hw 8 sol - ins5-2h22-28 In Problems find the area of the...

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Unformatted text preview: ins5-2h22-28 In Problems ??, find the area of the regions between the curve and the horizontal axis 5-2h22 5-2h23 5-2h24 5-2h25 5-2h26 5-2h27 5-2h28 5-2h29 2 f (x) 1 x 22. Under y = 6x3 − 2 for 5 ≤ x ≤ 10. −1 Assignment 8 (28 points) for Econ II, Written 23. Under the curve y = cos t for 0 ≤Math t ≤ π/2. Please write neat solutions for the problems below. Show 5-2h32fig −2 all your work. If you only write the answer with no work, 24. Under y = ln x for 1 ≤ x ≤ 4. 2 4 6 8 10 you will not be given any credit. 25. Under y = 2 cos(t/10) for 1 ≤ t ≤ 2. Figure 5.39: Graph consists of a semicircle and √ 26. Under the curve y =name cos and x forrecitation 0 ≤ x ≤ 2. section number. • Write your line segments 27. Under the curve y = 7 − x2 and above the x-axis. • Staple your homework if you have multiple pages! 28. Above the curve y = x4 − 8 and below the x-axis. 1. Figure (2 pts??total; ptvalues each)ofUse the figure below to find the values of 29. Use to find1the 5-2h33 33. (a) Graph f (x) = x(x + 2)(x − 1). "b R "c f (x)bdx (b) f (x) dx (a) (b) Find the total area between the graph and the x-axis f (x) dx " a(a) "bc c c R between (c) f (x)cdx (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. f (x) Area = 13 ! a b 5-2h29fig c √ 34. Compute the definite integral the result in terms of areas. 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 "4 5-2h34 ! (a) Figure 5.36 "1 0 0 cos x dx and interpret " 2π 2 e−x dx using n = 5 rectangles to form a Left-hand sum (b) Right-hand sum 5-2h37 "0 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 Rb " R 2 out the terms in the sum, but do not evaluate it. f (x)dx (b) c f−2 (a) (a) f (x) dx = − (x)f (x)dx dx = −(−2) = 2. 0 c b (b) On another sketch, represent " 2the right Riemann sum (c) The total shaded area. (b) The graph of |f (x)| is the same as the graph of with f (x)n except that the part the x-axis lnbelow x dx. Write out the is reflected to be = 2 approximating 1 above it. Thus f (x) Z c terms in the sum, but do not evaluate it. 5.4 THEOREMS DEFINITE INTEGRALS 311 (c)(x)| Which sum Which sum is an un|f dx = 13is+anABOUT 2overestimate? = 15. 2 derestimate? a 5-4h43 43. Using the graph of f in Figure ??, arrange the x ins5-4h47-48 following In Problems ??, evaluate the expression, if possible, or −2 2 of f in Figure, 5-2h38 2. (4 pts) Using the graph in give increasing order, from 38.the (a)following Draw the quantities rectangles the left-hand ap- least to greatest. quantities in increasing−2 order, from least to greatest. arrange say additional information is needed, givensum that " π that ! 4 what proximation R2!2 R2 R3 Rx2dx with n = 2. !1 R1 sin to g(x) 0 "− i)f (x)0 dx f (x) dx ii)(ii) 1 f (x) dxdx iii) 0 f (x) dx iv)dx 2=f12. (x) dx v) f (x) dx vi) The number 0 f (x) (i)5-2h30fig −4 0 1 0 !2 !13 (b) Repeat part (a) for −π sin x dx. (iii) f (x) dx (iv) f (x) dx " (c) " 4 (a) and (b), what is From your answers to parts 4 0 Figure 5.37 2 !2 5-4h47 5-4h48 Thedxnumber (vi) 20 The viii)number The number −10 the sum approximation to 47. g(x) 48. g(−x) dx (v) − vii) f (x) 0 " πdxvalue of the left-hand 1 "0 0 sin x dx with n = 4? −4 5-2h31 31.(vii) The number Thedx. number −10 (a) Using Figure20 ??, find (viii) f (x) −π −3 " (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 f (x) 0 f (x) dx. 10 −3 ins5-4h49-52 value as the areaifunder a curve. In ProblemsRepresent ??, evaluate the expression possible, or say "this ! 6 2 x + 1) dx using7afleft-hand sum (b) information Estimate 0 is(xneeded, (x) dx = 25. with what extra given 11 2 3f (x) 0 n = 3. Represent this sum graphically on a sketch 5-4h43fig −10 4 #" of f (x) = x2 + 1. Is this x an overestimate or ! " sum 7 3.5 −4 −3 2 −1 1 2 3 5 underestimate of the true value found in part (a)? Figure 5.71 5-4h49 49. 5-4h50 5-2h31fig f (x) dx " 6 50. f (x) dx −1 (c)0 Estimate 0 (x2 +1) dx using 0 a right-hand sum with 5-4h44 44. (a) Using Figures ?? and ??, find the average value on " 5 n = 3. Represent this sum " 7on your sketch. Is this Solution: Figure 5.38 0 ≤ x ≤ 2Rof R2 R 3sum an overestimate or underestimate? 5-4h51 51. 5-4h52 1 2)dx dx = A3 , we52. (x) + 2) dx f (x) dx = A and(iii)1 ff(x) dx = −A2 and f2(xf+ (x) know (f that Since (i) f (x) 0 (ii) g(x) 1 (x)·g(x) −2 0 Z Z 2 Z 3 (b) Is the following statement true? Explain your an-1 0 < f (x) dx < − f (x) dx < f (x) dx. swer. 5-4h53 0 1 a graph of f (x) 2 = sin(x2 ) and mark on it 53. (a) Sketch √ √ √ √ Average(f ) · Average(g) = Average(f · g) R2 R2 the points x = π, 2π, 3π, 4π. In addition, 0 f (x) dx = A1 − A2 , which is negative, but smaller in magnitude than 1 f (x) dx. Thus (b) Use your graph to decide which of the four numbers 1 1 f (x) g(x) 5-2h30 30. GivenSolution: f (x)dx = 4 and Figure ??, estimate: −2 Z Z 2 f (x) dx < 5-4h44figa The1area x 5-4h44figb A32 lies 1 x 1 2of height rectangle 0 2 " √ nπ f (x) dx <2 0. sin(x ) dx n = 1, 2, 3, 4 0 inside a 20 and base 1, so A3 < 20. The area A2 lies inside a rectangle below largest. Which is smallest? How many of the numthe x-axis of height 10 and width 1, so −10 < A2 . isThus: Figure 5.72 Figure 5.73 bers are positive? 5-4h45 5-4h46 (viii) < (ii) < (iii) < (vi) < (i) < (v) < ( iv) < (vii). 45. (a) Without computing any integrals, explainins5-4h54-56 why the 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 statistics, which is given by 2 1 √ e−x /2 . 2π For Problems ??, assuming F # = f , mark the quantity on a copy of Figure ??. F (x) R −1 3. (3 pts) Suppose f is even, enough information to find R −1 Solution: Because f is even, −2 Z R4 f (x) dx = 3, and R−2 4 f (x) 1 2 f (x) dx = 5. Either find R4 1 f (x) dx, or show there is not dx. f (x) dx = R2 Z 4 f (x) dx, so 1 Z 2 f (x) dx = 4 f (x) dx + 1 f (x) dx = 3 + 5 = 8 . 1 2 4. (5 pts) Evaluate the following integrals: 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. Z 3 (b) (2 pts) (x9 − 4x3 + x) dx −3 Solution: (a) Adding the first 2 integrals gives us 3 Z Z 1 f (x) dx − 0 R1 0 f (x) = 15, so Z 1 f (x) = 5. We use the first integral again: Z −2 g(x) dx = 6, 0 0 Thus, Z Z 1 (f (x) − g(x)) dx = 0 Z 0 1 5 − 2 2g(x) dx = 6, R1 Z 1 g(x) dx = 1, 0 Z 1 0 1 f (x) − g(x) dx = 5 − (−1/2) = 0 0 1 1 g(x) dx = − . 2 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 Z 1 −1 1 x4 from x = −1 to x = 1: h 1 1 1 1 i 1 2 dx = − 3 = − − = − . 4 x 3x −1 3 3 3 Solution: We cannot apply the evaluation theorem because x14 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. Z Z a ex dx = 6 2e2x dx. 0 Z a 0 0 a e dx = e = ea − 1, x b x 0 Z 6 b h ib 2e2x dx = 6 e2x = 6e2b − 6. 0 0 Thus ea − 1 = 6e2b − 6, ea = 6e2b − 5, Z 5 a = ln(6e2b − 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 7. (4 pts) Use the Evaluation Theorem to show (The actual value of ln(5) is 1.6094 . . .. For fun, plug your approximation into a calculator and compare) Z 5 Solution: By the Evaluation Theorem, 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, 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 . Z Z Q∗ (f (Q) − P ∗ ) dQ = CS = 0 50 ( 0 50 6000 − 60) dQ = [6000 ln(Q + 50) − 60Q] Q + 50 0 = (6000 ln(100) − 60(50)) − 6000 ln(50) = 6000 ln(2) − 3000 . Also, Z Z Q∗ (P ∗ − g(Q)) dQ = PS = 0 Z 50 (60 − (Q + 10)) dQ = 0 50 (50 − Q) dQ = [50Q − 0 2500 Q2 50 = = 1250 . 2 0 2 Some extra practice (not to be handed in) Z 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, h0 (1) = 3, h00 (1) = 4, h(2) = 6, h0 (2) = 5, h00 (2) = 13, and h00 is R2 continuous everywhere. Find 1 h00 (u) du. Solution: By the Evaluation Theorem, R2 1 h00 (u) du = h0 (2) − h0 (1) = 5 − 3 = 2. ...
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