CHAPTER 13. Definite Integrals. Since integration can be used in a practical sense in many applications it is often


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1 7 CHAPTER Definite Integrals Since integration can be used in a practical sense in many applications it is often useful to have integrals evaluated for different values of the variable of integration. Frequently we wish to integrate an epression between some limits. The practical significance of this will be eplained later in this chapter. For eample, if we wish to integrate + between = and = then we write + d which is " # + $. % & Mathematically this means find the value of + when = and then subtract the value of + when =. i.e. d ( ). " # $ + = ( + = % & + ' = 6 ),  * +. Similarly & d " $ # % ( ) = = = 7 + " % & And also ' d = '# + $ d = [ + ln ] b = ( + ln ) ( + ln) = + ln. In general " f ( ) d = F ( b) F ( a) where F '( ) f ( ) a =.
2 8 This is called the Fundamental Theorem of Calculus. Note that it is not necessary to include the constant of integration since the subtraction cancels out that constant. Such integrals are called DEFINITE INTEGRALS because we are substituting definite values of. Worksheet DEFINITE INTEGRALS. Evaluate the following definite integrals. You may use a calculator. a) 5 " d b) " 8 sin d c) + d d) d e) + d f) d g) d h) d i) 8 + d j) 8 d + k) + + d l) + + e + e d # " sin d n) m) cos e d " o) e " + + d p) " sin d q) d r) ( + ) " + sin d
3 9 " cos d t) sin " cos e d u) s) sin " sec + tan d v) + d w) + " d Answers to Worksheet. a) b).5 c).86 d).69 e).657 f) g).685 h).696 i).79 j).655 k).986 l).6 m).69 n) 9 o).656 p) q) r).95 s).55 t).78 u) v).87 w) 6
4 Worksheet. Evaluate the following integrals without using a calculator. a) d " b) " sin ( ) d c) e) e " d d) + d f) e " e d ln + d. dy d = and 5 y = when =. Find y when =. dy. d = + and y = when =. Find y when =.. A particle starts with an initial velocity of feet per second. Its acceleration is ( t + ) feet/sec, where t is the number of seconds from the start. Find the velocity after seconds. 5. Evaluate the following integrals without a calculator: a) log d b) e " + + d c) + d ( + ) Answers to Worksheet. a) + ln b) c) d) + e) e 7 f) e a) ln b) e c)
5 Applications of Definite Integrals to Area Eample Question: Find the area between the ais, the graph of y = and =. Answer: We are trying to find the area of "#$. By elementary geometry: Area OA AB 8 "#$% = = = 6. Note also that & d = $ " % = # = 6.
6 Eample Question: Find the area bounded by y =, =, = 7, and the ais. Answer: We wish to find the area of AD + BC " 6 + " ABCD = $ %# AB = $ %# =. & ' & ' Now consider 7 7 & d = $ " % = 9 # 9 = also. It appears as though area is related to the definite integral.
7 Area Under a Curve as a Definite Integral Let f ( ) be a positive continuous function as shown below. We will try to find area under the curve bounded by the y ais, y = f ( ) and the ais from to i.e. we are trying to find Area O AB C. Let A( ) represent the area. To evaluate A( ) for different values of it is helpful to investigate the derivative of A( ). By First Principles, the derivative of A( ) ( + )# ( ) A A = A' ( ) = lim " Area O AE D # Area O ABC = lim " = lim " Area C BE D
8 Now consider the region C BE D. Remember that CD = is considered as a small change in. If we think of C BE D as a body of sand whose upper edge is BE then it is clear that BE can be smoothed out horizontally so that area of C BE D = area of rectangle CFGD (see above). This is true also by the Intermediate Value Theorem. It is also clear that FG intersects BE at some point P whose coordinate lies between and +. Let P be (t, f ( t )) where t + ". Area C BE D = Area CFGD = CD ( y coordinate of P ) = " f ( t). Substituting into on the previous page, it follows that A' ( ) ( ) " # f t = lim " " ( ) = lim f t In the limiting case as ", t. A' ( ) = f ( ). It follows that ( ) ( ) A = f d.
9 5 To find the area under a curve y = f ( ), bounded by = a, = b and the ais we note that this is represented by region S in the diagram: b Area of region S is therefore ( ) f d. a Comment a Students often think that f ( ) d is the area of region R and ( ) of region R + S but strictly speaking this is not true. Area region ( ) a R = f d + k for some constant k And similarly Area region ( ) It does however remain true that b Area ( ) a b b f d is the area R + S = f d + the same constant k S = f d since the constant k cancels out.
10 6 Eample Find the area under y = from = to = above the ais. Area " = ( d = $ % = # = = 8 & ' Eample Find the area bounded by y =, =, = and the ais i.e. find + the area of shaded region R in the picture below:
11 7 Area = " d = #$ Arctan% & + = Arctan Arctan ( ) " # = $ % $ & ' ( 7 =. To find the area of a region whose boundaries involve more comple curves it is helpful to consider a thin strip procedure as follows. Eample Find the area bounded by y = and y =. Note that the intersection points O and A are (,) and (,) respectively. Draw a thin vertical strip in the region whose area is to be evaluated.
12 8 Consider the thin strip as though it were a rectangle whose area is ( = )# ( = ) y coordinate on y y coordinate on y " times d $ % ( Y ) = y d = ( )d To evaluate the area of the region we need to add up an infinite number of thin strips whose width d tends to the limit of. This is effected by " d because integration is a process of adding an infinite number of values. Note that the thin strips vary from = to = which are the limits of integration for the variable. Note also that d has changed its role from width of the strip to the variable of integration for reasons beyond the scope of this tet. Area of required region = ( )d " = ( % " = # & % $ ' % % * ) +,  = ( ) Note that in a later chapter a similar process will be used to evaluate volumes.
13 9 Sometimes it is more convenient (and even perhaps required) to draw thin horizontal strips to evaluate an area. Eample Find the area (in the first quadrant) bounded by y =, y =, and y =. Note that point A, the intersection of y = and y = is (,). In this eample it is better to draw a thin horizontal strip because using vertical strips would necessitate that the region be divided into two separate parts. The area of the thin horizontal strip is ( X ) dy where X denotes the coordinate of point P on y = and denotes the coordinate of point Q on y =.
14 5 Area of region is ( ) X dy = y dy y " " " 6 # $ # $ = & y % ln y' = ( % ln ) % ( % ln) &, ' * + * + = ln =.8 (appro.) Note also that the thin strips vary from y = to y = and hence are the limits of integration. It is sometimes assumed that area under a curve is simply obtained by integrating the function. Care however and understanding are required as illustrated by the following eample. Eample Consider the function f ( ) = + 6 = ( )( 8) as shown below.
15 5 Let s obtain the area A + A i.e. the area on either side of the ais from = to =. If we evaluate " + 6 d we get " $ % ( ) ( ) # = 8# # # = ( 57) ( 57) = It is clear that the area of region A + A is not zero. In fact for the function chosen, (,) is the point of inflection and the function is symmetric about (,). In this case area A = area A. In fact the required area is. Another eample follows which illustrates how to calculate areas of regions similar to those in the preceding eample. Eample Find the area, both above and below the ais bounded by the ais and a graph of y ( )( ) =.
16 5 Note that it is not correct to say that the area R + S = y d because in region R the thin strip has height y whereas in region S the thin strip has a height of y. It must be remembered that and y refer to the and y coordinates of a point on the graph. The area of region R + the area of region S = y d + " y ( ) d " " = d d " " = $ # + % + $ # + # % & ' & ' ( 6 6) ( ) ( 6 8 6) ( 6 6) = " $ + + #% + " $ + + #% = + = 8 Note that, using a graphing calculator, the area can be evaluated by " d. Sometimes a region will have to be divided into two distinct parts.
17 5 Eample Find the area bounded by y =, y = and y =. It is clear from the graph that the region R cannot be evaluated in one step by using horizontal or vertical strips. Point A is (,), point B (,) and point C can be found by graphing calculator to be (.6,.56) (appro). Considering thin vertical strips it follows that the area of region R.6 " " R = d + d " " = $ # % + $ ln # % & ln ln ' & ln '.6 ".6 " # $ # $ # $ # $ = & ( % ln.6 ln ln ln ) % ( % )' + &( % ) % % ' * ln ln + ln ( ln ) &, * + ' &, * + * + ' = (.76) + (.895) (using a calculator) =.67 (appro.)
18 5 Worksheet. Find the area bounded by the ais, the y ais and + y =.. Find the area bounded by y = +, =, the ais and the y ais.. Find the area under one halfperiod of y = sin.. Find the area under one halfperiod of y = sin. 5. Find the area cut off by the ais, above the ais, from the parabola ( )( ) y = Find the area bounded by y =, the ais, the y ais, and =. 7. Find the area cut off both above and below the ais by ( )( )( ) y =. 8. Find the area bounded by y = sin and y = for " ". 9. Find the area bounded by y = sin, y =, =, and =.. Find the area between. Find the area between y = and y = 8. y = and y =.. Find the area bounded by y = e, y = e, and =.. Find the area between the graphs of y = sin and y = cos between two consecutive points of intersection.. Find the area enclosed by the positive ais and the curve =. y 5. Find the smaller area bounded by = and y =. y
19 55 6. Consider the region R in the first quadrant bounded by the ais, the y ais, and y =. i) Find the area of region R. ii) Find the value of k so that regions of equal area. iii) Find the value of k so that y regions of equal area. iv) Find the value of k so that y of equal area. = k divides the region R into two equal = k divides the region R into two equal = k divides the region R into two regions v) Find the value of k so that y = k divides the region R into two regions of equal area. 7. Find the area between y =, y =, the ais and =. 8. Find the area bounded by y = + and y = Find the area between y = ( ) and y ( ) =.. Find the area between the graphs of y =, and y = ( ).. Find the area in the first quadrant bounded by the graphs of y = and y =.. Find the area lying above the ais of y = sin + cos for " # #.. Find the area bounded by + y =, y = +, and =.
20 56. a) Find the area of the region R, in the first quadrant, bounded by y = 8, the ais and the y ais. b) Find the value of k such that y = k divides the region R into two equal areas. Answers to Worksheet " " 6 6. i) 6 ii) k = iii) k =.8 iv) k =.6 v) k = a) b) 7
21 57 Average Value The average value of a finite set of data is of course the sum of the data divided by the number of items. In the case of average value of a function over a range of values of we are considering an infinite number of values and hence we use a graphical approach. F G To find the average value of f ( ) as varies between a and b we think of region R as though it were sand. The wavy top of the sand from C to D could be smoothed out horizontally so that the area of region R is rearranged into a rectangle indicated by the dotted line. i.e. area AFGB = area ACD B The height of the dotted line i.e. AF is the average value of the function as varies from a to b.
22 58 area AFGB = area ACD B b ( ) ( ) AF b " a = # f d a AF = (average value ) = b " a ( ) f d b a Note that this result would apply even if f ( ) took on negative values. Eample The average value of sin as varies between = and = = = # sin d " [" cos ] = =
23 59 Worksheet AREAS UNDER A CURVE/AVERAGE VALUE. Find the area bounded by y( y) = and + y =.. Find the area between y =, y =, the ais and =.. Find the average value of n as varies from to, where n >.. Find the area bounded by the parabola = 8 + y y, the y ais, y = and y =. 5. Find the area bounded by the parabola y = and the line y =. 6. Find the area enclosed by the graphs = 6 and y =. y 7. Find the average height of a semicircle of radius. 8. Find the finite area enclosed by the three graphs y =, y + =, and y = Find the average value of. Find the average value of as varies between and. as varies between  and.. Find the area bounded by the parabola = + y and the line =.. Find the average value of as varies between and.. Find the average value of sin as varies between and.. Find the area in the first quadrant bounded by y =, y = and =. 5. Find the area bounded by the parabolas = 6 and y y =. y y
24 6 6. Find the area bounded by y =, y =, the ais and 9 + y =. 7. Find the area bounded by = + and y = +. y 8. Find the area bounded by calculator). y 9. Sketch y ( )( 5) = and = (use your graphing y 5 8 =. The line y = cuts this curve at R and at A and B where A is between R and B. Find the area bounded by the curve from R to A and the line y =. Answers to Worksheet n " ln
25 6 Functions Defined As Definite Integrals Sometimes functions are defined in terms of another integral. For eample, f ( ) could equal ( + t ) dt = " $ + % = + # This means that f ( ) t t When the integral can be evaluated easily then f ( ) is defined eplicitly. However f ( ) might be defined as (say) t + t dt In this eample f ( ) cannot easily be defined eplicitly and hence graphing calculators, numerical methods, etc. need to be used to evaluate f for different values of. However, note that it is possible to define f '( ) In fact, f '( ) = +. This can be shown as follows: eplicitly. Let t t dt F ( t + = ) for some function F where F '( t) t t ". It follows that t + t dt = F ( ) F ( ) = +. If we now differentiate both sides of this equation with respect to we get
26 6 " ( ) D t + t dt = F ' = +. Note that the numerical value of the lower limit does not change f '( ). Similarly if ( ) g = t + t dt Then g '( ) F '( ) = times (by Chain Rule) = + Sometimes, f ( ) may be defined as (say) f ( ) = g( t) dt where g( t ) is as shown below. For eample, f and f ( ) = g ( t) dt =, ( ) ( ) ( ) = g( t) dt f = g t dt = Area ABCD = ( ) dt = " g t = "Area #ABO = "
27 6 Worksheet 5. Evaluate the following integrals without the aid of a calculator. a) ln e d b) e + 5 " d d d) c) ( + ) + + d. Find the average value of as varies between and.. Is the average value of f ( ) as varies between a and b the square root of the average value of f ( ) as varies between a and b? Eplain. Calculators are permitted for the rest of the questions on Worksheet 5.. The graph of a differentiable function f on the closed interval [6,6] is shown on the net page. Let ( ) ( ) a) Find G ( ). b) Find G '( ). G = " f t dt for 6 " " 6. 6 c) On which interval or intervals is the graph of G concave down? Justify your answer. d) Find the value of at which G has its maimum on the closed interval [,].
28 6 5. a) Find the area, A, as a function of k, of the region in the first quadrant enclosed by the y ais and the graphs of b) What is the value of A when k =? y = and y = k for k >. c) If the line y = k is moving upward at the rate of units per second, at what rate is A changing when k =? Suppose f ( ) d = 6 and f ( ) d = 7. Find ( ) Suppose that f ( ) d =9, f ( ) d = and ( ) following: 5 a) f ( ) + g( ) d b) 5 ( ) 5 7 % # f + " $ d. g d =. Evaluate the f d c) g( ) + d d) ( ) 7 5 f d
29 65 8. Let ( ) ( ) F = " f t dt, " ", where f is the function whose graph is shown below. The curved paths are semicircles. a) Complete the following table:   F ( ) b) For what intervals is F concave down? c) Does F have an inflection point? If so, at what value(s) of? d) In what interval(s) is F increasing? 9. Let ( ) sin # for " f = $ # % for = and A( ) ( ) = f t dt. a) Sketch the graph of the function f. b) At what values of does the function A have a local minimum? c) Find the coordinates of the first inflection point where > in the graph of A.
30 66. The graph of a function f is shown. Let ( ) ( ) F = " f t dt. a) Find F ( ), F ( ), F ( ) and F ( 5). b) Find F (). c) If ( ) ( ) K = " f t dt, determine K ( ) F ( ). d) When is the graph of F concave down? Answers to Worksheet 5. a) " ln # $ % & b) ln9 c) d) ln. b. No since f ( ) d is not equal to ( ) a b f d.. a) b) c) 6 < < or < < 5 d) = a 5. a) k A = b) c) square units/sec a) 5 b) 95 c) d) 8 8. a)   F ( ) " b) < < OR < < c) = or = d) < <
31 67 9. a) b) Since A' ( ) = f ( ), A will have a local minimum when A' ( ) = f ( ) changes from negative to positive. So there is a minimum at = ". But also, when >, sin will have the desired sign changes at ( k ) = ", ", " 5,..., " +, where k is any positive integer or =,,..., k, where k is any positive integer. c) A" ( ) = d d " # sin $ % &. The smallest positive at which A" ( ) = cos ' sin changes sign occurs when = tan ie.9..9 sin t y A(.9 ) " dt.656. So the approimate coordinates of the t inflection point are (.9,.656).. a),, 7, 5 b) c) d) < < 5
32 68 Worksheet 6 Calculators are allowed. The approimate area of the region enclosed by the graphs of y = e and y = e is (A).7 (B).6 (C).9 (D).97 (E).8. The rate at which ice is melting in a pond is given by = + ( ln( t + ) ) where V is the volume of ice in cubic feet and t is the time in minutes. What amount of ice has melted in the first minutes? (A) 6.7 ft (B) 5.85 ft (C).9 ft (D) 5. dv dt. Consider the function F defined so that F ( ) cos of F ( ) F '( ) + is ft (E).7 ft " t # $ = ) % & dt. The value ' ( (A).5 (B).7 (C).9 (D).7 (E) If f ( ) d = 5 and f ( ) d = then % ( ) # f + " $ d (A) 6 (B) 9 (C) (D) 5 (E) 8 5. Which of the following is a true statement? (A) If f in increasing on the interval [a,b], then the minimum value of f on [a,b] is f ( a ). b (B) The value of ( ) b (C) If f ( ) d > a f d must be positive. a then f is nonnegative for all in [a,b].
33 69 b b b & & & (D) " ( ) ( ) # = ( ) ( ) $ f g % d f d g d a a a (E) None of these. (A) A (B) B (C) C (D) D (E) E 6. Suppose F ( ) = dt for all real ; then '( ) + t F = (A) (B) (C) (D)  (E) 7. If f and g are continuous functions such that g '( ) f ( ) ( ) f d = = for all, then (A) g '( ) g '( ) (B) g '( ) g '( ) (C) g( ) g( ) (D) f ( ) f ( ) (E) f '( ) f '( ) G = # g t dt, " ", where g is 8. Let ( ) ( ) the function graphed in the figure. The value of G ( ) is (A) 5.5 (B).5 (C) (D) (E) None of these
34 7 9. Let f ( ) if < = " $ + if # and let F ( ) = f ( t) dt ". Which of the following statements is true? I. F ( ) = II. F '( ) = 5 III. F "( ) = (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, III 5 5. If % # f ( ) + " $ d =, then ( ) f d = (A) (B) (C) (D) (E) None of these. Let f ( ) =. The area of the region between the ais and the graph of f is: (A) (B) (C) 6 (D) (E) none of these. The area of the region enclosed by the graphs of y = k and k is a positive constant is: y =, where (A) 5k 6 (B) k 6 (C) k 6 (D) 5k 6 (E) none of these. ln " e d (A) (B) e (C) (D) 8.85 (E) 6.8
35 7. Suppose f is a continuous function such that ( ) the following statements is true? I. If f is an even function then ( ) II. If f is an odd function then ( ) 5 III. ( ) " f d =. " f d =. Which of " f d =. " f d =. (A) I only (B) II only (C) I and II only (D) II and II only (E) I, II, III 5 5. If ( ) " + k d =, then k = (A)  (B) (C)  (D) (E) none of these 6. The position of a particle moving along a straight line is given by ( ) = s t t du. The velocity at t = is u + (A) (B) (C) (D) (E)
36 7 7. The figure shows the graph of f. Determine a so that ( ) a " f t dt will be as small as possible. a = (A)  (B)  (C) (D) (E) none of these 8. Suppose that f is a continuous function defined on the interval [,5] and that 5 f ( )d =. Then which of the following is true? 5 5 I. % # f ( ) " $ d = 9 II. # ' f ( ) d$ 9 III. ( ) 5 " = #% $ & f d = (A) I, III only (B) II only (C) I, II, III (D) I only (E) They are all false 9. Consider the linear function f with the following properties: 5 i) f ( ) d = ii) ( ). F If f ( ) = a + b, then a + b = f d = 8 (A) (B) (C)  (D) (E) ( ) = sin tdt ". F() has an inflection point when = (A) (B) 6 (C) (D) (E) none of these
37 7 Answers to Worksheet 6. C 8. E 5. D. B 9. E 6. D. E. D 7. C. D. D 8. B 5. A. C 9. E 6. E. B. A 7. C. E
38 7 Riemann Sums To find the area under a curve whose equation is not capable of being integrated, ecept by graphing calculator, we can use an approimation technique known as Riemann Sums named after the 9 th century mathematician Georg Riemann. For eample, suppose we wished to find the area of the region R bounded by =, = 5, f ( ) = 6 ln, and the ais. To find the area by hand methods, i.e. without a calculator, is impractical but an approimation method is available as follows.
39 75 Let us split up the region R into 6 sections formed by drawing vertical lines at = 5, = 7, = 9, =, =, and = 5 as shown below. We wish to add up the 6 roundedtopped sections to find the area of region R. There are many ways to approimate the areas of these roundedtopped sections.
40 76 ) We can use rectangles whose height is the lefthand edge of each section. i.e. in which case the area is approimately ( ) ( ) ( ) ( ) ( ) ( ) = times # f + f 5 + f 7 + f 9 + f + f " $ = times # ln + ln ln ln ln+ ln" $ = 6.59 ) We can use rectangles whose height is the righthanded edge of each section. i.e. in which case the area is approimately = " f 5 ( ) + f ( 7) + f ( 9) + f ( ) + f ( ) + f ( 5) = ln5 + 9 ln7 + 7 ln9 + 5 ln+ ln+ ln5# = 59. " $ # $
41 77 ) We can use rectangles whose heights are the y values of the function at the mid point of each section. i.e. in which case the area is approimately = " f ( ) + f ( 6) + f ( 8) + f ( ) + f ( ) = ln + ln6 + 8 ln8 + 6 ln + ln + ln# " $ = 6.8 # $ ) We can use trapezoids formed by joining top points of each section. i.e.
42 78 in which case the area is approimately ( ) + f ( 5) = # f "# = times " f + f 5 ( ) + f ( 7) + f 9 ( ) + f ( ) ( ) + f ( ) + f ( ) + f ( 5) + f ( ) + f ( 5) + f ( 7) + f ( 9) + f ( ) + f ( ) + f ( 5) = ln + ln ln ln ln+ ln+ ln5" # $ = 6.56 In fact, using a graphing calculator we can ascertain that the area of region R is 6.. # $ $ & %& This shows that in the four cases, the four approimations are quite good with the midpoint approimation being the best in this eample. Of course the more sections into which we split up the region R the greater the accuracy. The accuracy and the fact of whether the trapezoidal approimation is an overestimate or an underestimate depends upon the concavity of the curve. For eample, the trapezoidal rule approimation is an underestimate because the curve is concave down for 5. For the Riemann Sums using rectangles, attention must be paid to the slope of the curve to gauge whether it is an underestimate or an overestimate.
43 79 Worksheet 7. The graph of the function f over the interval [,7] is shown below. Using values from the graph, find the Trapezoid rule estimates from the integral 7 ( ) f d using the indicated number of subintervals. a) n = b) n = 6. Use (a) the Trapezoidal Rule and (b) the Midpoint rule with equal subdivisions to approimate the definite integral " d.. Estimate cos( )d using (a) the Trapezoidal Rule and (b) the Midpoint rule with n =. From a graph of the integrand, decide whether your answer to the Trapezoidal Rule is an underestimate or overestimate.
44 8. The graph of g is shown in the figure. Estimates of ( ) g d were computed using the left, right, trapezoid and midpoint rules, each with the same number of subintervals. The answers recorded were.56,.76,.75, and.98. a) Match each approimation with the corresponding rule. b) Between which two approimations does the true value of the integral lie? 5. The graph of f over the interval [,9] is shown in the figure. Using the data in the figure, find a midpoint approimation with equal subdivisions for 9 ( ) f d. (A) (B) (C) (D) (E)
45 8 6. Let R be the region in the first quadrant bounded by the graph of f ( ) e = +, the line =, and the  ais. (a) Write an integral that gives the area of R. (b) Use the Trapezoidal Rule with n = to approimate the area. You must show the numbers that lead to your answer. 7. Answer the following questions about the function f, whose graph is shown below. (a) Find lim f ( ). (b) Find lim h" ( + ) ( ) f h f h. (c) Find lim f '( ). (d) Find f ( ) (e) Approimate f ( ) subdivisions. " d. " d using the Trapezoidal Rule with n =
46 8 8. The Mean Value Theorem guarantees the eistence of a special point on the graph of y point? (A) (,) (B) (, ) (C) (, ) (D) (,) = between (,) and (9,). What are the coordinates of this (E) None of the above 9. Let R be the first quadrant region bounded by the graph of f ( ) = + from = to =. Use the Trapezoidal Rule with equal subdivisions to approimate the area of the region R. (A) 8.5 (B) 8.7 (C) 8.9 (D) 8. (E) 8.. The following table lists the known values of a function f. 5 f ( )....5 If the Trapezoidal Rule is used to approimate f ( ) 5 d the result is (A).9 (B). (C).5 (D).8 (E).5
47 8. A cedar log 6 m long is cut at m intervals and its crosssectional areas A (at a distance from the end of the log) are recorded in the following table. (m) 5 6 A (m ) Find an approimate volume of the log in cubic metres. Answers to Worksheet 7. a) 5 b).5. a) 9 b) 8. a).99 underestimate b).99. a) Right.56 Midpoint.76 Trapezoid.75 Left.98 b) between.76 and E 6. a) " + e d b).7 (actual value is.9) 7. a) b) c) does not eist d).5 e) 8 8. D 9. B. C..7 cubic metres.
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