Math for Econ II Homeowkr 8 Solutions - (c 0(c Are the answers to parts(a and(b the same Explain ins5-2h22-28 2(d f(x dx 5 f(x dx 2 In Problems nd the

Math for Econ II Homeowkr 8 Solutions - (c 0(c Are the...

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Unformatted text preview: (c) 0 (c) Are the answers to parts (a) and (b) the same? Explain. ins5-2h22-28 2 (d) f (x) dx 5 f (x) dx 2 In Problems ??, find the area of the regions between the curve Math for Econ II, Written Assignment 8 (28 points) f (x) and the horizontal axis 1 5-2h22 5-2h23 5-2h24 5-2h25 5-2h26 5-2h27 5-2h28 5-2h29 Due Friday, November 7 x 22. Under y = 6x3 − 2 for 5 ≤ x ≤ 10. Jankowski, Fall 2014 −1 23. Please the curve y = solutions ≤ t ≤ π/2. Under write neat cos t for 0 for the problems below. Show all your work. If you only write the answer with no work, 5-2h32fig −2 24. you will = ln be given x ≤ 4. Under y not x for 1 ≤ any credit. 2 4 6 8 10 25. Under y = 2 cos(t/10) for 1 ≤ t ≤ 2. √ • Write your name and recitation section number. Figure 5.39: Graph consists of a semicircle and 26. Under the curve y = cos x for 0 ≤ x ≤ 2. line segments 27. Under the curve y =homework above the x-axis. • Staple your 7 − x2 and if you have multiple pages! 28. Above the curve y = x4 − 8 and below the x-axis. 1. (2 pts total; 1 pt each) 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 c f bf (b) f (x) dx (a) (b) Find the total area between the graph and the x-axis (a)(x) dx(x) dx a b c c c between x = −2 and x = 1. (c) f (x)c |f (x)| dx (d) dx |f (x)| dx a a 1 (b) a (c) Find −2 f (x) dx and interpret it in terms of areas. f (x) 5-2h34 Area = 13 34. Compute the definite integral the result in terms of areas. © 4 0 cos √ x dx and interpret 5-2h35 a b 5-2h29fig 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 2π x c sArea = 2 ! (a) Figure 5.36 5-2h30 5-4h43 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. f b (b) c (a) 0 (a)(x)dx (x) dx = − b f−2 f (x)dx −(−2) = 2. (b) On another sketch, represent the right Riemann sum f (x) dx = c (c) The total shaded area. 2 (b) The graph of |f (x)| is the same as the graph of with n except that the part belowWrite x-axis is reflected to be f (x) = 2 approximating 1 ln x dx. the out the f (x) terms in the sum, but do not evaluate it. above it. Thus c (c) Which sum is an overestimate? Which sum is an un2 5.4 THEOREMS 2 = 15. 311 |f (x)| dx = 13 +ABOUT DEFINITE INTEGRALS derestimate? a 0 30. GivenSolution: = 4 and Figure ??, estimate: f (x)dx −2 x 5-2h38 2 43. Using the graph −2 f in Figure ??, arrange the ins5-4h47-48 In 38. (a) Draw the rectangles that give the left-hand sumor of following Problems ??, evaluate the expression, if possible, ap2. (4 pts) Using the graph of f in Figure, arrange the following quantities in increasing order, from least to greatest. π −2 quantities in increasing order, from least to greatest. say what proximationinformation dx with n = 2. additional to sin x is needed, given that 0 1 2 2 3 2 5-2h30fig 4 1 i) 2 f (x) dx ii) 1 f (x)(x) dx iii) 0 f (x) dx g(x) dx 2 f (x) dx(a) for − 1 f x dx. dx iv) Repeat part v) 0 sin (x) dx vi) The number 0 = 12. (b) f (x) dx (ii) f (i) 0 −4 0 (iii) 2 0 Figure 5.37 (iv) f (x) dx 1 3 2 −π (c) From your answers to parts (a) and (b), what is 4 the value of the left-hand 4 sum approximation to 5-4h48 48. π g(x) dx x dx with n = 4? g(−x) dx sin −π f (x) dx 20 viii) The 0 5-4h47 47. (v) − vii) (x) dx f The number (vi) The numbernumber −10 0 1 31. (a) Using Figure ??, find −3 f (x) dx. 0 −4 (vii) The number 20 (viii) The number −10 (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 f (x) Represent this value as the area under a curve. 10 ins5-4h49-52 6 In Problems ??, evaluate 2the 1) dx using if left-hand sumsay (b) Estimate 0 (x + expression 7a possible, or with 1 x f (x) what extra information is needed, given graphically on 25. 1 2 3 n = 3. Represent this sum 0 f (x) dx = a sketch 4 x 5-4h43fig −10 of f (x) = x2 + 1. Is this sum an overestimate or −4 −3 −2 −1 1 2 3 ! 5 underestimate of the true value found in part (a)? 7 3.5 5-2h31fig −1 6 5-4h50 Figure 5.71 5-4h49 49. f (x) dx 50. (c) Estimate 0 (x2 +1) dx using a f (x) dx sum with right-hand 0 n = 3. Represent this sum on your sketch. Is this 0 Figure 5.38 5-4h44 44. (a) Using Figures ?? and ??, find the average value on sum an overestimate or underestimate? Solution: 5 7 0 ≤ x ≤ 2 of 5-4h51 51. 5-4h52 52. 1 2 3 f (x + 2)dx = A , we know (f (x) + 2) dx Since 0 f (x) dxg(x) 1 and 1 ff (x)·g(x) −A2 and 2 f (x) dx =A (x) dx = that 3 (i) f (x) (ii) (iii) −2 0 (b) Is the following statement true? Explain your an1 2 3 swer. 0 < 5-4h53f (x) dx < − f (x) dx < f (x) dx. 53. (a) Sketch a graph√ f√2 = sin(x2 ) and mark on it of (x) √ 1 √ Average(f ) · Average(g) = Average(f · g) 0 the points x = π, 2π, 3π, 4π. 2 (b) Use your graph to in magnitude than 2 f (x) 1 1 In addition, 0 f (x) dx = A1g(x) 2 , which is negative, but smaller decide which of the four numbersdx. Thus −A f (x) 1 5-2h31 2 2 5-4h44figa x 1 5-4h44figb Figure 5.72 5-4h45 5-4h46 2 f (x) dx < x 1 2 Figure 5.73 45. (a) Without computing any integrals, explain why the ins5-4h54-56 average value of f (x) = sin x on [0, π] must be between 0.5 and 1. (b) Compute this average. 46. Figure ?? shows the standard normal distribution from 2 1 0 √ nπ 2 f (x) sin(x 0. dx < ) dx n = 1, 2, 3, 4 0 is largest. Which is smallest? How many of the numbers are positive? For Problems ??, assuming F = f , mark the quantity on a copy of Figure ??. F (x) The area A3 lies inside a rectangle of height 20 and base 1, so A3 < 20. The area A2 lies inside a rectangle below the x-axis of height 10 and width 1, so −10 < A2 . Thus: (viii) < (ii) < (iii) < (vi) < (i) < (v) < ( iv) < (vii). −1 f (x) dx −2 4 f (x) dx. 1 3. (3 pts) Suppose f is even, enough information to find −1 −2 Solution: Because f is even, 4 2 = 3, and f (x) dx = 4 2 1 f (x) dx = 5. Either find 4 f (x) dx + 1 f (x) dx, or show there is not f (x) dx, so 2 f (x) dx = 4 1 f (x) dx = 3 + 5 = 8 . 1 2 4. (5 pts) Evaluate the following integrals: 1 (f (x) 0 (a) (3 pts) If 1 (2f (x) 0 − 2g(x)) dx = 6 and + 2g(x)) dx = 9, find 1 (f (x) 0 − g(x)) dx. 3 (x9 − 4x3 + x) dx (b) (2 pts) −3 Solution: (a) Adding the first 2 integrals gives us 3 1 1 0 1 f (x) dx − 0 1 0 f (x) = 15, so f (x) = 5. We use the first integral again: 1 5 − 2 2g(x) dx = 6, 1 −2 g(x) dx = 6, 0 0 1 g(x) dx = 1, 0 0 1 g(x) dx = − . 2 Thus, 1 1 (f (x) − g(x)) dx = 0 1 f (x) − g(x) dx = 5 − (−1/2) = 0 0 11 . 2 3 (x9 − 4x3 + x) dx = 0 because f (x) = x9 − 4x3 + x is an odd continuous function and the interval is (b) −3 symmetric. (check f is odd: f (−x) = (−x)9 − 4(−x)3 + (−x) = −x9 + 4x3 − x = −(x9 − 4x3 + x) = −f (x)). 5. (2 pts) Explain what is wrong with the following calculation of the area under the curve 1 −1 1 1 dx = − 3 x4 3x 1 −1 1 x4 from x = −1 to x = 1: 1 1 2 =− − =− . 3 3 3 1 Solution: We cannot apply the evaluation theorem because x4 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 zero, because 1/x4 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 six 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. a b ex dx = 6 2e2x dx. 0 a ex dx = ex 0 a 0 b = ea − 1, 0 2e2x dx = 6 e2x 6 0 b = 6e2b − 6. 0 Thus ea − 1 = 6e2b − 6, ea = 6e2b − 5, a = ln(6e2b − 5) . 5 1 dx = ln(5). Now find a fraction which approximates ln(5), by x 1 5 1 using M4 (midpoint sum with 4 rectangles) to approximate dx. 1 x (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 5 Solution: By the Evaluation Theorem, 1 1 dx = ln(x) x 5 = ln(5) − ln(1) = ln(5). 1 (the formulas for ln |x| and ln(x) are identical since x > 0 on the interval [1, 5]). M4 = 1 · f 3 5 7 9 2 2 2 2 496 +1·f +1·f +1·f = + + + = 2 2 2 2 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, 6000 = (Q∗ )2 + 60Q∗ + 500, (Q∗ − 50)(Q∗ + 110) = 0, 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 . Q∗ 50 (f (Q) − P ∗ ) dQ = CS = 0 ( 0 6000 − 60) dQ = [6000 ln(Q + 50) − 60Q] Q + 50 50 0 = (6000 ln(100) − 60(50)) − 6000 ln(50) = 6000 ln(2) − 3000 . Also, Q∗ 50 (P ∗ − g(Q)) dQ = PS = 0 50 (60 − (Q + 10)) dQ = 0 (50 − Q) dQ = [50Q − 0 Q2 2 50 = 0 2500 = 1250 . 2 Some extra practice (not to be handed in) 3 1. Estimate f (x) dx using R5 and L5 . −2 −1 1 2 3 −5 −4 −3 −2 −3 −1 1 2 3 4 5 −2 2. Suppose h is a function such that h(1) = −2, h (1) = 3, h (1) = 4, h(2) = 6, h (2) = 5, h (2) = 13, and h is 2 continuous everywhere. Find 1 h (u) du. Solution: By the Evaluation Theorem, 2 1 h (u) du = h (2) − h (1) = 5 − 3 = 2. ...
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