Chapter Overview: Definite Integrals In this chapter, we will study the Fundamental Theorem of Calculus, which establishes the link between the algebra and the geometry, with an emphasis on the mechanics of how to find the definite integral. We will consider the differences implied between the contet of the definite integral as an operation and as an area accumulator. We will learn some approimation techniques for definite integrals and see how they provide theoretical foundation for the integral. We will revisit graphical analysis in terms of the definite integral and view another typical AP contet for it. Finally, we will consider what happens when trying to integrate at or near an asymptote. 5
. The Fundamental Theorem of Calculus As noted in the overview, Anti-derivatives are known as Indefinite Integrals and this is because the answer is a function, not a definite number. But there is a time when the integral represents a number. That is when the integral is used in an Analytic Geometry contet of area. Though it is not necessary to know the theory behind this to be able to do it, the theory is a major subject of Integral Calculus, so we will eplore it briefly. We know, from Geometry, how to find the eact area of various polygons, but we never considered figures where one side is not made of a line segment. Here we want to consider a figure where one side is the curve y = f () and the other sides are the -ais and the lines = a and = b. y 3 y = f () a 3 b As we can see above, the area can be approimated by rectangles whose height is the y-value of the equation and whose width we will call Δ. The more rectangles we make, the better the approimation. The area of each rectangle would be n f ( ) Δ and the total area of n rectangles would be A = f ( i ) Δ. If we could make an infinite number of rectangles (which would be infinitely thin), we would have the eact area. The rectangles can be drawn several ways--with the left side at the height of the curve (as drawn above), with the right side at the curve, with i= 6
the rectangle straddling the curve, or even with rectangles of different widths. But once they become infinitely thin, it will not matter how they were drawn--they will have virtually no width (represented by d instead of the Δ ) and a height equal to the y-value of the curve. We can make an infinite number of rectangles mathematically by taking the Limit as n approaches infinity, or n Lim f ( i ) Δ. n i= This limit is rewritten as the Definite Integral:! b a f ( )!d b is the "upper bound" and a is the "lower bound," and would not mean much if it were not for the following rule: The Fundamental Theorem of Calculus If f ( )is a continuous function on a,b, then ) d d c f ( t) dt = f ( ) or d d u c f ( t) dt = f ( u)id u ) If F' ( ) = f ( ), then f ( ) d b a = F( b) F( a). The first part of the Fundamental Theorem of Calculus simply says what we already know--that an integral is an anti-derivative. The second part of the Fundamental Theorem says the answer to a definite integral is the difference between the anti-derivative at the upper bound and the anti-derivative at the lower bound. This idea of the integral meaning the area may not make sense initially, mainly because we are used to Geometry, where area is always measured in square units. But that is only because the length and width are always measured in the same kind of units, so multiplying length and with must yield square units. We are epanding 7
our vision beyond that narrow view of things here. Consider a graph where the - ais is time in seconds and the y-ais is velocity in feet per second. The area under the curve would be measured as seconds multiplied by feet/sec--that is, feet. So the area under the curve equals the distance traveled in feet. In other words, the integral of velocity is distance. Objectives Evaluate Definite Integrals Find the average value of a continuous function over a given interval Differentiate integral epressions with the variable in the boundary Let us first consider Part of the Fundamental Theorem, since it has a very practical application. This part of the Fundamental Theorem gives us a method for evaluating definite integrals. 8 E Evaluate ( + 3)d 8 ( + 3)d = 8 + 3 ( ) ( 8 + 6) = 8 + The antiderivative of + 3 is + 3. We use this notation when we apply the Fundamental Theorem Plug the upper limit of integration into the antiderivative and subtract it from the lower limit when plugged into the antiderivative = 38 Check with Math 9. E Evaluate d d = = ( ) ( ) = 8
E 3 Evaluate sin d sin d = cos = = E Evaluate + u u 3 du + u du = ( u 3 + u )du u 3 = u + ln u = + ln ( + ln) = 3 + ln E 5 Evaluate 5 5 3 d This is a trick! We use the Fundamental Theorem on this integral because the curve is not continuous on a, b. When =, does not eist, so 3 the Fundamental Theorem of Calculus does not apply. One simple application of the definite integral is the Average Value Theorem. We all recall how to find the average of a finite set of numbers, namely, the total of the numbers divided by how many numbers there are. But what does it mean to take the average value of a continuous function? Let s say you drive from home to school what was your average velocity (velocity is continuous)? What was the average temperature today (temperature is continuous)? What was your average height for the first 5 years of your life (height is continuous)? Since the integral is the sum of infinite number of function values, the formula below answers those questions. 9
Average Value Formula: The average value of a function f on a closed interval a,b f avg = b f ( )d b a a is defined by If we look at this formula in the contet of the Fundamental Theorem of Calculus, it can start to make a little more sense. b a f ( ) d = F( b) F( a) The Fundamental Theorem of Calculus b b a f d F b F a a b a ( ) = ( ) ( ) F( b) F( a) = b a Notice that this is just the average slope for F( ) on a, b slope of F( ) would be the average value of F' ( ). But since the definition in the Fundamental Theorem of Calculus says that F' ( ) f ( ) this is actually just the average value of f ( ). E 6 Find the average value of f f avg = b a f avg = 5 b a 5 f ( )d ( ) = + on,5.. The average ( +)d Substitute in the function and interval =, f avg = 8 3 Use Math 9 (or integrate analytically) to calculate answer 3
( ) = secθ tanθ on, E 7 Find the average value of h θ h avg = b h( θ )dθ b a a = secθ tanθdθ =.5. Now let s take a look at the First Part of the Fundamental Theorem of Calculus: d d a ( ) f t dt = f ( ) or d d u a f ( t) dt = f ( u) du d This part of the Fundamental Theorem of Calculus says that derivatives and integrals are inverses operations of each other. Confusion usually pops up with the seemingly etra variable t showing up. It is a dummy variable, somewhat like the parameter in parametric mode. It would disappear in the process of integrating. In terms of the AP Test, the symbology is what is important and usually only comes into a process with L Hopital s Rule. Therefore, we will discuss this further in a later chapter. For now, the key idea is that the First Part of the Fundamental Theorem of Calculus simply tells us that an integral is an antiderivative. 3
. Homework Set A Use Part II of the Fundamental Theorem of Calculus to evaluate the integral, or eplain why it does not eist. 3. 5 7 d. ( 5 )d 3. 5 d. 5 3 3 7 3 e d 5. 3 t dt 6. 3 csc y cot ydy 7. sec ydy 8. 9 3 z dz 9. If F() = f (t)dt, where f (t) = + u du, find F ''(). u Find the Average Value of each of the following functions.. F( ) = ( 3) on 3, 7. H. F ( ) = on, 3 ( ) = sec on, t 3. F( ) = on, 3. If a cookie taken out of a 5 F oven cools in a 6 F room, then according to Newton s Law of Cooling, the temperature of the cookie t minutes after it has been taken out of the oven is given by T t ( ) = 6 + 39e.5t. What is the average value of the cookie during its first minutes out of the oven? 3
5. We know as the seasons change so do the length of the days. Suppose the length of the day varies sinusoidally with time by the given equation t L( t) = 3cos 8, where t the number of days after the winter solstice (December, 7). What was the average day length from January, 8 to March 3, 8? 6. During one summer in the Sunset, the temperature is modeled by the function T t ( ) = 5 +5sin t, where T is measured in F and t is measured in hours after 7 am. What is the average temperature in the Sunset during the sihour Chemistry class that runs from 9 am to 3 pm? Use Part I of the Fundamental Theorem of Calculus to find the derivative of the function. 7. g( y) = t sintdt 8. g() = + tdt y 9. F() = cos(t )dt. h() = arctantdt. y = 3 cost t dt. Homework Set B. e + d 5. sin y dy 3. Find the average value of f t 5. ( + 5 + 6 )d 5. answer. 3 e + 3 ( ) = t t + 5 on t, d, show your work, you may use the calculator to verify your 33
3 6. cos y dy, show your work, you may use the calculator to verify your answer. e 3 + 7. d, show your work, you may use the calculator to verify your 5 answer. 8. Find the average value of f t 3 ( ) = t t + 5 on t,9 9. sin y dy, show your work, you may use the calculator to verify your answer.. + 7 d. d d e ln( t +)dt. If h ( ) = e 5t dt, find h' ( ) 3. d d t ln( t)dt. h m cos m ( ) = t cos t 5 ( )dt, find h' ( m) 5. h( y) = ln y 5 e t t dt, find h' ( y) 6. d d 5 e ( t 3 + t +)dt 3
. Definite Integrals and The Substitution Rule Let s revisit u subs with definite integrals and pick up a couple of more properties for the definite integral. Objectives Evaluate definite integrals using the Fundamental Theorem of Calculus. Evaluate definite integrals applying the Substitution Rule, when appropriate. Use proper notation when evaluating these integrals. ( ) dt E Evaluate t t 3 +. ( t t 3 +) dt = 3t t 3 + 3 = 9 u 3 ( )du = u 3 3 3 ( ) dt 9 = 9 93 3 Let u = t 3 + du = 3t dt ( ) = ( ) = 9 u u = 5 9 ( ( )) d E Evaluate cos. ( cos( )) d = = = cosu du Let u = du = d u( ) = u( ) = = sinu = 35
E 3 e d e d = e u du Let u = du = d ( ) = ( ) = u u = e u = e + e E + a d a > + a d = + a d = + a d = +a a = = u 3 3 a u du +a ( ) 3 a 3 = 3 + a Let u = + a du = d E 5 Find the derivatives of the following functions. (a) F (b) y = ( ) = tanθdθ e sin 3 tdt ( ) = tanθdθ F F ( ) = tan y = sin 3 tdt y = sin 3 e e e 36
(c) F ( ) = tanθdθ ( ) = tanθdθ F = 37
. Homework Set A Evaluate the definite integral, if it eists.. (+ 3 ) 5 d. sec (t / )dt 3. 5. e d. e d 6. e ln 3 sinθ cos θ dθ d + 5 7. sin d 8. cos d Ln 9. Ln e d. cos 5 sin d + e 6. cos θ dθ. f () d where f () =, if < 5, if Find the Average Value of each of the following functions. 3. f ( ) = cos sin on,. g( ) = e on, 5 5. G( ) = ( + 3 on, ) 6. h ( ) = ( + on, ) 38
. Homework Set B 8. sec ( )d. cos d + sin 3. 5 3 6 ( + 5 + 6) d. cos d + sin 6 5. sec ( )d 6. sin y dy + cos y 7. e e ( ) y csc ln y dy 8. msec( m )tan( m )dm 9. cos 9 ( )sin( )d. sec tan 3 d. ye y 3dy 3. If a( t) = cos( t), find v( t) and ( t) if v( ) = 6 and =. 3. If a t ( ) =. ( ) = e 3t + cos( t), find v( t) when v( ) = 3. Then find ( t) when. If f '( θ ) = tan 6 ( θ )sec ( θ ), find f ( θ ) when f 3 8 =. 39
5. If v( t) = sec( 3t )tan( 3t ), find a( t) and ( t) when 6. If f '( ) = cos, find f ( ) when f = = 3 6 7. sec d 8. Find the eact value of 8 e + d 5 9. sin y dy. e 5 dm m + 3m +. e 3 + ( ln( ) ) d. cos 6 sin d 3. e e ( ) y csc ln y dy. msec( m )tan( m )dm 5. cos 9 ( )sin( )d 6. Find the average value of f ( ) = csccot on 8,
.3 Contet for Definite Integrals: Area, Displacement, and Net Change Here are a couple of more properties for the definite integral. Properties of Definite Integrals b. f ( )d = a a b f ( )d a. f ( )d = a b ( ) 3. f d = f d + f d, where a < c < b a c a ( ) b c ( ) Since we originally defined the definite integrals in terms of area under a curve, we need to consider what this idea of area really means in relation to the definite integral. ( ( )) Let s say that we have a function, y = cos like this: on,. The graph looks ( ( )) In the last section, we saw that cos d =. But we can see there is area under the curve, so how can the integral equal the area and equal? Remember that the integral was created from rectangles with width d and height f(). So the area below the -ais would be negative, because the f()-values are negative.
E What is the area under y = ( cos( )) ( ( )) on, We already know that cos d =, so this integral cannot represent the area. As with eample b, we really are looking for the positive number that represents the area (total distance), not the difference between the positive and negative areas (displacement). The commonly accepted contet for area is a positive value. So, ( ) ( ) Area = cos. d? ( ) = cos d cos d. = We could have used our calculator to find this answer: When we use the phrase area under the curve, we really mean the area between the curve and the -ais. CONTEXT IS EVERYTHING. The area under the curve is only equal to the definite integral when the curve is completely above the -ais. When the curve goes below the -ais, the definite integral is negative, but the area, by definition, is positive. Objectives: Relate definite integrals to area under a curve. Understand the difference between displacement and total distance. Etend that idea to understanding the difference between the two concepts in other contets.
E Find the area under y = 3 5 + 6 on,. A quick look at the graph reveals that the curve crosses the -ais at =. 8 y If we integrate y on,, we will get the difference between the areas, not the sum. To get the total area, we need to set up two integrals: ( ) Area = 3 5 + 6 = 3 3 5 + 6 = 3 3 + 9 = 57 ( ) d + 3 5 + 6 3 3 5 + 6 d 3
E 3 Find the area under y = 3 on,. Note that there are three regions here, therefore there will be three integrals: Area = ( 3 ) d + = = 3 ( )+ = 3 ( ) d + + ( 3 ) d F could be a function that describes anything volume, weight, time, height, temperature. F represents its rate of change. The left side of that equation accumulates the rate of change of F from a to b and the right side of the equation says that accumulation is difference in F from a to b. Imagine if you were leaving your house to go to school, and that school is 6 miles away. You leave your house and halfway to school you realize you have forgotten your calculus homework (gasp!). You head back home, pick up your assignment, and then head to school. There are two different questions that can be asked here. How far are you from where you started? And how far have you actually traveled? You are si miles
from where you started but you have traveled miles. These are the two different ideas behind displacement and total distance. Vocabulary: Displacement How far apart the starting position and ending position are. (It can be positive or negative.) Total Distance how far you travel in total. (This can only be positive.) Displacement = Total Distance= b a b a vdt v dt E A particle moves along a line so that its velocity at any time t is v t ( ) = t + t 6 (measured in meters per second). (a) Find the displacement of the particle during the time period t. (b) Find the distance traveled during the time period t. b ( a) vdt = ( t + t 6)dt a = t3 3 + t 6t =.5 b ( b) v dt = ( t + t 6)dt a = t ( + t 6)dt + ( t + t 6)dt = t3 3 t + 6t + t3 3 + t 6t = 6 Note that we used the properties of integrals to split the integral into two integrals that represent the separate positive and negative distanced and then 5
made the negative one into a positive value by putting a in front. We split the integral at t = because that would be where v t ( ) =. E 5 AB 997 # 6
.3 Homework Set A Find the area under the curve of the given equation on the given interval.. y = 3 3 on,. y = 3 + on, 3. y = 3 + on 3, 3. y = ( ) on, cos sin( + sin ) 5. y = + 6. y = + 7. y = sin on, on 3, 3 on., 8. y = 8 on, 9. y = 3sin cos on, 3. y = e 3 on,.5. The velocity function (in meters per second) for a particle moving along a line is v(t) = 3t 5 for t 3. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval.. The velocity function (in meters per second) for a particle moving along a line is v(t) = t t 8 for t 6. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval. ( ) = e on, 3. Find the area under the curve f absolute values in your setup, break it into multiple integrals). (do not use 7
( ) = e on,. Find the area under the curve f absolute values in your setup, break it into multiple integrals). (do not use 5. Find the area under the curve f ( ) = 6. Find the area under the curve g + + cos ( ) on, ( ) = sin on, 8
. Graphical Analysis II An important part of calculus is being able to read information from graphs. Earlier, we looked at graphs of first derivatives and answered questions concerning the traits of the original function. This section will answer the same questions, even with similar graphs but this time the original function in will be defined as an integral in the form of the FTC. This is a major AP topic. Objectives Use the graph of a function to answer questions concerning maimums, minimums, and intervals of increasing and decreasing Use the graph of a function to answer questions concerning points of inflection and intervals of concavity. Use the graph of a function to answer questions concerning the area under a curve. As we saw before, the graphs of a function and its derivatives are related. We summarized that relationship in a table that we will now epand. G()= F( t) dt The y-values of G() Increasing EXTREME Decreasing Concave up POI Concave down F() Area under F() Positive ZERO Negative Increasing EXTREME Decreasing F () Positive ZERO Negative ( ) = f t dt ( ) = f ( ), and the arguments about etremes, concavity and NB. The AP Test will always require the statement that g means that g' increasing and decreasing be made from the point of view of the derivatives. ( ) 9
E The graph of f is shown below. f consists of three line segments. 3 3 Let g ( ) = f t ( )dt. a) Find g( ), g ( ), and g ( ). b) For what values on, c) For what values on, d) Sketch a graph of g [ ]. ( ) is g( ) increasing? Eplain your reasoning. ( ) is g( ) concave up? Eplain your reasoning. ( ) on, ( ) = 3. g( ) would equal the area of the triangle from - to, but the a) g integral goes from to -, therefore the integral equals the negative of the area. g ( ) = f ( ) =. g ( ) = f ( ) = 3. This is the slope of f () at =-. ( ) is increasing when g ( ) is positive. g ( ) = f ( ). Therefore, g( ) is [ ] ( ) concave up when f ( ) is increasing. Therefore, g( ) is concave up on ( ) ( ) = g( ) = g( ) = ( ) has a relative ma at = and a relative min at =. ( ) is increasing on [, ] and decreasing on [, ] [, ] b) g increasing on, c) g, d) Because of the equal triangle areas, g g g 5
g( ) is concave up on (, ) and concave down on (, ). The graph looks something like this: E The graph of f is shown below. f consists of three line segments. 6 - -5 5 ( ). (b) Find g 6 (c) Find g ( ), g ( ), and g ( ). (d) On what intervals is g decreasing? (e) Where is g concave up? (f) Make a sketch of the original function. (g) How do the answers in (a) (e) change if m - - -6 Let g ( ) = f ( t)dt. 3 (a) Find g ( 3). ( ) = + f t ( )dt? 3 (a) Find g ( 3). 3 ( ) = f t g 3 ( )dt We must split this integral into four parts 3 the two triangles above the ais and the two triangles below the - ais. 5
3 ( ) = f t g 3 9 ( )dt = f ( t)dt + f ( t)dt + 8 3 f ( t)dt + f ( t)dt 3 3 9 8 Where did = 9 8 come from? That is the intercept of that line segment. g ( 3) = 6 3 9 6 + 5 = Use the area formula for a triangle to 6 calculate the different areas. Area below the (b) Find g( 6). 6 3 ais must be subtracted. g( 6) = f ( t)dt = f ( t)dt 3 Use the properties of integrals. 6 = Area of triangles triangle above the ais is added to the integral, triangle below the ais is subtracted from the integral. =-3 (c) Find g ( ), g ( ) = f t g ( ), and g ( ). ( )dt = f ( t)dt 3 + f ( t)dt 3 Again, we must split the integral into two parts one for each separate triangle. g ( ) = 6 3 = How did we conclude that 3 as the area of that second triangle? Find the equation of the line making up the hypotenuse of the triangle, plugged in =, and use that number as the height of the triangle. Subtract this number because that triangle was below the ais. Now in order to find g ( ), we must find g ( ). But since g was defined as an integral, the FTC kicks in when we take the derivative of g. 5
g ( ) = d d 3 g ( ) = f ( ) = 3 f ( t)dt = f ( ) Derivatives and integrals are inverse operations the FTC You can read the value from the graph. In order to find g ( ), we must find g ( ). But this follows from what we just did. Since g ( ) = f ( ), we can see that g ( ) = f ( ). g ( ) = f ( ) = 3 The derivative is the slope of f and the slope of that line segment is 3. (d) On what intervals is g decreasing? g is decreasing when g < -- but as stated above g graphically that f is negative from ( 7, 5), 9 8 (e) Where is g concave up? g is concave up when f is increasing. So g is concave up from ( 7, 3) (,3). (f) Make a sketch of the original function. ( ) = f ( ). And we can see. 53
6 - -5 5 - - -6 (g) How do the answers in (a) (e) change if m ( ) = + f t ( )dt? 3 The answers in (a) and the first part of (b) would both need a added to them. The graph in part (e) would be shifted up units. And all other answers would remain the same. 5
E 3 Let h ( ) = g t ( )dt. The graph of g is shown below. 5 5 (a) Find h( ). (b) Find h( ), h ( ), and h ( ). (c) What is the instantaneous rate of change of h() at =? (d) Find the absolute minimum value of h on 6, 8. (e) Find the coordinate of all points of inflection of h on 6, 8 (f) Let F ( ) = g t ( )dt. When is F increasing? ( ). (a) h( ) = g( t)dt would equal the area from to. This is one quarter of the circle with radius. So, h( ) = g( t)dt =. (b) h ( ) = g t ( ) = d g t d ( ) = g' ( ) = h h ( )dt = ( )dt = g ( ) = 55
(c) The instantaneous rate of change of h() at = would be g(). To find g(), we would need the equation of the circle. Since I t has its center at the origin and radius =, the equation is + y = 6 or y = 6. g ( ) = 6 = 5 (d) Since g ( ) = h' ( ), the critical values of h would be the zeros and endpoints of g, namely, = -6, -6,, 9. The graph of g() yields a sign pattern: g + + 6 8 From this, we can see that = - and cannot be minimums. = is a maimum because the sign of g ( ) = h' ( ) switches from + to - and = - is not an etreme because the sign does not change. Now all we need are h(-6) and h(-9). Whichever is smaller is the absolute minimum. 6 ( ) = g t h 6 ( )dt = g ( t)dt 6 = ( + ) 9 h( 9) = g ( t)dt = h(-6) is the smaller value so the absolute minimum is + (e) g' ( ) ( ) = h" ( ) so the points of inflection on h would be where the slope of g changes sign, namely at = -, and 6. (f) ( ) = g t F namely on, 8 ( )dt. F '( ) = g ( ). So F is increasing when g is negative, 56
. Homework Handout of AP Questions. BC B #. BC 999 #5 3. BC 3 #5. BC # 5. BC 9B #5 6. Let g ( ) = f ( t)dt for t 7, where the graph of the differentiable function f is shown below. 5 a) Find g ( 3), g ( 3), and g ( 3). b) Find the average rate of change of g c) For how many values of c, where < c < 3, is g c found in part (b)? Justify your answer. ( ) on c 3? ( ) equal to the average rate 57
d) Find the -coordinate of each point of inflection of the graph of g on the interval < t < 7. Justify your answer. 7. Let g( ) = f ( t)dt for 7 t 7, where the graph of the differentiable function f is shown below. 5 5 a) Find g( ), g ( ), and g ( ). b) Find the average rate of change of g ( ) on 7 t? c) At what -values is g( ) decreasing and concave up? Justify your answer. d) Find the -coordinate of the absolute minimum of g( ). Justify your answer. 8. Let g( ) = f ( t)dt for 7 t 7, where the graph of the differentiable function f is shown below. 3 8 6 6 3 58
a) Find g( ), g ( ), and g ( ). b) Find the equation of the line tangent to g( ) at =. c) At what -values is g( ) decreasing and concave up? Justify your answer. d) Find the -coordinate of the absolute maimum of g( ). Justify your answer. 9. Let g( ) =+ f ( t)dt for t 8 and let f ( t) be the differentiable function (shown below ) comprised of two horizontal line segments and one cycle of the cosine wave y = cos ( t ). 6 8 3 a) Find g( ), g ( ), and g ( ). b) Find the average rate of change of g ( ) on t 8? c) Find the average value of f ( ) on t 8. d) Find the absolute minimum of g( ). Justify your answer. 59
.5 Approimate Integration Riemann and Trapezoidal Sums We have been focusing on anti-derivatives of functions where the equation is known. But let us suppose we need to evaluate an integral where either the function is unknown of cannot be anti-differentiated such as 3 e d. If we had some eact y-values, we could approimate the area geometrically. Mathematicians came up with this technique of dividing the area in question into rectangles, and then finding the area of each rectangle. Why rectangles? Because their area formula was so simple. Objectives: Find approimations of integrals using different rectangles. Use proper notation when dealing with integral approimation. E Use a left end Riemann sum with four equal subintervals to approimate 5 f ( )d given the table of values below. 3 5 f ( ) 9 6 5 Let s first figure out the width of our rectangles. If we need four rectangles in between = and = 5, then our rectangles will be unit wide. The height will be determined by how high the function is for each rectangle. y 5 3 3 5 6 6
5 d + + 3 + = 3 **Notice the use of and = E Use a right end Riemann sum with four rectangles to approimate from E. Again, our rectangles would be unit wide. 5 f ( )d y 5 3 3 5 6 5 d + 3 + + 5 = 5 **Notice the use of and = These two approimations are very different, and both aren t too close to the true value of the integral you remembered how to find 5 d using the FTC 5 d =.333 3 How could we find a better approimation? Let s try using midpoints. 6
E 3 The function described in E is actually f ( ) =. Use a midpoint Riemann sum with four rectangles to approimate d. 5 y 5 3 3 5 6 Because we now have the function, we can determine the y-values that would be the midpoint values if they had been on the table. They would be: y = (.5), (.5), ( 3.5), and (.5) Yet again, our rectangles would be unit wide and the approimation would be: 5 d.5 ( ) + (.5) + ( 3.5) + (.5) = This is a much better approimation than either left-hand or right-hand rectangles. Would our approimation get more accurate or less accurate if we added in more rectangles? Think about this graphically. 6
Steps to Approimating an Integral with Rectangles:. Determine the width of your rectangle.. Determine the height of your rectangle whether that be using the left endpoint, right endpoint, midpoint, or any other height requested. 3. Calculate the areas of each rectangle.. Add the areas together for your approimation. 5. State answer using proper notation. E Use the midpoint rule and the given data to approimate the value of.6 f ( )d. f ( ) f ( ) 7.5.6 7.7. 7.3. 7.9.8 7..6 8.. 7.3.6 f ( )d.8 7.3 ( ) +.8( 7.3) +.( 7.9) = 9.58 A geometric alternative to rectangles would be to use trapezoids. E 5 Using the table of data from E, approimate trapezoids..6 f ( )d with 6 Keep in mind the area formula for a trapezoid is A Trap = ( b + b ) h, where b and b are the parallel sides of the trapezoid. Each trapezoid must have a height equal to the change in and the bases must be the y-values (see the illustration below). 63
8 y 7 6 5 3.5..5..5 These look much more like trapezoids if you turn the page sideways then it becomes obvious that the height is Δ, and that the y-values are the bases: Δ =. b = 7.5 b = 7.3.6 f ( )d (. )( 7.5+ 7.3)+ (. )( 7.3+ 7.)+. ( )( 7. + 7.3) + (. )( 7.3+ 7.7)+ (.5 )( 7.7+ 7.9)+ (.5 )( 7.9+8.) = 9.635 This isn t far off from the true answer for the integral the FTC way. Notice what happened when we did our last calculation do you see how each trapezoid has a and a. in its area formula, the outer endpoints showed up once where the inner endpoints showed up twice. 6
This can be summed up in the Trapezoidal Rule formula feel free to memorize this formula, or just derive it when you come across these problems. The Trapezoidal Rule* b a f ( )d b a n f ( ) + f ( ) + f ( ) +...+ f ( n ) + f ( n ) *The Trapezoidal Rule requires equal sub-intervals E 6 Estimate the area under the graph in the figure by first using Midpoint Rule and then by the Trapezoidal Rule with n = 5. 6 y 5 3 3 5 6 7 A Mid (.5) +.5 ( ) + ( 3) +.5 ( ) + (.5) = 8.75 A Trap ( + ) + ( + ) + ( + ) + ( +.5 ) + (.5+ ) = 9 65
E 7 The following table gives values of a continuous function. Approimate the average value of the function using the midpoint rule with 3 equal subintervals. Repeat the process using the Trapezoidal Rule. 3 5 6 7 f ( ) 3.69.78 6.8 9.389.8.86 35.5 Midpoint Rule: f avg = 7 7 f ( )d 7 (.78) + ( 9.389) + (.86) =.6 Trapezoidal Rule: f avg = 7 7 =.77 7 f ( )d 7 ( ) + ( 6.8) + ( ) + (.8) + (.86) + 35.5 3.69 +.78 9.389 E 8 Let us assume that the function that determined the values in the chart above is f ( ) = + e.5. Calculate the average value of the function and compare it to your approimations. (Can you figure out ahead of time whether our approimation will be an overestimate or an underestimate?) f avg = 7 7 ( + e.5 )d =.89 As noted in the Trapezoidal Rule footnote, the Trapezoidal Rule in the stated form only works for equal subintervals. This is because the Trapezoidal Rule as stated is a factored form of setting up the problem using the trapezoidal area formula from Geometry: Trapezoid Area: A = h ( b + b ) 66
E 8 Let us assume that the function that determined the values in the chart above.5 is f ( ) = + e. Calculate the average value of the function and compare it to your approimations. (Can you figure out ahead of time whether our approimation will be an overestimate or an underestimate?).5 ( ) 7 favg = + e d =.89 7 If we want to figure out if our approimations are overestimates or underestimates, we have to look at the graph of the function. y 3 3 5 6 7 Since this is concave up throughout, the area of the trapezoid would be larger than the area under the curve (see the trapezoid above for an eample, it is just slightly above the curve). In looking at whether an approimation is an over- or under- estimate, we always look at the concavity and compare whether our rectangles (or trapezoids) are above the curve or below it. If they are above the curve, we have overestimates, below the curve, we have underestimates. 67
As noted in the Trapezoidal Rule footnote, the Trapezoidal Rule in the stated form only works for equal subintervals. This is because the Trapezoidal Rule as stated is a factored form of setting up the problem using the trapezoidal area formula from Geometry: Trapezoid Area: A = h ( b + b ) E 9 Given the table of values below, find an approimation for 6 vtdtusing 3 midpoint rectangles. a) ( ) 8 v t dt using trapezoids. b) ( ) vtdtusing left Riemann rectangles. c) ( ) Make sure you indicate the units for each. t (in sec.) v(t) (in m/sec.) 3 5 6 7 8 8 7 3 8 6 6 a) ( ) ( ) ( ) ( ) 8 vtdt + + 8 = 6 meters b) ( ) ( ) ( ) ( ) ( ) vtdt 3+ 8 + 8+ 6 + 6+ + + = 3.5 meters c) ( ) ( ) ( ) ( ) ( ) vtdt 8 + + 7 + = 7 meters 68
The units were in meters for all of them by virtue of the multiplication: the widths were always in seconds, and the heights were in meters per second. When we multiply these quantities, the seconds cancel. E The function g is continuous on the closed interval [,] and has values shown on the table below. Using the subintervals [,5], [5,], and [,], what is the approimation of ( ) sum? g d found by using a right Riemann 5 g()..8 3. 3. (A) 9.6 (B) 3. (C) 3.3 (D) 37. (E) 39. Remember from Chapter : Tangent line approimations are an overestimate if the curve is concave down (since your tangent lines will be above the curve). Tangent line will be an underestimate if the curve is concave up (since your tangent lines will be below the curve). Add to that: If a function is increasing, Right-hand Riemann Rectangle approimations are an overestimate, while Left-hand Rectangle are an under-approimation. If a function is decreasing, Right-hand Riemann Rectangle underestimate are an underestimate, while Left-hand Rectangle are an overestimate. 69
.5 Homework Set A. The velocity of a car was read from its speedometer at -second intervals and recorded in the table. Use the Midpoint Rule to estimate the distance traveled by the car. t(s) v (mi/h) t(s) v (mi/h) 6 56 38 7 53 5 8 5 3 58 9 7 55 5 5 5. The following table gives values of a continuous function. Estimate the average value of the function on, 8 using (a) Right-Hand Riemann rectangles, (b) Left-Hand Riemann rectangles, and (c) Midpoint Riemann rectangles. 3 5 6 7 8 F() 5 7 3-5 8-3. Below is a chart showing the rate of a rocket flying according to time in minutes. Use this information to answer each of the questions below. t (in minutes) 3 5 6 v(t) (in km/min) 3 8 3 8 5 8 8 7
6 a) Find an approimation for v( t) dtusing midpoint rectangles. Make sure you epress your answer in correct units. 3 b) Find an approimation for v( t) dtusing trapezoids. Make sure you epress your answer in correct units. 6 c) Find an approimation for v( t) dtusing left rectangles. Make sure you epress your answer in correct units. 3 d) Find an approimation for v( t) dtusing right rectangles. Make sure you epress your answer in correct units.. Below is a chart showing the rate of water flowing through a pipeline according to time in minutes. Use this information to answer each of the questions below. t (in minutes) 8 6 3 8 V(t) (in m 3 /min) 6 3 3 9 6 8 a) Find an approimation for ( ) epress your answer in correct units. 6 b) Find an approimation for ( ) you epress your answer in correct units. V t dt using midpoint rectangles. Make sure you V t dt using right Riemann rectangles. Make sure 5. Below is a chart of your speed driving to school in meters/second. Use the information below to find the values in a) and b) below. 7
t (in seconds) 3 9 3 36 v(t) (in m/sec) 3 38 3 36 a) Find an approimation for ( ) you epress your answer in correct units. b) Find an approimation for ( ) your answer in correct units. v t dt using left Riemann rectangles. Make sure v t dt using trapezoids. Make sure you epress 6. Below is a chart showing the velocity of the Flash as he runs across the country. Use this information to answer each of the following. t (in seconds) 8 6 W(t) (in km/second) 5 9 8 7 a) Find an approimation for ( ) v t dtusing midpoint rectangles. Make sure you epress your answer in correct units. Describe what this integral means. 6 b) Find an approimation for ( ) epress your answer in correct units. v t dtusing Trapezoids. Make sure you 7
7. The graph below shows energy use in kilowatts per hour for a community for June 5. Which of the following best approimates the number of kilowatts that were used on that day? 5 Kilowatts per Hour 5 3 6 9 5 8 7 Hours a) 3 b) 35 c) 36 d) e) 7 8. Below is a chart showing the rate of sewage flowing through a pipeline according to time in minutes. Use this information to answer each of the questions below. t (in minutes) 6 3 5 V(t) (in gallons/min) 83 68 8 38 3 68 a) Find an approimation for ( ) your answer in correct units. b) Find an approimation for ( ) you epress your answer in correct units. V t dt using trapezoids. Make sure you epress V t dt using left Riemann rectangles. Make sure 73
9. AP Handout: AB 998 #3, AB #, BC7 #5. Use (a) the Trapezoidal Rule and (b) the Midpoint Rule to approimate the given integral with the specified value of n. + d,n = 8. Use (a) the Trapezoidal Rule and (b) the Midpoint Rule to approimate the ln + d with n =.. If w (t) is the rate of growth of a child in pounds per year, what does 5 w'(t) dt represent? 3. If oil leaks from a tank at a rate of r(t) gallons per minute at time t, what does r(t)dt represent?. If is measured in meters and f() is measured in newtons, what are the units for f ()d?.6 Homework Set B. Use the trapezoidal rule with n = to approimate the value of do you think that approimations would be necessary for evaluation of this integral? e d. Why. Given the table of values below for velocities of a glider, find an approimation for the displacement of the glider using a left Riemann sum, a right Riemann sum, and a trapezoidal sum (make sure you include units in your answer). Which do you think is the most accurate? Eplain. 7
t (minutes) 5 3 9 V(t) (m/min) 5 55 35 5 3. The following table is for a continuous function, f ( ). Use the information in the table to find each of the following: 3 5 6 7 8 9 f ( ).7 3..... -.3 -.5 -.. 5 a. Approimate f ( ) d using a right approimation 5 b. Approimate f ( ) d using a left approimation 6 ( ) c. Approimate f d using a midpoint approimation 6 ( ) d. Approimate f d using a trapezoidal approimation 9 ( ) e. Approimate f d using a right approimation 9 ( ) f. Approimate f d using a left approimation 9 ( ) g. Approimate f d using a midpoint approimation 9 ( ) h. Approimate f d using a trapezoidal approimation. Given the table of values below, find each of the following: t in seconds R(t) in meters/second 5 5 7 5 3 6 5 8 3 75
3 a) Find a left Riemann approimation for ( ) R t dt 5 b) Find a left Riemann approimation for ( ) R t dt 3 c) Find a right Riemann approimation for ( ) R t dt 5 d) Find a right Riemann approimation for ( ) R t dt 3 e) Find a midpoint Riemann approimation for ( ) R t dt 3 f) Find a midpoint Riemann approimation for ( ) 3 g) Find a trapezoidal approimation for ( ) R t dt R t dt 76
.6 The Mean Value and Rolle s Theorems The Mean Value Theorem is an interesting piece of the history of Calculus that was used to prove a lot of what we take for granted. The Mean Value Theorem was used to prove that a derivative being positive or negative told you that the function was increasing or decreasing, respectively. Of course, this led directly to the first derivative test and the intervals of concavity. Mean Value Theorem If f is a function that satisfies these two hypotheses. f is continuous on the closed interval a,b. f is differentiable on the closed interval a,b ( ) Then there is a number c in the interval ( a,b) such that f '( c) = f b ( ) f ( a) b a. Again, translating from math to English, this just says that, if you have a smooth, continuous curve, the slope of the line connecting the endpoints has to equal the slope of a tangent somewhere in that interval. Alternatively, it says that the secant line through the endpoints has the same slope as a tangent line. a c b MEAN VALUE THEOREM 77
E Show that the function f ( ) = 3 +,, 3 conditions of the Mean Value Theorem and find c. satisfies all the Polynomials are continuous throughout their domain, so the first condition is satisfied. Polynomials are also differentiable throughout their domain, so the second condition is satisfied. f '( c) = 3c 8c 3c 8c = 3c 8c += c = 8 ± 8 3 3 f ( 3) f ( ) = 3 ( ) ( )( ) ( ) c =.535 or.3 8 ( ) = Rolle s Theorem is a specific a case of the Mean Value theorem, though Joseph- Louis Lagrange used it to prove the Mean Value Theorem. Therefore, Rolle s Theorem was used to prove all of the rules we have used to interpret derivatives for the last couple of years. It was a very useful theorem, but it is now something of a historical curiosity. Rolle s Theorem If f is a function that satisfies these three hypotheses. f is continuous on the closed interval a,b. f is differentiable on the closed interval a,b ( ) 3. f(a) = f(b) Then there is a number c in the interval ( a,b) such that f '( c) =. 78
Written in this typically mathematical way, it is a bit confusing, but it basically says that if you have a continuous, smooth curve with the initial point and the ending point at the same height, there is some point in the curve that has a derivative of zero. If you look at this from a graphical perspective, it should be pretty obvious. 6 a b E Show that the function f ( ) = +,, of the Mean Value Theorem and find c. satisfies all the conditions Polynomials are continuous throughout their domain, so the first condition is satisfied. Polynomials are also differentiable throughout their domain, so the second condition is satisfied. ( ) f ( a) ( ) f ( ) f ' c f ' c = E 3 BC B # ( ) = f b b a f ' c ( ) = f ( ) = c = c = 79
E Suppose that f possible value for f ( )? ( ) = 3, and f '( ) 5 for all values of. What is the largest We know that f is differentiable for all values of and must, therefore, also be continuous. So we just make up an interval to look at using the mean value theorem. The interval will be [, ] because we are looking at f ( ). and f f '( c) = f ( ) f ( ) = f ( ) 3 f '( c) = f ( )+ 3 f '( c) 3= f ( ) Since we know the maimum value for f ' c we get ( ) 3= f ( ) ( ) = 7 ( ) is 7. 5 f Therefore the maimum possible value for f ( ) ( ) ( ) is 5, plug in 5 for f '( c), and Other Theorems that can be confused with The Mean Value Theorem The Average Value Theorem: The average value of a function f on a closed interval [a, b] is defined as f avg = b f ( )d b a. a Average Rate of Change: The average value of a function f on a closed interval [a, b] is defined as = f ( b ) f ( a). b a The Intermediate Value Theorem: If f is continuous on the closed interval [a, b] and f (b) and f (a) are opposite signs, there is a zero in the closed interval. 8
.6 Homework Verify that the following functions fit all the conditions of Rolle s Theorem, and then find all values of c that satisfy the conclusion of Mean Value Theorem.. f ( ) = 3 3 + + 5,,. f ( ) = sin 3. g t ( ),, ( ) = t t + 6,, Given the graph of the function below, estimate all values of c that satisfy the conclusion of the Mean Value Theorem for the interval [,8].. y 3 3 5 6 7 8 9 3 8
5. Given the graph of the function below, estimate all values of c that satisfy the conclusion of the Mean Value Theorem for the interval [,9]. y 3 3 5 6 7 8 9 3 AP Handout 6. AP Calc AB/BC B #3 7. AP Calc AB #6 8
.7 Accumulation of Rates Beginning in, AP shifted emphasis on understanding of the accumulation aspect of the Fundamental Theorem to a new kind of rate problem. Previously, accumulation of rates problems were mostly in contet of velocity and distance, though the 996 Cola Consumption problem hinted at the direction the test would take. The Amusement Park Problem of caught many students and teachers off guard. Almost every year since then, the test has included this kind of problem. Here is an eample similar to the Amusement Park Problem: E The rate at which people enter a park is given by the function 56 Et ( ) = t t + 6, and the rate at which they are leaving is given by 989 Lt ( ) = 76 t 38t+ 37. Both Et ( ) and Lt ( ) are measured in people per hour where t is the number of hours past midnight. The functions are valid for when the park is open, 8 t. At t = 8 there are no people in the park. a) How many people have entered the park at pm (t = 6)? Round your answer to the nearest whole number. b) The price of admission is $36 until pm (t = 6). After that, the price drops to $. How much money is collected from admissions that day? Round your answer to the nearest whole number. t c) Let H ( t) = E( ) L( )d for 8 t. The value of ( 6) 8 nearest whole number is 53. Find the value of '6 ( ) of H ( 6) and H '6 ( ) in the contet of the amusement park. H to the H and eplain the meaning d) At what time t, for 8 t, does the model predict the number of people in the park is at a maimum. a) Since ( ) Et is a rate in people per hour, the number of people who have entered the park will be an integral from 8 to 6. 83
Total entered= 6 8 E( t)dt = 66 people b) Since there is an entry fee per person, and we saw in the first part that the integral gave us the number of people the total revenue that the park gets from admissions just means multiplying the number of people by the admission charge. Total entered at $36/person = Total entered at $/person = 6 8 6 E( t)dt E( t)dt = 66 people =88 people Total revenue = $36(66) + $(88) = $56,696 c) Since ( ) = ( ) ( ) t H t E L d, we can use the Fundamental Theorem of 8 Calculus to determine the derivative. H t ( ) = d dt t E( ) L( )d 8 = E( t) L( t) = E( 6) L( 6) = But we still need to interpret the meaning of the numbers. ( ) H 6 = 53 people since it was defined as an integral, we knew that integrating the rates gave us people. This is how many people are in the park at that moment. H '6 ( ) = people per hour since the original equations we were provided were rates, and H' ( t) = E( t) L( t), so this is how many people are entering the park at t = 6. Since the value is positive, this is the rate at which the number of people in the park is increasing. d) To find the maimum number of people, we have to set the derivative, H'( t ), equal to zero. We should also check the endpoints, which are also 8
critical values (we could just do a sign pattern instead to verify that the zero is the maimum). ( ) ( ) ( ) H' t = E t L t = Using a graphing calculator, we find the zero is at t = 6.6 Therefore we have critical values at t =, 6.6, and. When t =, H = (this was given at the beginning of the problem) When t = 6.6, H = 53 When t =, H = 53 Objective Analyze the interplay between rates and accumulation in contet. E A certain industrial chemical reaction produces synthetic oil at a rate of 5t St ( ) =. At the same time, the oil is removed from the reaction vessel by a + 3t skimmer that has a rate of Rt ( ) = + 5sin t 5. Both functions have units of gallons per hour, and the reaction runs from t = to t = 6. At time of t =, the reaction vessel contains 5 gallons of oil. a) How much oil will the skimmer remove from the reaction vessel in this si hour period? Indicate units of measure. b) Write an epression for Pt, ( ) the total number of gallons of oil in the reaction vessel at time t. c) Find the rate at which the total amount of oil is changing at t =. d) For the interval indicated above, at what time t is the amount of oil in the reaction vessel at a minimum? What is the minimum value? Justify your answers. 85
a) Amountremoved = +5sin 5 t b) P t t ( ) = 5 + S 6 ( ) R ( ) c) S ( ) R( ) =.99 gal/hr. d d) Critical values occur when S t graphing calculator shows the time when S t t dt = 3.86 gallons ( ) = R( t) and at the endpoints. Our ( ) = R( t) is t = 5.7. P( t) 5 5.7 9.367 6 93.77 The minimum is 9.367 gallons and occurs at t = 5.7. 86
.7 Homework Handout of AP Questions. AB/BC B #. AB 5B # 3. AB 6 #. BC # 5. BC 5 # 6. The number of parts per million (ppm), C(t), of chlorine in a pool changes at. t the rate of P' ( t) = 3e ounces per day, where t is measured in days. There are 5 ppm of chlorine in the pool at time t =. Chlorine should be added to the pool if the level drops below ppm. a) Is the amount of chorine increasing or decreasing at t = 9? Why or why not? b) For what value of t is the amount of chlorine at a minimum? Justify your answer. c) When the value of chlorine is at a minimum, does chlorine need to be added? Justify your answer. 7. The basement of a house is flooded, and water keeps pouring in at a rate of wt ( ) = 95 tsin t ( 6) gallons per hour. At the same time, water is being pumped out at a rate of r ( t ) = 75sin t ( ). When the pump is started, at time t =, there is 3 gallons of water in the basement. Water continues to pour in and be pumped out for the interval t 8. a) Is the amount of water increasing at t = 5? Why or why not? b) To the nearest whole number, how many gallons are in the basement at the time t = 8? 87
c) At what time t, for t 8, is the amount of water in the basement at an absolute minimum? Show the work that leads to this conclusion. d) For t > 8, the water stops pouring into the basement, but the pump continues to remove water until all of the water is pumped out of the basement. Let k be the time at which the tank becomes empty. Write, but do not solve, an equation involving an integral epression that can be used to find a value of k. 8. A tank at a sewage processing plant contains 5 gallons of raw sewage at time t =. During the time interval t hours, sewage is pumped into the tank at the rate Et ( ) = +. During the same time interval, sewage is + ln t + ( ) t pumped out at a rate of Lt ( ) = sin. 7 a) How many gallons of sewage are pumped into the tank during the time interval t hours? b) Is the level of sewage rising or falling at t = 6? Eplain your reasoning. c) How many gallons of sewage are in the tank at t = hours? d) At what time t, for t, is the volume of the sewage at an absolute maimum? Show the analysis that leads to your answer. If the sewage level ever eceeds 5 gallons, the tank overflows. Is there a time at which the tank overflows? Eplain. 9. At an intersection in San Francisco, cars turn left at the rate t Lt ( ) = 5 tsin cars per hour for the time interval t 8. 3 a) To the nearest whole number, find the total number of cars turning left on the time interval given above. b) Traffic engineers will consider turn restrictions if Lt ( ) equals or eceeds 5 cars per hour. Find the time interval where Lt ( ) 5, and find the average value of Lt ( ) for this time interval. Indicate units of measure. 88
c) San Francisco will install a traffic light if there is a two hour time interval in which the product of the number of cars turning left and the number of cars travelling through the intersection eceeds 6,. In every two hour interval, 8 cars travel straight through the intersection. Does this intersection need a traffic light? Eplain your reasoning. 89
Definite Integral Chapter Test. Find 6 d a. b. 8 c. 7 d. e. 8. 5 d 3 + a. b. 3 c. d. e. 6 3. t dt a. b. 3 c. 6 d. e.. The solution to the differential equation dy = 8y with the initial condition d y() = 5 is a. ln( + 5) b. e + 5 c. e + d. 5ln( ) e. 5e 9
5. Which of the following is equal to cos d? a. sin d b. cos d c. sin d d. sin d e. cos d 3 6. Consider the function f ( ) is shown below. If the Trapezoidal Rule is used with n = to approimate ( ) f d the result is 9 a. b. c. 3 d. e. 5. The figure below shows the graph of f, the derivative of a twicedifferentiable function f, on the closed interval 8. The graph of has horizontal tangent lines at =, = 3, and = 5. The areas of the regions between the graph of and the -ais are labeled in the figure. The function is defined for all real numbers and satisfies f 8 ( ) =. 9