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Pre Calculus Intro to Integrals 2015 03 24 www.njctl.org 2
Riemann Sums Trapezoid Rule Table of Contents click on the topic to go to that section Accumulation Function Antiderivatives & Definite Integrals Fundamental Theorem of Calculus Substitution Method Area Between Curves Volume: Disk Method Volume: Washer Method Volume: Shell Method 3
Riemann Sums Return to Table of Contents 4
Reimann Sums Consider the following velocity graph: 30 mph How far did the person drive? 5 hrs The area under the velocity graph is the total distance traveled. Integration is used to find the area. 5
Reimann Sums But we seldom travel at a constant speed. 50 mph 5 hrs The area under this graph is still the distance traveled but we need more than multiplication to find it. 6
Reimann Sums George Riemann (Re mon) studied making these curves into a series of rectangles. So the area under the curve would be the sum of areas of the rectangles, this is called Riemann Sums. 7
Reimann Sums Riemann Sums, or Rectangular Approximation Method (RAM), is calculated by drawing rectangles from the x axis up to the curve. The question is: What part of the "top" of the rectangle should be used to determine the height of the rectangle? The right hand corner. (RRAM) The left hand corner. (LRAM) The middle. (MRAM) 8
Reimann Sums Example: Find the area between y = x 2, the x axis, and [0,1] using Riemann Sums and 4 partitions. LRAM 0 1/4 1/2 3/4 1 Found the width of the rectangle: (b a)/n= (1 0)/4 = 1/4 Is this approximation an overestimate or an underestimate? 9
Reimann Sums Example: Find the area between y = x 2, the x axis, and [0,1] using Riemann Sums and 4 partitions. RRAM 0 1/4 1/2 3/4 1 Is this approximation an overestimate or an underestimate? 10
Reimann Sums Example: Find the area between y = x 2, the x axis, and [0,1] using Riemann Sums and 4 partitions. MRAM 0 1/4 1/2 3/4 1 This value falls between LRAM and RRAM. 11
Reimann Sums *NOTE: MRAM LRAM + RRAM 2 12
Reimann Sums Q: What units should be used? A: Since the area is found by multiplying base times height, the units of the area are the units of the x axis times the units of the y axis. In our example at the beginning of the unit we had a velocity (mph) vs. time (hours) the units would then be 13
Reimann Sums 1 When finding the area between and the x axis [1,3] using four partitions, how wide should each interval be? 14
Reimann Sums 2 Find the area between and the x axis [1,3] using four partitions and LRAM. 15
Reimann Sums 3 Find the area between and the x axis [1,3] using four partitions and RRAM. 16
Reimann Sums 4 When finding the area between and the x axis [1,3] using four partitions and MRAM, when in the third rectangle, what x should be used to find the height? 17
Reimann Sums 5 Find the area between and the x axis [1,3] using four partitions and MRAM. 18
Reimann Sums We can write the four areas using where a k is the area of the k th rectangle. It is just another way of writing what we just did. Σ is the Greek letter sigma and stands for the summation of all the terms evaluated at starting with the bottom number and going through to the top. 19
Reimann Sums Selected Rules for Sigma 20
Reimann Sums Equivalent Formulas 1 st n integers: 1 st n squares: 1 st n cubes: 21
Reimann Sums 6 22
Reimann Sums 7 23
Reimann Sums 8 24
Reimann Sums 9 25
Trapezoid Rule Return to Table of Contents 26
Reimann Sums Example: Find the area y = x 2 and the x axis [0,1] using 4 partitions. 0 1/4 1/2 3/4 1 Why were areas found using RAM only estimates? How could we draw lines to improve our estimates? What shape do you get? 27
Reimann Sums Example: Find the area y = x 2 and the x axis [0,1] using 4 partitions and the trapezoids. Trapezoids 0 1/4 1/2 3/4 1 28
Reimann Sums *NOTE: Trapezoid Approximation = LRAM + RRAM 2 We could make our approximation even closer if we used parabolas instead of lines as the tops of our intervals. This is called Simpson's Rule but this is not on the AP Calc AB exam. 29
Reimann Sums 10 The area between and the x axis [1,3] is approximated with 4 partitions and trapezoids. What is the height of each trapezoid? 30
Reimann Sums 11 The area between and the x axis [1,3] is approximated with 4 partitions and trapezoids. What is the area of the 4 th trapezoid? 31
Reimann Sums 12 The area between y = and the x axis [1,3] is approximated with 4 partitions and trapezoids. What is the approximate area? 32
Reimann Sums 13 What is the approximate area using the trapezoids that are drawn? 4 y 3 2 1 0 1 2 3 4 5 6 7 8 9 x 33
Reimann Sums 14 What is the approximate fuel consumed using the trapezoids rule for this hour flight? Time (minutes) Rate of Consumption (gal/min) 0 0 10 20 25 30 40 40 60 45 34
Reimann Sums In the last 2 responder questions, the partitioned intervals weren't uniform. The AP will use both. So don't assume. 35
Reimann Sums So far we have been summing areas using Σ. Gottfried Leibniz, a German mathematician, came up with a symbol you're going to see a lot of:. It is actually the German S instead of the Greek. It still means summation. As a point of interest, we use the German notation in calculus because Leibniz was the first to publish. Sir Isaac Newton is now given the credit for unifying calculus because his notes predate Leibniz's. 36
Accumulation Function Return to Table of Contents 37
Accumulation Function V (m/s) y 2 Another way we can calculate area under a function is to use geometry. What's happening during t=0 to t=3? What is the area of t=0 to t=3? What does the area mean? 1 t (sec) x 0 1 2 3 4 5 6 1 What is the acceleration at t=3? What is happening during t=3 to t=6? What is the area of t=3 to t=6? What does this area mean? 2 Where is the object at t=6 in relation to where it was at t=0? 38
Accumulation Function The symbol notation for the area from zero to 3: "The area from t=0 to t=3 is the integral from 0 to 3 of the velocity function with respect to t." In general: 39
Accumulation Function 5 4 3 2 y What is the area from x= 4 to x=0? 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 Note: 4 5 Def: where 40
Accumulation Function When solving an accumulation function: (direction)(relation to x axis)(area) 5 y 4 3 2 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 4 5 41
Accumulation Function 15 4 y 3 2 semicircle 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 4 42
Accumulation Function 16 (round to two decimal places) 4 y 3 2 semicircle 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 4 43
Accumulation Function 17 (round to two decimal places) 4 y 3 2 semicircle 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 4 44
Accumulation Function 18 4 y 3 2 semicircle 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 4 45
Accumulation Function 19 4 y 3 2 semicircle 1 x 5 4 3 2 1 0 1 2 3 4 5 1 2 3 4 46
Antiderivatives & Definite Integrals Return to Table of Contents 47
Antiderivatives Area under the curve of f(x) from a to b is We have been using geometry to find A. The antiderivative of f(x) can also be used. 48
Antiderivatives Properties of Definite Integrals 49
Antiderivatives 20 50
Antiderivatives 21 51
Antiderivatives 22 52
Antiderivatives 23 53
Antiderivatives Antiderivative Rules Why + C? 54
Antiderivatives Where F(x) is the anti derivative of f(x). 55
Antiderivatives Examples: Notice how C always disappears? We don't need C when we do definite integrals. 56
Antiderivatives 24 57
Antiderivatives 25 58
Antiderivatives 26 59
Antiderivatives 27 60
Antiderivatives 28 61
Antiderivatives 29 62
Antiderivatives 30 63
Antiderivatives You can do definite integrals on your graphing calculator. For the TI 84: use the MATH key 9: fnint( For example: Depending on which version of the operating system you have: 64
Antiderivatives The graphing calculator also has a built in integration function. MATH > 9:fnInt( depending on the version of the operating system: fnint( or For example, integrate x 2 3 from 1 to 4 with respect to x. fnint(x 2 3,x,1,4) 65
Fundamental Theorem of Calculus Return to Table of Contents 66
Fundamental Theorem of Calculus There are 2 parts to the Fundamental Theorem of Calculus, depending on the book, the order will change. Fundamental Theorem of Calculus (F.T.C.) Part 1 If f(x) is continuous at every point of [a,b] and F(x) is the antiderivative of f(x) then (which is what you've been doing) 67
Fundamental Theorem of Calculus Fundamental Theorem of Calculus (F.T.C.) Part 2 If F(x) is continuous at every point of [a,b] then has a derivative at every point on [a,b],and 68
Fundamental Theorem of Calculus Example: It looks easy, but be aware. When the derivative of the bounds in anything other that 1, need to multiply f(x) by the derivative. 69
Fundamental Theorem of Calculus 31 A B C D 70
Fundamental Theorem of Calculus 32 A B C D 71
Fundamental Theorem of Calculus 33 A B C D HINT 72
Fundamental Theorem of Calculus 34 A B C D HINT 73
Substitution Method Return to Table of Contents 74
Substitution Method Just like with differentiation, there are many integrals that are more complicated to evaluate. In situations like these, we use the Substitution Method to turn a difficult integral into a much simpler one. 75
Substitution Method The Substitution Method When we are given the initial function substitution where u is a function of x,, we make a, and 76
Substitution Method Ex: Let 77
Substitution Method Ex: Let 78
Substitution Method 35 What is the value of u? A B C D 79
Substitution Method 36 What will the integral be after the substitution is made? A B C D 80
Substitution Method 37 Evaluate the integral A B C D 81
Substitution Method The Substitution Method If you have a definite integral, you have two options for plugging in the bounds to get the final answer. 1. Plug a and b into the integrated function AFTER you have re substituted the x's back into the function 2. Plug a and b into to create two new bounds, and plug these into the integrated function 82
Substitution Method Ex: Let 83
Substitution Method 38 What is the value of u? A B C D 84
Substitution Method 39 What is the new upper bound? 85
Substitution Method 40 What is the new lower bound? 86
Substitution Method 41 What will the integral be after the substitution is made? A B C D 87
Substitution Method 42 What is the value of the integral? 88
Substitution Method 43 What is the value of u? A B C D 89
Substitution Method 44 What is the new upper bound? 90
Substitution Method 45 What is the new lower bound? 91
Substitution Method 46 What will the integral be after the substitution is made? A B C D 92
Substitution Method 47 What is the value of the integral? 93
Area Between Curves Return to Table of Contents 94
Area Between Curves The area between a curve and the x axis is But what about the area between two curves? Such as the area between and from the intersection to x=1. x=.30054197 (don't round till the end) 95
Area Between Curves Area Between Curves Where a and b are the left and right bounds of the region. f(x) is upper curve on a graph and g(x) the lower. So for our example: 96
Area Between Curves 48 When finding the area between and, what is the left bounds of x? 97
Area Between Curves 49 When finding the area between and, what is the right bounds of x? 98
Area Between Curves 50 When finding the area between and, what is integral used? A B C D 99
Area Between Curves 51 What is the area between and? 100
Area Between Curves Consider: Integrating in terms of y would be easier. f(x) g(x) is right function minus left and the bounds are now the least value of y to the greatest. 101
Area Between Curves 52 When finding the area between and the y axis, what is the lower bound of y? 102
Area Between Curves 53 When finding the area between and the y axis, what is the upper bound of y? 103
Area Between Curves 54 What is the area between and the y axis? 104
Example: Find the area between the curves in the first quadrant. Notice that the lower function isn't the same for the entire region. We could find the area in terms of y or divide the region into 2 separate integrals and add their areas. 105
Area Between Curves 55 When finding the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1, what would be the area to the left of the y axis? A B C D 106
Area Between Curves 56 When finding the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1, what is the left bounds of h(x) f(x)? 107
Area Between Curves 57 When finding the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1, what is the right bound of h(x) f(x)? 108
Area Between Curves 58 When finding the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1, what would be the area to the right of the y axis? A B C D 109
Area Between Curves 59 When finding the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1, what is the right bound of h(x) g(x)? 110
Area Between Curves 60 When finding the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1, what is the left bound of h(x) g(x)? 111
Area Between Curves 61 Find the area between f(x) = x, g(x) = 1/2x, and h(x) =1/2x 1. 112
Volume: Disk Method Return to Table of Contents 113
Volume: Disk Method Another way to make a 3 D object is to take a region and rotate it about an axis. When this rectangle is rotated a cylinder is formed. We could use geometry, but can we use calculus? From the section on known cross sections: And since the cross sections are circles: 114
Volume: Disk Method Volume: Disk Method When rotating about a horizontal axis: When rotating about a vertical axis: 115
Volume: Disk Method Rotate about x axis from x=1 to x=4. 116
Volume: Disk Method Rotate about y=2 from x=1 to x=4. Since y=2 is a horizontal axis of rotation, integral is in terms of x. r=2 y Why? From the x axis to the curve is y and to axis of rotation is 2, we want upper minus lower. 117
Volume: Disk Method Rotate about y= 2 from x=1 to x=4. Since y= 2 is a horizontal axis of rotation, integral is in terms of x. 2 y r=y 2=y+2 Why? upper minus lower 118
Volume: Disk Method 62 Rotate y=2x 2 about the x axis over [0,5]. What is the lower bound? 119
Volume: Disk Method 63 Rotate y=2x 2 about the x axis over [0,5]. What is the upper bound? 120
Volume: Disk Method 64 Rotate y=2x 2 about the x axis over [0,5]. What is integral? A B C D 121
Volume: Disk Method 65 Rotate y=2x 2 about the x axis over [0,5]. What is the volume? 122
Volume: Disk Method 66 Rotate y=2x 2 about the line x=4 over [0,5]. What is the lower bound? 123
Volume: Disk Method 67 Rotate y=2x 2 about the line x=4 over [0,5]. What is the upper bound? 124
Volume: Disk Method 68 Rotate y=2x 2 about the line x=4 over [0,5]. What is the radius? 125
Volume: Disk Method 69 Rotate y=2x 2 about the line x=4 over [0,5]. What is integral? A B C D 126
Volume: Disk Method 70 Rotate y=2x 2 about the line x=4 over [0,5]. What is the volume? 127
Volume: Disk Method Rotate for x=0 to x=2 about the y axis. Since this is a vertical axis,the problem should be rewritten in terms of y: Rotate for y=0 to y=16 about the y axis. 16 128
Volume: Disk Method Rotate for x=0 to x=2 about the x=2. Since this is a vertical axis, the problem should be rewritten in terms of y: 16 Rotate for y=0 to y=16 about the x=2. 129
Volume: Disk Method 71 Rotate about the y axis over. What is the lower bound? 130
Volume: Disk Method 72 Rotate about the y axis over. What is the upper bound? 131
Volume: Disk Method 73 Rotate about the y axis over. What is the radius? 132
Volume: Disk Method 74 Rotate about the y axis over. What is integral? A B C D 133
Volume: Disk Method 75 Rotate about the y axis over. What is the volume? 134
Volume: Disk Method 76 Rotate about the over. What is the lower bound? 135
Volume: Disk Method 77 Rotate about the over. What is the upper bound? 136
Volume: Disk Method 78 Rotate about the over. What is the radius? 137
Volume: Disk Method 79 Rotate about the over. What is integral? A B C D 138
Volume: Disk Method 80 Rotate about the over. What is the volume? 139
Volume: Washer Method Return to Table of Contents 140
Volume: Washer Method For the washer method, there is a gap between the region being rotated and the axis. Rotating this rectangle we get a tube, or a cylinder with a smaller cylinder taken away. Our cross section would be: R r 141
Volume: Washer Method Volume: Washer Method When rotating about a horizontal axis: Where R is the greater distance from the axis, not necessarily the upper function. When rotating about a vertical axis: *Caution: π can be factored out but not the square. 142
Volume: Washer Method Find the volume when f(x) and g(x) are rotated about the x axis [0,2]. 143
Volume: Washer Method Rotate the region bound by y=x 2, x=2, and y=0 about the y axis. Since a vertical axis of rotation, integration is done in terms of y. (2,4) Hint 144
Volume: Washer Method Rotate the region bound by x axis, y=x 2, x=1, and x=2 about y= 1. Since y= 1 is horizontal integration is done in terms of x. R=x 2 +1 r=1 145
Volume: Washer Method 81 Rotate the region between and about the x axis over [0,1]. What is the lower bound? 146
Volume: Washer Method 82 Rotate the region between and about the x axis over [0,1]. What is the upper bound? 147
Volume: Washer Method 83 Rotate the region between and about the x axis over [0,1]. What is integral? A B C D 148
Volume: Washer Method 84 Rotate the region between and about the x axis over [0,1]. What is the volume? 149
Volume: Washer Method 85 Rotate the region between and about the y axis over [0,1]. What is integral? A B C D 150
Volume: Washer Method 86 Rotate the region between and about the y axis over [0,1]. What is the volume? 151
Volume: Washer Method 87 Rotate the region between and about the y=1 over [0,1]. What is R? A B C D 152
Volume: Washer Method 88 Rotate the region between and about the y=1 over [0,1]. What is integral? A B C D 153
Volume: Washer Method 89 Rotate the region between and about the y=1 over [0,1]. What is the volume? 154
Volume: Washer Method 90 Rotate the region between and about the x= 1 over [0,1]. What is integral? A B C D 155
Volume: Washer Method 91 Rotate the region between and about the x= 1 over [0,1]. What is the volume? 156
Volume: Shell Method Return to Table of Contents 157
Volume: Shell Method Volume: Shell Method When rotating about a horizontal axis: When rotating about a vertical axis: 158
Volume: Shell Method Find the volume of the solid obtained by rotating the area under the graph of over about the y axis. 159
Volume: Shell Method Find the volume of the solid obtained by rotating the area between the graph of and about the y axis. 160
Volume: Shell Method Find the volume of the solid obtained by rotating the area under the graph of over about 161
Volume: Shell Method 92 Find the volume of the solid generated by rotating the area under the graph of from about the y axis. A B C D 162
Volume: Shell Method 93 Find the volume of the solid generated by rotating the area between the graphs of and about the y axis. A B C D 163