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Unformatted text preview: Math for Econ II, Written Assignment 4 (28 points) Due Monday, July 31, in class Summer 2017 1 5-4h43 5.4 THEOREMS ABOUT DEFINITE INTEGRALS Part I (28 points) 43. Using the graph of f in Figure ??, arrange the ins5-4h47-48 following In Problems ??, evaluate the expression, if possible, or 1. (4 pts) Using the graph of f below, arrange the following quantities in increasing order, from least to greatest. quantities in increasing order, from least to greatest.R say information R R R 2 is needed, given that ! 4 what Radditional 2 3 ! 1 i) 1 f (x) dx ii) 2 f! (x) 2 dx iii) f (x) dx iv) f (x) dx v) − f (x) dx vi) The number 0 g(x) dx = 12. 0 dx 0 2 1 f (x) (ii) 1 f (x) dx (i) −4 0 (iii) !2 0 (iv) f (x) dx !13 f (x) dx 2 !2 5-4h47 Thedxnumber (vi) 20 The viii)number The number −10 47. (v) − vii) f (x) 0 1 (vii) The number 20 (viii) 1 5-4h43fig The number −10 f (x) 10 ins5-4h49-52 2 3 −10 5-4h46 5-4h46fig 4 5-4h48 g(x) dx 48. 0 " 4 g(−x) dx −4 ! 5-4h49 49. #" 7 5-4h50 f (x) dx 50. 0 " 3.5 f (x) dx 0 44. (a) Using Figures ?? and ??, find the average value on " 5 " 7 Solution: 0 ≤ x ≤ 2Rof R2 Rf (x 5-4h51 51. 5-4h52 52. 1 3 + 2) dx (x) + 2) dx = A1 and(iii)1 ff(x) dx = −A2 and 2 f (x) dx = A3 , we know (f that (i) Since f (x) 0 f (x) (ii)dxg(x) (x)·g(x) −2 0 (b) Is the following statement true? Explain your Zan-1 Z 2 Z 3 swer. 0 < 5-4h53f (x) dx < − f (x) dx < f (x) dx. 53. (a) Sketch a graph√ of f√ (x) =√sin(x√2 ) and mark on it 1 2 Average(f ) · Average(g) = Average(f · g) 0 the points x = π, 2π, 3π, 4π. R2 R2 (b) Use your smaller graph to decide which of the four numbers 1 1 − A2 , which is negative, but in magnitude than f (x) dx. Thus fIn (x)addition, 0 f (x) dx = A1g(x) 1 5-4h44figa 5-4h45 " In Problems ??, evaluate the expression ! 7if possible, or say what extra information is needed, given 0 f (x) dx = 25. x Figure 5.71 5-4h44 311 x x 5-4h44figb 1 2 1 2 Z 2 f (x) dx < 1 Z 0 " 2 √ nπ sin(x2 ) dx f (x) dx < 0. 0 n = 1, 2, 3, 4 is largest. Which is smallest? How many of the num- Figurearea 5.72 A3 lies inside Figure 5.73 of height 20 andbers The a rectangle base so A3 < 20. The area A2 lies inside a rectangle below are1, positive? the x-axis of height 10 and width 1, so −10 < A2 . Thus: 45. (a) Without computing any integrals, explainins5-4h54-56 why the For Problems ??, assuming F # = f , mark the quantity on a average value of f (x) = sin x on (viii) [0, π] must be be-< copy < (ii) (iii) of<Figure (vi) ??. < (i) < (v) < ( iv) < (vii). tween 0.5 and 1. (b) Compute this average. R4 R −1 R4 2. (3 pts) Suppose f is even, −2 f (x)from dx = 3, and 2 f (x) dx = 5. Either find 1 f (x) dx, or show there is not F (x) 46. Figure ?? shows the standard normal distribution R4 enough to find 1 f (x) dx. statistics, which isinformation given by 2 1 √ e−x f/2 .is even, R −1 f (x) dx = R 2 f (x) dx, so Solution: Because 2π −2 1 Statistics books often contain tables such as followZ the Z Z 4 4 2 ing, which show the area under the curve from f0 (x) to b dx for = fins5-4h54-56fig (x) dx + fa(x) dx = 3 + 5x= 8 . b various values of b. 1 1 2 Figure 5.75 ! bintegrals: 2 3. (5 pts) Evaluate following ! the Area = √1 e−x /2 dx 2π 0 R1 R1 R1 (a) (3 pts) If 0 (f (x) − 2g(x)) dx = 6 and 0 (2f (x) + 2g(x)) dx = 9, find 0 (f (x) − g(x)) dx. 5-4h54 54. A slope representing f (a). Z 1 x " b 17 5 3 0 b (b) (2 pts) (x − x + x − x) dx 5-4h55 55. A length representing −1 Figure 5.74 Solution: b 1 √1 2π !b 0 2 e−x /2 0.3413 2 0.4772 3 0.4987 4 0.5000 f (x) dx. a dx 5-4h56 5-4h57 Use the information given in the table and the symmetry of the curve about the y-axis to find: " 3 " 3 2 2 1 1 e−x /2 dx (b) √ e−x /2 dx (a) √ 2π 1 2π −2 56. A slope representing 1 b−a " b f (x) dx. a 57. In Chapter ??, the average velocity over the time interval a ≤ t ≤ b was defined to be (s(b) − s(a))/(b − a), where s(t) is the position function. Use the Fundamental Theorem of Calculus to show that the average value of the velocity function v(t), on the interval a ≤ t ≤ b, is also (s(b) − s(a))/(b − a). R1 R1 (a) Adding the first 2 integrals gives us 3 0 f (x) = 15, so 0 f (x) = 5. We use the first integral again: Z 1 Z 1 Z 1 Z 1 Z 1 1 f (x) dx − 2g(x) dx = 6, 5 − 2 g(x) dx = 6, −2 g(x) dx = 1, g(x) dx = − . 2 0 0 0 0 0 Thus, Z 1 (f (x) − g(x)) dx = 0 (b) Z Z 1 f (x) − 0 Z 1 g(x) dx = 5 − (−1/2) = 0 11 . 2 1 (x17 − x5 + x3 − x) dx = 0. −1 This is because f (x) = x17 − x5 + x3 − x is an odd continuous function and our interval is [−1, 1]. (check f is odd: f (−x) = (−x)17 −(−x)5 +(−x)3 −(−x) = −x17 +x5 −x3 +x = −(x17 −x5 +x3 −x) = −f (x)). 4. (2 pts) A honeybee population starts with 100 bees and increases at a rate of n0 (t) bees per week. What does R 15 100 + 0 n0 (t) dt represent? R 15 Solution: 0 n0 (t) dt is the net change in honeybee population from t = 0 to t = 15 weeks. Therefore, R 15 100 + 0 n0 (t) dt is the number of bees in the colony 15 weeks after the start. 5. (2 pts) Explain what is wrong with the following calculation of the area under the curve Z 1 h 1 i 1 1 dx = − = −1 − 1 = −2. 2 x −1 −1 x 1 x2 from x = −1 to x = 1: Solution: We cannot apply the evaluation theorem because x12 is not continuous on [−1, 1] (it has an infinite discontinuity when x = 0). Whatever the integral is (if it even exists!), it can’t be negative, because 1/x2 is a strictly positive function with plenty of positive area. 6. (3 pts) Suppose the area under the curve ex from x = 0 to x = a is twenty times the area under the curve 2e2x from x = 0 to x = b. Solve for a in terms of b. (in other words, write a =(some formula involving b)) Solution: We use the Evaluation Theorem. Z a ex dx = 20 0 Z a Thus b 2e2x dx. 0 a ex dx = ex = ea − 1, 20 0 0 Z 0 ea − 1 = 20e2b − 20, Z Z b h ib 2e2x dx = 20 e2x = 20e2b − 20. ea = 20e2b − 19, 0 a = ln(20e2b − 19) . 5 1 dx = ln(5). Now find a fraction which approximates ln(5), by 1 x Z 5 1 using M4 (midpoint sum with 4 rectangles) to approximate dx. x 1 (The actual value of ln(5) is 1.6094 . . .. For fun, plug your approximation into a calculator and compare) 7. (4 pts) Use the Evaluation Theorem to show Solution: By the Evaluation Theorem, Z 5 1 5 1 dx = ln(x) = ln(5) − ln(1) = ln(5). x 1 (the formulas for ln |x| and ln(x) are identical since x > 0 on the interval [1, 5]). M4 = 1 · f 3 2 +1·f 5 2 +1·f 7 2 +1·f 9 2 = 2 2 2 2 496 + + + = 3 5 7 9 315 How good is our approximation? ln(5) = 1.6094.... Our approximating fraction is 496 315 = 1.5746... 6000 and the supply curve is given by P = Q + 10. Find Q + 50 the equilibrium price and quantity, and compute the consumer and producer surplus. 8. (5 pts) Suppose the demand curve is given by P = Solution: To find Q∗ we set both values of P equal: 6000 = Q∗ + 10, Q∗ + 50 6000 = (Q + 50)(Q + 10), (Q∗ )2 + 60Q∗ − 5500 = 0, 290 5-2h21 ins5-2h22-28 5-2h23 5-2h24 5-2h25 5-2h26 5-2h27 5-2h28 5-2h29 (Q∗ − 50)(Q∗ + 110) = 0, Chapter Five KEY CONCEPT: THE DEFINITE INTEGRAL Z Z Q∗ 50 6000 Z Z Q∗ 50 2 In Problems ??, find the area of∗ the regions between the curve PS = (P − g(Q)) dQ = (60 − (Q + 10)) dQ = and the horizontal axis 1 0 Z 50 (x)[50Q − (50 − Q) dQf= 0 2500 Q2 50 = = 1250 . 2 0 2 x 22. Under y = 6x − 2 for 5 ≤ x ≤ 10. Some extra practice (not to be handed in) −1 23. Under the curve y = cos t for 0 ≤ t ≤ π/2. 5-2h32fig0 −2 24. Under y = ln x for ≤ 4. 1. Suppose h is1 ≤ ax function such that h(1) = −2, h (1) = 3, h002(1) = 44, h(2)6 = 6, 8h0 (2) 10 = 5, h00 (2) = 13, and h00 is R 2 00 25. Undercontinuous y = 2 cos(t/10) for 1 ≤ t ≤ 2. everywhere. Find 1 h (u) du. Figure 5.39: Graph consists of a semicircle and √ 26. Under the curve y = cos x for 0 ≤ x ≤ 2. line segments R 2 00 0 0 27. UnderSolution: the curve y = x2Evaluation and above theTheorem, x-axis. By7 − the h (u) du = h (2) − h (1) = 5 − 3 = 2. 1 28. Above the curve y = x4 − 8 and below the x-axis. 2. Use the figure below to find the values of 29. Use Figure ?? to find the values of 5-2h33 33. (a) Graph f (x) = x(x + 2)(x − 1). "b Rb "c f (x) dx (b) f (x) dx (a) (a) f (x) dx (b) Find the total area between the graph and the x-axis " ac Rc "bc between cdx (c) f (x) (d) |f (x)| dx " 1 x = −2 and x = 1. a a (b) a |f (x)| dx (c) Find −2 f (x) dx and interpret it in terms of areas. 3 5-2h34 Area = 13 ! a b 5-2h29fig c "4 0 cos √ x dx and interpret " 2π 5-2h35 35. Without computation, decide if 0 e−x sin x dx is positive or negative. [Hint: Sketch e−x sin x.] 5-2h36 36. Estimate x "Area = 2 34. Compute the definite integral the result in terms of areas. ! (a) Figure 5.36 "1 0 2 e−x dx using n = 5 rectangles to form a (b) Left-hand sum Right-hand sum 5-2h37 37. (a) On a sketch of y = ln x, represent"the left Riemann 2 sum with n = 2 approximating 1 ln x dx. Write "2 "2 out the terms in the sum, but do not evaluate it. Rb f (x)dx (b)R c f (x)dx (a) 0 (x) dx = −(−2) = 2. (b) On another sketch, represent the right Riemann sum (a) c f (x) dx = − b f−2 "2 (c) The total shaded area. lnbelow x dx. Write out the is reflected to be = 2 approximating (b) The graph of |f (x)| is the same as the graph of with f (x)n except that the part the x-axis 1 f (x) terms in the sum, but do not evaluate it. above it. Thus Z c (c) Which sum is an overestimate? Which sum is an un2 |f (x)| dx = 13 + 2 = 15. derestimate? "0 30. GivenSolution: f (x)dx = 4 and Figure ??, estimate: −2 x −2 5-2h30fig 2 −2 Figure 5.37 5-2h31 50 ∗ 5-2h32 21. (a) Find the total area between x3 −−x P and finddQ the = values of ln(Q + 50) − 60Q] (f (Q) ) the dQ =32. Use( Figure ??−to60) CS = f (x) = [6000 Q"+ 50 0 x-axis! for 0 ≤ x ≤ 3. 0 0 "7 2 3 (a) f (x) dx (b) f (x) dx 0 3 f (x)dx. (b) Find "7 " 8 − 3000 . = (6000 ln(100) − 60(50)) = 6000 (c) − 26000 f (x)ln(50) dx (d) ln(2) f (x) dx 0 5 (c) Are the answers to parts (a) and (b) the same? ExAlso, plain. f (x) 5-2h30 Q∗ = 50 , Q∗ =-110. For practical purposes we ignored the negative equilibrium quantity Q∗ = −110 above. Thus P ∗ = Q∗ + 10 = 50 + 10 = 60, so P ∗ = 60 . 0 5-2h22 6000 = (Q∗ )2 + 60Q∗ + 500, "0 a 5-2h38 38. (a) Draw the rectangles " π that give the left-hand sum approximation to 0 sin x dx with n = 2. "0 (b) Repeat part (a) for −π sin x dx. (c) From your answers to parts (a) and (b), what is the " π value of the left-hand sum approximation to sin x dx with n = 4? −π 31. (a) Using Figure ??, find −3 f (x) dx. " (b) " If the area of the shaded region is A, estimate 5-2h39 39. (a) Use a calculator or computer to find 6 (x2 + 1) dx. 4 0 f (x) dx. −3 Represent"this value as the area under a curve. 6 (b) Estimate 0 (x2 + 1) dx using a left-hand sum with 1 f (x) n = 3. Represent this sum graphically on a sketch 4 x of f (x) = x2 + 1. Is this sum an overestimate or 2 Part I I(28 points) • Write your name and recitation section number. • Staple your homework if you have multiple pages! 1. (4 pts) Find a function f on [3, 8] such that the limit below is equal to lim n→∞ n X i=1 s  5i 1 + ln 3 + n  5i R8 3 · e−3− n · f (x) dx. 5 . n Solution: Regardless of what our f is, the above looks like lim Rn . Indeed, with ∆x = n→∞ xi = 3 + i∆x = 3 + 8−3 n = 5 n, we get 5i n p for i = 0, 1, 2, . . . n. The terms in the original sum are just 1 + ln(xi )e−xi ∆x. One suitable candidate for f is p therefore f (x) = 1 + ln(x)e−x . One can check that for this choice of f , we get the limit stated in the problem; Z 3 8 n X p 1 + ln(x)e−x dx = lim Rn = lim f (xi )∆x n→∞ n→∞ s i=1   5i 5i 5 1 + ln 3 + · e−(3+ n ) · n→∞ n n i=1 s   n X 5i 5i 5 1 + ln 3 + · e−3− n · . = lim n→∞ n n i=1 = lim n X Z x i d h e p 2. (3 pts) Find 1 + r2 dr dx x2 Solution: By the Chain Rule and part 1 of the FTC, d h dx Z b(x) a(x) i f (t) dt = f (b(x))b0 (x) − f (a(x))a0 (x). a(x) = x2 and b(x) = ex , so d h dx 3. (4 pts) Find Z ex x2 Z p i p p p p d d 1 + r2 dr = 1 + (ex )2 (ex ) − 1 + (x2 )2 (x2 ) = ex 1 + e2x − 2x 1 + x4 . dx dx 2x (2x + 1)4/5 dx Solution: Let u = 2x + 1. Then du = 2x ln 2, so Z 2x (2x +1)4/5 dx = 4. Let F (x) = Z x Z u4/5 · 1 ln(2) du = 2x dx, and 9/5 1 1 u9/5 5 5 du = · +C = ·u9/5 +C = 2x + 1 +C ln(2) ln(2) 9/5 9 ln(2) 9 ln(2) 2 e−t dt 0 (a) (1 pt) Find F 0 (x). (b) (3 pts) Find the intervals where F is concave up and the intervals where F is concave down. (hint: differentiate the formula for F 0 (x) to get F 00 (x)) (C constant). 2 Solution: (a) By part 1 of the FTC, F 0 (x) = e−x . (b) For concavity, we take F 00 (x). 2 d h −x2 i F 00 (x) = e = −2xe−x . dx e raised to any power is positive, so the sign of F 00 is the same as the sign of −2x. Thus, F 00 (x) > 0 if x < 0, F 00 (0) = 0, F 00 (x) < 0 if x > 0. Thus, F is concave upward on (−∞, 0) and concave downward on (0, ∞). It is also fine if the student says concave up on (−∞, 0] and concave down on [0, ∞). Z x 5. (8 pts) A function f is graphed below. Let g(x) = f (t) dt. −2 4 y3 2 1 -2 -1 1 −1 t 2 −2 −3 −4 (a) (2 pts) Find the interval or intervals where g is decreasing. (b) (4 pts) Find g(−1), g(0), g(1), and g(2). (c) (2 pts) Find the inflection points of g. Solution: (a) By the FTC, g 0 (x) = f (x). Therefore, g is increasing on [−1, 1.5], but g is decreasing on [−2, −1] and [1.5, 2]. (it is fine whether the student includes or excludes the endpoints of each interval) R −1 (b) We use areas of triangles and rectangles. g(−1) = −2 f (t) dt = − 21 (2)(1) = −1. g(0) = Z Z 0 f (t) dt = −2 Z 1 −1 f (t) dt + −2 Z 0 Z Z 0 1 f (t) dt = −1 + (1)(2) = 0. 2 −1 1 1 · 1 · 2) = 3. 2 −2 −2 0   Z 2 Z 1 Z 2 1 1 1 1 · · 4 − · · 4 = 3. g(2) = f (t) dt = f (t) dt + f (t) dt = 3 + 2 2 2 2 −2 −2 1 g(1) = f (t) dt = f (t) dt + f (t) dt = 0 + (1 · 2 + (c) g 00 (x) = f 0 (x). Therefore, g is concave up on (−2, 1) since f 0 > 0 there, while g is concave down on (1, 2) since f 0 < 0 there. Thus, g has an inflection point when x = 1. This POINT is (1, g(1)), which is (1, 3) . Z √ 6. (5 pts) Find (z + 2) 1 − z dz Solution: Let u = 1 − z (so du = −dz). Then z = 1 − u, so z + 2 = 3 − u, hence Z √ (z+2) 1 − z dz = Z √ (3−u) u (−du) = Z Some additional problems (not to be handed in): √ 2 2 2 (−3 u+u3/2 ) du = −3 u3/2 + u5/2 = −2(1 − z)3/2 + (1 − z)5/2 + C . 3 5 5 1. Using your answers from problem 5, graph the function g from problem 5. 4 3 2 1 -2 1 -1 2 -1 2. Let F (x) = Z x (1 + t4 )1/4 dt. Is F even or odd? Justify your answer. 0 Solution: F is odd. Since (1 + t4 )1/4 is even, the area under the curve from −x to 0 is the same as the area R0 Rx under the curve from 0 to x, so (in shorthand) −x = 0 . From the properties of the definite integral, we also R0 R −x have −x = − 0 . Putting these together we get: F (x) = Z Z x (1 + t4 )1/4 dt = 0 (1 + t4 )1/4 dt = − −x 0 Z −x (1 + t4 )1/4 dt = −F (−x) . 0 The boxed formulas are the very definition of an odd function. Z r √ 1+ x 3. Find dx x √ Solution: Let u = 1 + x. Then du = Z r Z 1 √ 2 x dx so 2du = Z 4 0 4 √ x √ dx = 1 + 2x Z = Z 1+2(4) (u−1) 2 √ 1+2(0) 1 = 4 5. Find dx, and √ √ Z p Z √ √ 2 4 1+ x 1+ x √ dx = u 2 du = 2 u3/2 + C = (1 + x)3/2 + C . dx = x 3 3 x x dx. 1 + 2x 0 Solution: Let u = 1 + 2x. Then du = 2 dx, so dx = 4. Find √1 x  u du = 2 Z 1 9 du 2 , u−1 2 . and 2x = u − 1, so x = 1u−1 1 √ du = 4 4 u Z 9 1 u−1 1 √ du = 4 u Z 9 (u1/2 − u−1/2 ) du 1      √ 9 2 3/2 1 1 1 3 1 1 3/2 1 1/2 9 1 3/2 1 √ u −2 u = u − u = 9 − 9 − − = (27) − + 3 6 2 6 2 6 2 6 2 3 1 1 27 9 2 20 10 − + = = . 6 6 6 6 3 √ (x + 7) 3 3 − 2x dx Solution: Let u = 3 − 2x, so −2x = u − 3, hence x = − u2 + 3 2 and x + 7 = − u2 + 17 2 . Also, dx = − 12 du, and we find Z √ (x + 7) 3 3 − 2x dx = = = Z (− u 17 1/3 du −1 + )u · − = 2 2 2 4 Z (−u4/3 + 17u1/3 ) du 1 3 7/3 3 3 7/3 51 4/3 ( u − 17 u4/3 + C = u − u +C 4 7 4 28 16 51 3 (3 − 2x)7/3 − (3 − 2x)4/3 + C . 28 16 ...
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