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Solve the 5 question attached in the word doc ( please send answers in a word doc not PDF)

Attached PDF section 4.0 ( Solve problem 6)

Attached PDF section 4.7 ( Solve problem 7)

In problems 7 sketch the graph of each function and find the area between the graphs of f and g for x in the given interval. 7. f(x) = x 2 + 3 , g(x) = 1 and –1 ≤ x ≤ 2.

Chapter 4, Section 4.0: Introduction to Integrals, remixed by Jeff Eldridge from work by Dale Hoffman, is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. © Mathispower 4u. 4.0 Introduction to Integration Contemporary Calculus 1 Chapter 4: THE INTEGRAL Previous chapters dealt with Differential Calculus. They started with the "simple" geometrical idea of the slope of a tangent line to a curve, developed it into a combination of theory about derivatives and their properties, techniques for calculating derivatives, and applications of derivatives. This chapter begins the development of Integral Calculus and starts with the "simple" geometric idea of area. This idea will be developed into another combination of theory, techniques, and applications. One of the most important results in mathematics, The Fundamental Theorem of Calculus, appears in this chapter. It unifies the differential and integral calculus into a single grand structure. Historically, this unification marked the beginning of modern mathematics, and it provided important tools for the growth and development of the sciences. The chapter begins with a look at area, some geometric properties of areas, and some applications. First we will see ways of approximating the areas of regions such as tree leaves that are bounded by curved edges and the areas of regions bounded by graphs of functions. Then we will find ways to calculate the areas of some of these regions exactly. Finally, we will explore more of the rich variety of uses of "areas". The primary purpose of this introductory section is to help develop your intuition about areas and your ability to reason using geometric arguments about area. This type of reasoning will appear often in the rest of this book and is very helpful for applying the ideas of calculus. AREA The basic shape we will use is the rectangle; the area of a rectangle is (base).(height). If the units for each side of the rectangle are "meters," then the area will have the units ("meters").("meters") = "square meters" 2 = m . The only other area formulas needed for this section are for triangles, area = b.h/2, and for circles, 2 area = π.r . Three other familiar properties of area are assumed and will be used: Addition Property: The total area of a region is the sum of the areas of the non–overlapping pieces which comprise the region. (Fig. 1) Inclusion Property: If region B is on or inside region A, then the area of region B is less than or equal to the area of region A. (Fig. 2) Location–Independence Property: The area of a region does not depend on its location. (Fig. 3) 4.0 Introduction to Integration Example 1: Contemporary Calculus 2 Determine the area of the region in Fig. 4a Solution: The region can easily be broken into two rectangles, Fig. 4b, with areas 35 square inches and 3 square inches respectively, so the area of the original region is 38 square inches. Practice 1: Determine the area of the region in Fig. 5 by cutting it in two ways: (a) into a rectangle and triangle and (b) into two triangles. We can use the three properties of area to get information about areas that are difficult to calculate exactly. Let A be the region bounded by the graph of f(x) = 1/x, the x–axis, and vertical lines at x = 1 and x = 3. Since the two rectangles in Fig. 6 are inside the region A and do not overlap each, the area of the rectangles, 1/2 + 1/3 = 5/6, is less than the area of region A. Practice 2: Build two rectangles, each with base 1 unit, outside the shaded region in Fig. 6 and use their areas to make a VALID statement about the area of region A. Practice 3: What can be said about the area of region A in Fig. 6 if we use both inside and outside rectangles with base 1/2 unit? 4.0 Introduction to Integration Example 2: Contemporary Calculus 3 In Fig. 7, there are 32 dark squares, 1 centimeter on a side, and 31 lighter squares of the same size. We can be sure that the area of the leaf is smaller than what number? Solution: The area of the leaf is smaller than 2 32 + 31 = 63 cm . Practice 4: We can be sure that the area of the leaf is at least how large? Functions can be defined in terms of areas. For the constant function f(t) = 2, define A(x) to be the area of the rectangular region bounded by the graph of f, the t–axis, and the vertical lines at t=1 and t=x (Fig. 8a). A(2) is the area of the shaded region in Fig. 8b, and A(2) =2. Similarly, A(3)= 4 and A(4)=6. In general, A(x)= (base)(height) = (x–1)(2) = 2x – 2 for any x≥1. The graph of y=A(x) is shown in Fig. 8c, and A'(x) = 2 for every value of x > 1. Practice 5: For f(t)=2, define B(x) to be the area of the region bounded by the graph of f, the t–axis, and vertical lines at t = 0 and t = x. Fill in the table in Fig. 9 with the values of B. How are the graphs of y = A(x) and y = B(x) related? Sometimes it is useful to move regions around. The area of a parallelogram is obvious if we move the triangular region from one side of the parallelogram to fill the region on the other side and ending up with a rectangle (Fig. 10). At first glance, it is difficult to estimate the total area of the shaded regions in Fig. 11a . However, if we slide all of them into a single column (Fig. 11b), then it is easy to determine that the shaded area is less than the area of the enclosing rectangle = (base)(height) = (1)(2) = 2. 4.0 Introduction to Integration Practice 6: Contemporary Calculus 4 The total area of the shaded regions in Fig. 12 is less than what number? SOME APPLICATIONS OF "AREA" One reason "areas" are so useful is that they can represent quantities other than simple geometric shapes. For example, if the units of the base of a rectangle are "hours" and the units of the height are "miles/hour", then the units of the "area" of the rectangle are ("hours").("miles/hour") = "miles," a measure of distance (Fig. 13a). Similarly, if the base units are "centimeters" and the height units are "grams" (Fig. 13b), then the "area" units are "gram.centimeters," a measure of work. Distance as an "Area" In Fig. 14, f(t) is the velocity of a car in "miles per hour", and t is the time in "hours." Then the shaded "area" will be (base).(height) = (3 hours).(20 miles/hour ) = 60 miles, the distance traveled by the car in the 3 hours from 1 o'clock until 4 o'clock. Distance as an "Area" If f(t) is the (positive) forward velocity of an object at time t, then the "area" between the graph of f and the t–axis and the vertical lines at times t = a and t = b will be the distance that the object has moved forward between times a and b. 4.0 Introduction to Integration Contemporary Calculus 5 This "area as distance" can make some difficult distance problems much easier. Example 3: A car starts from rest (velocity = 0) and steadily speeds up so that 20 seconds later it's speed is 88 feet per second (60 miles per hour). How far did the car travel during those 20 seconds? Solution: We can answer the question using the techniques of chapter 3 (try it). But if "steadily" means that the velocity increases linearly, then it is easier to use Fig. 15 and the idea of "area as distance." The "area" of the triangular region represents the distance traveled, so 1 1 distance = 2 (base) .(height) = 2 (20 seconds) .(88 feet/second) = 880 feet. Practice 7: A train traveling at 45 miles per hour (66 feet/second) takes 60 seconds to come to a complete stop. If the train slowed down at a steady rate (the velocity decreased linearly), how many feet did the train travel while coming to a stop? Practice 8: You and a friend start off at noon and walk in the same direction along the same path at the rates shown in Fig. 16. (a) Who is walking faster at 2 pm? Who is ahead at 2 pm? (b) Who is walking faster at 3 pm? Who is ahead at 3 pm? (c) When will you and your friend be together? (Answer in words.) Total Accumulation as "Area" In the previous examples, the function represented a rate of travel (miles per hour), and the area represented the total distance traveled. For functions representing other rates such as the production of a factory (bicycles per day), or the flow of water in a river (gallons per minute) or traffic over a bridge (cars per minute), or the spread of a disease (newly sick people per week), the area will still represent the total amount of something. "Area" as a Total Accumulation If f(t) represents a positive rate (in units per time interval) at time t, then the "area" between the graph of f and the t–axis and the vertical lines at times t=a and t=b will be the total units which accumulate between times a and b. (Fig. 17) 4.0 Introduction to Integration Contemporary Calculus For example, Fig. 18 shows the flow rate (cubic feet per second) of water in the Skykomish river at the town of Goldbar in Washington state. The area of the shaded region represents the total volume (cubic feet) of water flowing past the town during the month of October: Total water = "area" = area of rectangle + area of the triangle 1 ≈ (2000 cubic feet/sec)(30 days) + 2 (1500 cf/s)(30 days) = (2750 cubic feet/sec)(30 days) 9 = (2750 cubic feet/sec)(2,592,000 sec) ≈ 7.128 x 10 cubic feet 5 (For comparison, the flow over Niagara Falls is about 2.12x10 cf/s.) Practice 9: Fig. 19 shows the number of telephone calls made per hour on a Tuesday. Approximately how many calls were made between 9 pm and 11 pm? PROBLEMS 1. Calculate the areas of the shaded regions in Fig. 20. 2. Calculate the area of the trapezoidal region in Fig. 21 by breaking it into a triangle and a rectangle. 3. Break Fig. 22 into a triangle and rectangle and verify h+H that the total area of the trapezoid is b.( 2 ). 4. (a) Calculate the sum of the rectangular areas in Fig. 23a. (b) What can we say about the area of the shaded region in Fig. 23b? 6 4.0 5. Introduction to Integration Contemporary Calculus 7 (a) Calculate the sum of the rectangular areas in Fig. 24a. (b) From part (a), what can we say about the area of the shaded region in Fig. 24b? 6. (a) Calculate the sum of the areas of the shaded regions in Fig. 24c. (b) From part (a), what can we say about the area of the shaded region in Fig. 24b? 7. Let A(x) represent the area bounded by the graph and the horizontal axis and vertical lines at t=0 and t=x for the graph in Fig. 25. Evaluate A(x) for x = 1, 2, 3, 4, and 5. 8. Let B(x) represent the area bounded by the graph and the horizontal axis and vertical lines at t=0 and t=x for the graph in Fig. 26. Evaluate B(x) for x = 1, 2, 3, 4, and 5. 9. Let C(x) represent the area bounded by the graph and the horizontal axis and vertical lines at t=0 and t=x for the graph in Fig. 27. Evaluate C(x) for x = 1, 2, and 3 and find a formula for C(x). 10. Let A(x) represent the area bounded by the graph and the horizontal axis and vertical lines at t=0 and t=x for the graph in Fig. 28. Evaluate A(x) for x = 1, 2, and 3 and find a formula for A(x). 11. A car had the velocity given in Fig. 29. How far did the car travel from t=0 to t=30 seconds? 12. A car had the velocity given in Fig. 30. How far did the car travel from t=0 to t=30 seconds? 13. The velocities of two cars are shown in Fig. 31. (a) From the time the brakes were applied, how many seconds did it take each car to stop? (b) From the time the brakes were applied, which car traveled farther until it came to a complete stop? 4.0 Introduction to Integration Contemporary Calculus 8 14. Police chase: A speeder traveling 45 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off after the speeder. If the police car speeds up steadily to 60 miles/hour in 10 seconds and then travels at a steady 60 miles/hour, how long and how far before the police car catches the speeder who continued traveling at 45 miles/hour? (Fig. 32) 15. What are the units for the "area" of a rectangle with the given base and height units? Base units miles per second hours square feet kilowatts houses meals Height units seconds dollars per hour feet hours people per house meals Section 4.0 "Area" units PRACTICE Answers Practice 1: 1 1 1 area = 3(6) + 2(4)(3) = 24 and area = 2(3)(10) + 2(6)(3) = 24. Practice 2: 1 outside rectangular area = (1)(1) + (1)( 2 ) = 1.5 Practice 3: Practice 4: Pr 5: x 0 1/2 1 2 Practice 6: Practice 7: 1 2 1 2 1 57 Using rectangles with base = 1/2. Inside: area = 2 ( 3 + 2 + 5 + 3 ) = 60 ≈ 0.95 1 2 1 2 72 Outside: area = 2 ( 1 + 3 + 2 + 5 ) = 60 ≈ 1.2 The area of the region is between 0.95 and 1.2 . 2 The area of the leaf is larger than the area of the dark rectangles, 32 cm . B(x) 0 1 2 4 y = B(x) = 2x is a line with slope 2 so it is parallel to y = A(x) = 2x – 2. See Fig. 33. Area < area of the rectangle enclosing the shifted regions = 5 1 See Fig. 34. Distance = area of shaded region = 2 ( base )( height ) 1 = 2( 60 seconds)( 66 feet/second) = 1980 feet. Practice 8: (a) At 2 pm both are walking at the same velocity. You are ahead. (b) At 3 pm your friend is walking faster than you, but you are still ahead. (The "area" under your velocity curve is larger than the "area" under your friend's.) (c) You and your friend will be together on the trail when the "areas" (distances) under the two velocity graphs are equal. Practice 9: Total calls = "area" under rate curve from 9 am to 11 am ≈ 300 calls.
Chapter 4, Section 4.7: First Applications of Definite Integrals , remixed by Jeff Eldridge from work by Dale Hoffman, is licensed under a Creative Commons Attribution-ShareAlike 3.0 Unported License. © Mathispower 4u. 4.7 4.7 Applications of Integration Contemporary Calculus 1 FIRST APPLICATIONS OF DEFINITE INTEGRALS The development of calculus by Newton and Leibniz was a vital step in the advancement of pure mathematics, but Newton also advanced the applied sciences and mathematics. Not only did he discover theoretical results, but he immediately used those results to answer important applied questions about gravity and motion. The success of these applications of mathematics to the physical sciences helped establish what we now take for granted: mathematics can and should be used to answer questions about the world. Newton applied mathematics to the outstanding problems of his day, problems primarily in the field of physics. In the intervening 300 years, thousands of people have continued these theoretical and applied traditions and have used mathematics to help develop our understanding of all of the physical and biological sciences as well as the behavioral sciences and business. Mathematics is still used to answer new questions in physics and engineering, but it is also important for modeling ecological processes, for understanding the behavior of DNA, for determining how the brain works, and even for devising strategies for voting effectively. The mathematics you are learning now can help you become part of this tradition, and you might even use it to add to our understanding of different areas of life. It is important to understand the successful applications of integration in case you need to use those particular applications. It is also important that you understand the process of building models with integrals so you can apply it to new problems. Conceptually, converting an applied problem to a Riemann sum (Fig. 1) is the most valuable step. Typically, it is also the most difficult. Area between f and g We have already used integrals to find the area between the graph of a function and the horizontal axis. Integrals can also be used to find the area between two graphs. If f(x) ≥ g(x) for all x in [a,b], then we can approximate the area between f and g by partitioning the interval [a,b] and forming a Riemann sum (Fig. 2). The height of each rectangle is th f(ci) – g(ci) so the area of the i rectangle is (height).(base) = {f(ci) – g(ci)}.∆x i . This approximation of the total area is n area ≈ {f(ci) – g(ci)}.∆xi , a Riemann sum. ∑ i=1 4.7 Applications of Integration Contemporary Calculus 2 The limit of this Riemann sum, as the mesh of the partitions approaches 0, is the definite integral b ⌠ ⌡ { f(x) – g(x) } dx . a We will sometimes use an arrow to indicate "the limit of the Riemann sum as the mesh of the partitions approaches zero," and will write n ∑ b {f(ci) – g(ci)}.∆xi → ⌠ ⌡ { f(x) – g(x) } dx . a i=1 If f(x) ≥ g(x) on the interval [a,b], then  area bounded by the graphs   of f and g and vertical    lines at x = a and x = b Example 1: b = ⌠ ⌡ { f(x) – g(x) } dx . a Find the area bounded between the graphs of f(x) = x and g(x) = 3 for 1 ≤ x ≤ 4. (Fig. 3) Solution: It is clear from the figure that the area between f and g is 2.5 square inches. Using the theorem, area between f and g for 1 ≤ x ≤ 3 is 3 2 x ⌠ ⌡ { 3 – x } dx = 3x – 2 3 |1 9 5 = ( 2 ) – ( 2 ) = 2 , and 1 area between f and g for 3 ≤ x ≤ 4 is 4 2 ⌠ { x – 3 } dx = x2 – 3x ⌡ 3 4 |3 –8 –9 1 =( 2 )–( 2 ) =2 . The two integrals also tell us that the total area between f and g is 2.5 square inches. 4 The single integral ⌠ { 3 – x } dx = 1.5 which is not the area we want in this problem. The value ⌡ 1 of the integral is 1.5, and the value of the area is 2.5 . Practice 1: Use integrals and the graphs of f(x) = 1 + x and g(x) = 3 – x to determine the area between the graphs of f and g for 0 ≤ x ≤ 3. Example 2: Two objects start from the same location and travel along the same path with velocities vA(t) = t + 3 and 2 vB(t) = t – 4t + 3 meters per second (Fig. 4). How far ahead is A after 3 seconds? After 5 seconds? Solution: Since vA(t) ≥ v B(t) , the "area" between the graphs of vA and v B represents the distance between the objects. 4.7 Applications of Integration Contemporary Calculus 3 3 0 0 3 ⌠ 2 After 3 seconds, the distance apart = ⌠ ⌡ vA(t) – vb(t) dt = ⌡ (t+3) – ( t –4t+3) dt 3 3 5 t =⌠ ⌡ 5t – t2 dt = 2 t2 – 3 0 3 |0 5 27 5 0 1 = ( 2 .9 – 3 ) – ( 2 .0 – 3 ) = 13 2 meters. 3 5 5 2 t After 5 seconds, the distance apart = ⌠ v (t) – v (t) dt = t – ⌡ A 2 3 b 0 5 5 = 20 6 meters. |0 b ⌠ f(x) dx and the area of ⌡ If f(x) ≥ g(x) , we can use the simpler argument that the area of region A is a b region B is ⌠ ⌡ g(x) dx , so the area of region C, the area between f and g , is a area of C = (area of A) – (area of B) = b b b a a a ⌠ ⌡ f(x) dx – ⌠ ⌡ g(x) dx = ⌠ ⌡ f(x) – g(x) dx . If the same function is not always greater, then we need to be very careful and find the intervals where f ≥ g and the intervals where g ≥ f. Example 3: Find the area of the shaded region in Fig. 5. Solution: For 0≤ x ≤5, f(x) ≥ g(x) so the area of A is 5 5 0 0 ⌠ ⌡ (x+3) – (x2–4x+3) dx ⌡ f(x) – g(x) dx = ⌠ 5 3 5 2 x ⌠ 2 = ⌡ 5x–x dx = 2 x – 3 5 |0 5 = 20 6 . 0 For 5≤ x ≤ 7, g(x) ≥ f(x) so the area of B is 7 7 7 3 ⌠ ⌡ (x2–4x+3) – (x+3) dx = ⌠ ⌡ x2 – 5x dx = x3 – 52 x2 ⌡ g(x) – f(x)dx = ⌠ 5 5 5 Altogether, the total area between f and g for 0≤ x ≤ 7 is 5 7 0 5 5 4 ⌠ ⌡ f(x) – g(x) dx + ⌠ ⌡ g(x) – f(x) dx = 20 6 + 12 6 1 = 33 2 . 7 |5 4 = 12 6 . 4.7 Applications of Integration Contemporary Calculus 4 Average Value of a Function 1 We know the average of n numbers, a1, a2, . . . , an , is their sum divided by n: average (mean) = n n ∑ ak . k=1 Finding the average of a function on an interval, an infinite number of values, requires an integral. To find a Riemann sum approximation of the average value of f on the interval [a,b], we can partition [a,b] into n equally long subintervals of length ∆x = (b–a)/n, pick a value ci of x in each subinterval, and find the average of the numbers f(ci). Then f(c1)+f(c2)+. . . +f(cn) 1 average of f ≈ = n n n n .1 f(c ) = ∑ i ∑ f(ci) n k=1 k=1 This last sum is not a Riemann sum since it does not have the form ∑ f(ci) .∆xi , but it can be manipulated into one: n 1 ∑ f(ci) . n = i=1 n ∑ f(ci) . b–a . 1 n b–a 1 = b–a i=1 Then {average of f} n 1 ≈ b–a ∑ f(ci) . ∆x n ∑ f(ci) . b–a n i=1 → k=1 1 b–a 1 = b–a n ∑ f(ci) . ∆x. i=1 b ⌠ ⌡ f(x) dx = {average of f} a as the number of points n gets larger and the mesh , (b–a)/n , approaches 0. Definition: Average (Mean) Value of a Function For an integrable function f on the interval [a,b], the average value of f on [a,b] is b 1 ⌠ b–a ⌡ f(x) dx a The average value of a positive f has a nice geometric interpretation. Imagine that the area under f (Fig. 6a) is a liquid that can "leak" through the graph to form a rectangle with the same area (Fig. 6b). If the height of the rectangle is H, then the area of the rectangle is H.(b–a). We know the area of the rectangle is the same as the area under f so b H.(b–a) = 1 H = b–a ⌠ ⌡ f(x) dx . Then a b ⌠ ⌡ f(x) dx , the average value of f on [a,b]. a The average value of positive f is the height H of the rectangle whose area is the same as the area under f. 4.7 Applications of Integration Example 4: Contemporary Calculus 5 Find the average value of f(x) = sin(x) on the interval [0,π] . (Fig. 7) Solution: 1 Average value = π–0 1 = π ( – cos(x)) π ⌠ ⌡ sin(x) dx 0 π |0 1 2 = π { –(–1) – (–1) } = π ≈ 0.6366 . A rectangle with height 2/π ≈ 0.64 on the interval [0,π] encloses the same area as one arch of the 2π 1 sine curve. The average value of sin(x) on the interval [0, 2π] is 0 since 2π ⌠ ⌡ sin(x)dx = 0. 0 Practice 2: During a 9 hour work day, the production rate at time t hours was r(t)= 5 + t cars per hour. Find the average hourly production rate. Function averages, involving means and more complicated averages, are used to "smooth" data so that underlying patterns are more obvious and to remove high frequency "noise" from signals. In these situations, the original function f is replaced by some "average of f." If f is rather jagged time data, then x+5 1 the ten year average of f is the integral g(x) = 10 ⌠ ⌡ f(t) dt .an average of f over 5 units on each side x–5 of x. For example, Fig. 9 shows the graphs of a Monthly Average (rather “noisy” data) of surface temperature data, an Annual Average (still rather “jagged), and a Five Year Average (a much smoother function). Typically the average function reveals the pattern much more clearly than the original data. This use of a “moving average” value of “noisy” data (weather information, stock prices) is a very common. Work The amount of work done on an object is the force applied to the object times the distance the object is moved while the force is applied (Fig. 10) or, more succinctly, work = (force).(distance) . If you lift a 3 pound book 2 feet, then the force is 3 pounds, the weight of the book, and the distance moved is 2 feet, so you have done (3 pounds).(2 feet) = 6 foot– pounds of work. When the applied force and the distance are both constants, then calculating work is simply a matter of multiplying. 4.7 Applications of Integration Practice 3: Contemporary Calculus 6 How much work is done lifting a 10 pound object from the ground to the top of a 30 foot building (assume the cable is weightless). If either the force or the distance is variable, then integration is needed. Example 5: How much work is done lifting a 10 pound object from the ground to the top of a 30 foot building if the cable weighs 2 pounds per foot. (Fig. 11) Solution: This is more difficult than the Practice problem. When the object is at ground level, a force of 70 pounds (10 pounds plus the weight of 30 feet of cable) must be applied, but when the object is 29 feet above the ground, only 12 pounds of force are needed. In general, if the object is x feet above the ground (Fig. 12), then 30 – x feet of cable, weighing 2(30–x) pounds, is used so the required force is f(x) = 10 + 2(30–x) = 70 – 2x pounds. Let's partition the height of the building into small increments so the force needed th in each subinterval does not change much. The force in the i subinterval will be approximately f(ci) for some ci in the subinterval, and the distance moved will be the length of the subinterval, ∆xi . The work done to move the object through the subinterval will be f(ci).∆xi , and the total work will be the sum of the work on each subinterval: work ≈ ∑ (subinterval work) = ∑ f(ci) ∆xi = ∑ {70 – 2ci} ∆xi (a Riemann sum) 30 → ⌠ ⌡ {70 – 2x } dx as the mesh approaches 0. 0 We have approximated the solution to an applied problem by a Riemann sum, and obtained an exact solution by taking the limit of the Riemann sum to get a definite integral. Now we just need to evaluate the definite integral: 30 30 2 2 2 ⌠ = {70.30 – (30) } – {70.0 – (0) } = (2100–900) – (0) = 1200 foot–pounds. ⌡(70 – 2x) dx = 70x – x 0 0 | Practice 4: Suppose the building in Example 5 is 50 feet tall and the cable weighs 3 pounds per foot. (a) How much force is needed when the 10 pound object is x feet above the ground? (b) Write an integral for the work done raising the object from the ground to a height of 10 feet. From a height of 10 feet to a height of 20 feet. 4.7 Applications of Integration Contemporary Calculus 7 In the previous Example and Practice problem, the force was variable and the distance was ∆x. In later sections we will examine situations where the force is constant and the distance changes. Summary These area, average and work applications simply introduce a few of the applications of definite integrals and illustrate the pattern of going from an applied problem to a Riemann sum, on to a definite integral and, finally, to a number. More applications will be explored in Chapter 5. The rest of this chapter will examine additional ways of finding antiderivatives and of finding the values of definite integrals when an antiderivative can not be found. PROBLEMS In problems 1 – 4, use the values in Table 1 to estimate the areas. 1. Estimate the area between f and g for 1 ≤ x ≤ 4. 2. Estimate the area between f and g for 1 ≤ x ≤ 6. 3. Estimate the area between f and h for 0 ≤ x ≤ 4. 4. Estimate the area between g and h for 0 ≤ x ≤ 6. 5. Estimate the area of the island in Fig. 13. 6. Estimate the area of the island in Fig. 13 if the distances x! f(x)! 0! 1! 2! 3! 4! 5! 6! 5! 6! 6! 4! 3! 2! 2! g(x)! h(x)"" 2! 1! 2! 2! 3! 4! 5! 5 6 8 6 5 4 2 Table 1 between the lines is 50 feet instead of 40 feet,. In problems 7 – 18, sketch the graph of each function and find the area between the graphs of f and g for x in the given interval. 7. 2 f(x) = x + 3 , g(x) = 1 and –1 ≤ x ≤ 2. 8. 2 f(x) = x + 3 , g(x) = 1 + x and 0 ≤ x ≤ 3. 9. 2 f(x) = x , g(x) = x and 0 ≤ x ≤ 2. 10. f(x) = 4 – x 2 , g(x) = x + 2 and 0 ≤ x ≤ 2. 1 11. f(x) = x , g(x) = x and 1 ≤ x ≤ e. 12. f(x) = x , g(x) = x and 0 ≤ x ≤ 4. 13. f(x) = x + 1 , g(x) = cos(x ) and 0 ≤ x ≤ π/4. 2 14. f(x) = (x –1) , g(x) = x + 1 and 0 ≤ x ≤ 3. x 15. f(x) = e , g(x) = x and 0 ≤ x ≤ 2. 16. f(x) = cos(x ), g(x) = sin(x ) and 0 ≤ x ≤ π/4. 4.7 Applications of Integration 17. f(x) = 3 , g(x) = 2 1–x Contemporary Calculus and 0 ≤ x ≤ 1. 18. f(x) = 2 , g(x) = 8 2 4–x and –2 ≤ x ≤ 2. In problems 19 – 22, use the values in Table 1 to estimate the average values. 19. Estimate the average value of f on the interval [0.5, 4.5]. 20. Estimate the average value of f on the interval [0.5, 6.5]. 21. Estimate the average value of f on the interval [1.5, 3.5]. 22. Estimate the average value of f on the interval [3.5, 6.5]. In problems 23 – 32, find the average value of f on the given interval. 23. f(x) in Fig. 14 for 0 ≤ x ≤ 2. 24. f(x) in Fig. 14 for 0 ≤ x ≤ 4. 25. f(x) in Fig. 14 for 1 ≤ x ≤ 6. 26. f(x) in Fig. 14 for 4 ≤ x ≤ 6. 27. f(x) = 2x + 1 for 0 ≤ x ≤ 4. 28. f(x) = x 29. f(x) = x 2 for 1 ≤ x ≤ 3. 31. f(x) = sin(x) for 0 ≤ x ≤ π. π. 2 30. f(x) = x for 0 ≤ x ≤ 2. for 0 ≤ x ≤ 4. 32. f(x) = cos(x) for 0 ≤ x ≤ 33. Calculate the average value of f(x) = x on the interval [0, C] for C = 1, 9, 81, 100. What is the pattern? 34. Calculate the average value of f(x) = x on the interval [0, C] for C = 1, 10, 80, 100. What is the pattern? 35. Fig. 15 shows the number of telephone calls per minute at a large company. (a) Estimate the average number of calls per minute from 8 am to 5 pm. (b) From 9 am to 1 pm. 36. Fig. 16 shows the velocity of a car during a 5 hour trip. (a) Estimate how far the car traveled during the 5 hours. (b) At what constant velocity should you drive in order to travel the same distance in 5 hours? 37. (a) How much work is done lifting a 20 pound bucket from the ground to the top of a 30 foot building with a cable which weighs 3 pounds per foot? (b) How much work is done lifting the same bucket from the ground to a height of 15 feet with the same cable? 4.7 Applications of Integration Contemporary Calculus 9 38. (a) How much work is done lifting a 60 pound chair from the ground to the top of a 20 foot building with a cable which weighs 1 pound per foot? (b) How much work is done lifting the same chair from the ground to a height of 5 feet with the same cable? 39. (a) How much work is done lifting a 10 pound calculus book from the ground to the top of a 30 foot building with a cable which weighs 2 pounds per foot? (b) From the ground to a height of 10 feet? (c) From a height of 10 feet to a height of 20 feet? 40. How much work is done lifting an 80 pound injured child to the top of a 20 foot hole using a stretcher weighing 14 pounds and a cable which weighs 1 pound per foot? 41. How much work is done lifting an 60 pound injured child to the top of a 15 foot hole using a stretcher weighing 10 pounds and a cable which weighs 2 pound per foot? 42. How much work is done lifting an 120 pound injured adult to the top of a 30 foot hole using a stretcher weighing 10 pounds and a cable which weighs 2 pound per foot? Section 4.7 PRACTICE Answers 1 1 Using geometry (Fig. 17): A = 2(2)(1) = 1 and B = 2(4)(2) = 4 so total area = A + B = 5. Practice 1: Using integrals: 1 1 0 3 0 3 1 1 ⌠ ⌠ 2 A = ⌡ (3–x) – (1+x) dx = ⌡ (2 – 2x) dx = 2x – x | ⌠ ⌠ 2 B = ⌡ (1+x) – (3–x) dx = ⌡ (2x – 2) dx = x – 2x | 3 3 0 0 1 0 = (2 – 1) – (0) = 1. 3 1 = (9 – 6) – (1 – 2) = 4. ⌠ ⌠ The single integral ⌡ (1+x) – (3–x) dx is not correct: ⌡ (1+x) – (3–x) dx = 3 . 4.7 Applications of Integration Practice 2: Average value = Contemporary Calculus 9 b 1 ⌠ 1 ⌠ ⌡ 5 + t dt b–a ⌡ f(x) dx = 9–0 0 a 9 1 = 9 10 ⌠ ⌡ 5 + t1/2 dt = 19 ( 5t + 23 t3/2 ) 0 9 | 0 1 2 3/2 1 1 = 9 (45 + 3 9 ) – 9 (0) = 9 (45 + 18) = 7 cars per hour. Practice 3: Work = (force).(distance) = (10 pounds).(30 feet) = 300 foot.pounds. Practice 4: (a) force = (force for cable) + (force for object) = (length of cable)(density of cable) + 10 pounds = (50 – x feet)(3 pounds/foot) + 10 pounds = 160 – 3x pounds (b) "from the ground to a height of 10 feet:" Work ≈ ∑ (subinterval work) = ∑ f(ci) ∆xi = ∑ {160 – 3ci} ∆x i → (a Riemann sum) 10 ⌠ ⌡ {160 – 3x} dx as the mesh approaches 0. 0 20 "From a height of 10 feet to a height of 20 feet:" work = ⌠ {160 – 3x} dx . ⌡ 10
MATH 140 Quiz 5 NAME___________________________ INSTRUCTIONS • • • • • The quiz is worth 100 points. There are five problems (each worth 20 points). This quiz is open book and open notes. This means you may refer to your textbook, notes, and online classroom materials, but you may not consult anyone. You may take as much time as you wish, provided you turn in your quiz no later than the due date posted in our syllabus. You must show all work in order to receive full credit. If you do not show your work, you may earn only partial or no credit at the discretion of the professor. Please carefully review How to submit quizzes and exams in our Syllabus. If you have any questions, please feel free to send me a PAGER message in LEO. Emailed quizzes and exams will not be accepted (they crash my email system). Thank you for understanding. Best of luck! ☺ MULTIPLE CHOICE Select the best answer choice. Write your answer choices below: (1) _____ (2) _____ (3) _____ (4) _____ (5) _____ 1) Evaluate (1) _____ ∫ (𝐴) 1 1 cos ( ) 𝑑𝑥 𝑥2 𝑥 1 1 sin ( ) 𝑥 𝑥 1 1 (𝐵) − sin ( ) + 𝐶 𝑥 𝑥 1 (𝐶) sin ( ) + 𝐶 𝑥 1 (𝐷) −sin ( ) + 𝐶 𝑥 (𝐸) 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 2) Solve the initial value problem (2) _____ 𝑑𝑦 1 = , 𝑑𝑥 √𝑥 + 2 (𝐴) 1 √𝑥 + 2 𝑦(2) = −1 +1 (𝐵) 2√𝑥 + 2 − 1 (𝐶) 2√𝑥 + 2 − 5 (𝐷) 1 2√ 𝑥 + 2 +2 (𝐸) 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 3) Given 𝑓(𝑥) = 𝑥 2 + 3, find the exact area 𝐴 of the region under 𝑦 = 𝑓(𝑥) on the interval [1, 3] by first computing the sum 𝑛 ∑ 𝑓(𝑥𝑘 )∆𝑥 𝑘=1 and then taking the limit as 𝑛 → ∞. Hint: ∑𝑛𝑘=1 𝑘 2 = 𝑛(𝑛+1)(2𝑛+1) 6 (3) _____ (Note: If limits are not used correctly, the maximum number of points possible for this problem will be 5 points out of 20 points.) (𝐴) 14 (𝐵) 44 3 (𝐶) 13 (𝐷) 56 3 (𝐸) 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 5 4) Evaluate ∫0 (2𝑥 3 − 4𝑥 2 + 1)𝑑𝑥 by computing the limit 𝑛 lim ∑ 𝑓(𝑥𝑘 )∆𝑥 𝑛→∞ 𝑘=1 (Note: If limits are not used correctly, the maximum number of points possible for this problem will be 5 points out of 20 points.) (4) _____ (𝐴) 905 6 (𝐵) 875 6 (𝐶) 145 3 (𝐷) 𝑥 4 4𝑥 3 − +𝑥+𝐶 2 3 (𝐸) 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 5) Evaluate (5) _____ 4 ∫ 1 (𝐴) − (𝐵) 1 2 1 2 (𝐶) − 2 (𝐷) 2 (𝐸) 𝑁𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 1 √𝑥 𝑑𝑥

Tutor Answer

JesseCraig
School: Purdue University

Attached.

Running head: CALCULAS

1

CALCULAS
Student name:
Course:
Institution affiliation:

CALCULAS

2
CALCULAS

Problem 6 (PDF section 4.0)
a) Area = area of trapezium + area of triangle
= ½ (a+b)h + ½ bh
= ½ (2+1)2 + ½ *1*1
= 3.5 in2
b) The area of the shaded region in Fig. 24b is greater than the area of the shaded region in
part a.

Problem 7 (PDF section 4.7)
f (x) = x2 + 3, g(x) = 1 and -1≤ x ≤ 2

f(x) ≥ g(x)
2
Thus,

2
∫−1[(𝑥 2

+ 3) − 1] = [(x /3 + 3x) – x]
3

-1
= [(2 /3 + 3*2) – 2] – [(-13/3 + 3*-1) - -1]
3

= 20/3 – (-7/3)
= 20/3 + 7/3
= 9 in2


Running head: MATH 140 Quiz 5

1

MATH 140 Quiz 5
Student name:
Course:
Institution affiliation:

MATH 140 Quiz 5
MATH 140 Quiz 5

2
NAME___________________________

INSTRUCTIONS







The quiz is worth 100 points. There are five problems (each worth 20 points).
This quiz is open book and open notes. This means you may refer to your textbook,
notes, and online classroom materials, but you may not consult anyone. You may take as
much time as you wish, provided you turn in your quiz no later than the due date posted
in our syllabus.
You must show all work in order to receive full credit. If you do not show your
work, ...

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