du = e u + C Example 31 Evaluate the following integrals 1 2 e x x dx e 3x+1 dx 6 7 e x + e x dx 2 x 2 e 2x3 dx 3 4 5 5xe x2 dx e 2x e x + 3 dx e 3x (1 2e 3x ) 2 dx 8 9 10 2 1 e 1 dx x(ln x) 2 e x e x + e dx ln x x dx 52