Homework 3 with Solution - First Year Interest Group Seminars | N 1, Assignments of Health sciences

Download Homework 3 with Solution - First Year Interest Group Seminars | N 1 and more Assignments Health sciences in PDF only on Docsity! Van Ligten (hlv63) – HW03 – Gilbert – (56650) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Decide which of the following regions has area = lim n→∞ n ∑ i = 1 π 4n cos iπ 4n without evaluating the limit. 1. { (x, y) : 0 ≤ y ≤ cos 2x, 0 ≤ x ≤ π 4 } 2. { (x, y) : 0 ≤ y ≤ cos 3x, 0 ≤ x ≤ π 4 } 3. { (x, y) : 0 ≤ y ≤ cos 3x, 0 ≤ x ≤ π 8 } 4. { (x, y) : 0 ≤ y ≤ cos x, 0 ≤ x ≤ π 4 } correct 5. { (x, y) : 0 ≤ y ≤ cos 2x, 0 ≤ x ≤ π 8 } 6. { (x, y) : 0 ≤ y ≤ cos x, 0 ≤ x ≤ π 8 } Explanation: The area under the graph of y = f(x) on an interval [a, b] is given by the limit lim n→∞ n ∑ i = 1 f(xi) ∆x when [a, b] is partitioned into n equal subin- tervals [a, x1], [x1, x2], . . . , [xn−1, b] each of length ∆x = (b − a)/n. If A = lim n→∞ n ∑ i = 1 π 4n cos iπ 4n , therefore, we see that f(xi) = cos iπ 4n , ∆x = π 4n . But in this case xi = iπ 4n , f(x) = cos x, [a, b] = [ 0, π 4 ] . Consequently, the area is that of the region under the graph of y = cos x on the interval [0, π/4]. In set-builder notation this is the region { (x, y) : 0 ≤ y ≤ cos x, 0 ≤ x ≤ π 4 } . 002 10.0 points Estimate the area under the graph of f(x) = 3 sin x between x = 0 and x = π4 using five approx- imating rectangles of equal widths and right endpoints. 1. area ≈ 1.023 2. area ≈ 1.043 correct 3. area ≈ 1.003 4. area ≈ 1.063 5. area ≈ 1.083 Explanation: An estimate for the area under the graph of f on [0, b] with [0, b] partitioned in n equal subintervals [xi−1, xi] = [ (i − 1)b n , ib n ] and right endpoints xi as sample points is A ≈ { f(x1) + f(x2) + . . . + f(xn) } b n . For the given area, f(x) = 3 sinx, b = π 4 , n = 5 , and x1 = 1 20 π, x2 = 1 10 π, x3 = 3 20 π, x4 = 1 5 π, x5 = 1 4 π . Van Ligten (hlv63) – HW03 – Gilbert – (56650) 2 Thus A ≈ 3 { sin ( 1 20 π ) + . . . + sin (1 4 π )} π 20 . After calculating these values we obtain the estimate area ≈ 1.043 for the area under the graph. keywords: area, sin function, estimate area, numerical calculation, 003 10.0 points Rewrite the sum 4 n ( 2 + 3 n )2 + 4 n ( 2 + 6 n )2 + . . . + 4 n ( 2 + 3n n )2 using sigma notation. 1. n ∑ i = 1 4 n ( 2 + 3i n )2 correct 2. n ∑ i = 1 3i n ( 2 + 4i n )2 3. n ∑ i = 1 3 n ( 2i + 4i n )2 4. n ∑ i = 1 3 n ( 2 + 4i n )2 5. n ∑ i = 1 4i n ( 2 + 3i n )2 6. n ∑ i = 1 4 n ( 2i + 3i n )2 Explanation: The terms are of the form 4 n ( 2 + 3i n )2 , with i = 1 , 2 , . . . , n. Consequently in sigma notation the sum becomes n ∑ i =1 4 n ( 2 + 3i n )2 . 004 10.0 points Cyclist Joe brakes as he approaches a stop sign. His velocity graph over a 5 second period (in units of feet/sec) is shown in 1 2 3 4 5 4 8 12 16 20 Compute best possible upper and lower es- timates for the distance he travels over this period by dividing [0, 5] into 5 equal subinter- vals and using endpoint sample points. 1. 58 ft < distance < 74 ft 2. 56 ft < distance < 76 ft 3. 56 ft < distance < 74 ft 4. 58 ft < distance < 76 ft 5. 58 ft < distance < 72 ft 6. 54 ft < distance < 72 ft 7. 54 ft < distance < 74 ft 8. 56 ft < distance < 72 ft correct 9. 54 ft < distance < 76 ft Explanation: Van Ligten (hlv63) – HW03 – Gilbert – (56650) 5 Explanation: Since [−3, 3] is subdivided into six equal subintervals, each of these will have length 1 and the six corresponding rectangles are shown as the shaded areas in -2 -1 0 1 2 3 4 5 6 7 8 9 10111213 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 1 2 3−1−2−3 2 4 6 8 −2 −4 −6 The heights of the rectangles are right end- point sample values of f that can be read off from the graph. Thus, with right endpoints, I ≈ 2 − 5 − 1 + 3 + 4 + 6 = 9 . 008 (part 2 of 3) 10.0 points (ii) Estimate the definite integral I = ∫ 3 −3 f(x) dx with six equal subintervals using left end- points. 1. I ≈ 9 2. I ≈ 12 3. I ≈ 10 4. I ≈ 11 5. I ≈ 8 correct Explanation: Since [−3, 3] is subdivided into six equal subintervals, each of these will have length 1 and the six corresponding rectangles are shown as the shaded areas in -2 -1 0 1 2 3 4 5 6 7 8 9 10111213 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 1 2 3−1−2−3 2 4 6 8 −2 −4 −6 The heights of the rectangles are left endpoint sample values of f that can be read off from the graph. Thus, with left endpoints, I ≈ 5 + 2 − 5 − 1 + 3 + 4 = 8 . 009 (part 3 of 3) 10.0 points (iii) Estimate the definite integral I = ∫ 3 −3 f(x) dx with six equal subintervals using midpoints. 1. I ≈ 10 2. I ≈ 9 3. I ≈ 11 correct 4. I ≈ 8 5. I ≈ 7 Explanation: Since [−3, 3] is subdivided into six equal subintervals, each of these will have length 1 and the six corresponding rectangles are shown as the shaded areas in Van Ligten (hlv63) – HW03 – Gilbert – (56650) 6 -2 -1 0 1 2 3 4 5 6 7 8 9 10111213 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 1 2 3−1−2−3 2 4 6 8 −2 −4 −6 The heights of the rectangles are midpoint sample values of f that can be read off from the graph. Thus, with midpoints, I ≈ 4 − 2 − 3 + 2 + 4 + 6 = 11 . 010 10.0 points If an nth-Riemann sum approximation to the definite integral I = ∫ b a f(x) dx is given by n ∑ i = 1 f(x∗i ) ∆xi = 6n2 − 2n + 1 2n2 , determine the value of I. 1. I = 5 2 2. I = 3 correct 3. I = 1 2 4. I = 2 5. I = −1 Explanation: By definition, ∫ b a f(x) dx = lim n→∞ n ∑ i = 1 f(x∗i ) ∆xi . Thus ∫ b a f(x) dx = lim n→∞ 6n2 − 2n + 1 2n2 . Consequently, I = 3 . 011 10.0 points Find the limit of n ∑ k =1 ( 5 n3 k2 − 1 n2 k + 4 n ) as n → ∞. 1. limit = 10 2. limit = 17 2 3. limit = 6 4. limit = 13 3 5. limit = 31 6 correct Explanation: It is known that n ∑ k =1 1 = n, n ∑ k =1 k = 1 2 (n + 1)n, and n ∑ k =1 k2 = 1 6 n(n + 1)(2n + 1). Thus n ∑ k = 1 4 n = 4, n ∑ k = 1 1 n2 k = 1 2n2 (n + 1)n Van Ligten (hlv63) – HW03 – Gilbert – (56650) 7 while n ∑ k = 1 5 n3 k2 = 5 6n3 n(n + 1)(2n + 1). But then 1 n2 n ∑ k = 1 k = 1 2 ( 1 + 1 n ) → 1 2 as n → ∞, while 5 n3 n ∑ k =1 k2 = 5 6 ( 2 + 3 n + 1 n2 ) → 5 3 as n → ∞. Hence lim n→∞ n ∑ k =1 ( 5 n3 k2 − 1 n2 k + 4 n ) = 31 6 . 012 10.0 points Evaluate the definite integral I = ∫ 2 4 (2f(x) − 4g(x))dx when ∫ 4 2 f(x) dx = 2 , ∫ 4 2 g(x) dx = 3 . 1. I = 5 2. I = 6 3. I = 4 4. I = 8 correct 5. I = 7 Explanation: Since ∫ a b F (x) dx = − ∫ b a F (x) dx for all a, b and F , we see that I = − ∫ 4 2 (2f(x) − 4g(x))dx . Using the Linearity of Integrals we can then write I = − ( 2 ∫ 4 2 f(x) dx− 4 ∫ 4 2 g(x) dx ) . Consequently, I = −(4 − 12) = 8 . 013 10.0 points Use properties of integrals to determine the value of I = ∫ 7 0 f(x) dx when ∫ 5 0 f(x) dx = 11, ∫ 7 5 f(x) dx = 6 . Correct answer: 17. Explanation: Since ∫ 7 0 f(x) dx = ∫ 5 0 f(x) dx + ∫ 7 5 f(x) dx , we see that I = 11 + 6 = 17 . 014 10.0 points If f and g are continuous functions such that f(x) ≥ 0 for all x, which of the following must be true? I. ∫ b a f(x) g(x) dx = ( ∫ b a f(x) dx )( ∫ b a g(x) dx ) II. ∫ b a { f (x) + g(x) } dx = ∫ b a f(x) dx + ∫ b a g(x) dx Van Ligten (hlv63) – HW03 – Gilbert – (56650) 10 3. not enough information given 4. x = 3.5 5. x = 6 correct 6. x = 2 Explanation: By the Fundamental theorem of calculus, if g(x) = ∫ x 2 f(t) dt, then g′(x) = f(x). Thus the critical points of g occur at the zeros of f , i.e., at the x- intercepts of the graph of f . To determine which of these gives a local maximum of g we use the sign chart g′ + − 2 6 8 for g′. This shows that the maximum value of g occurs at x = 6 since the sign of g′ changes from positive to negative at x = 6. 017 10.0 points Find g′(x) when g(x) = ∫ x π (6 + cos t)dt . 1. g′(x) = 6x + sinx 2. g′(x) = − sin x 3. g′(x) = 6 − sin x 4. g′(x) = 6x − cos x 5. g′(x) = 6 + cos x correct Explanation: By the Fundamental theorem of Calculus, if g(x) = ∫ x a f(t) dt , then g′(x) = d dx ∫ x a g(t) dt = f(x) . In the given example, therefore, g′(x) = 6 + cos x . 018 10.0 points Determine F ′(x) when F (x) = ∫ x2 4 2 √ 1 + t3 dt 1. F ′(x) = 2x√ 1 + x3 2. F ′(x) = 4x √ 1 + x3 3. F ′(x) = 2x2 √ 1 + x6 4. F ′(x) = 4x√ 1 + x3 5. F ′(x) = 4x √ 1 + x6 correct 6. F ′(x) = 2x2√ 1 + x6 Explanation: By the Fundamental Theorem of Calculus and the Chain Rule, d dx ( ∫ g(x) a f(t) dt ) = f(g(x))g′(x) . When F (x) = ∫ x2 4 2 √ 1 + t3 dt , Van Ligten (hlv63) – HW03 – Gilbert – (56650) 11 therefore, F ′(x) = 4x √ 1 + x6 . 019 10.0 points Evaluate the definite integral I = ∫ 2 0 ( 2 + 4y − y2 ) dy . 1. I = 22 3 2. I = 31 3 3. I = 28 3 correct 4. I = 34 3 5. I = 25 3 Explanation: By the Fundamental Theorem of Calculus, I = [ F (x) ]2 0 = F (2) − F (0) for any anti-derivative F of f(y) = 2 + 4y − y2 . Taking F (y) = 2y + 2y2 − 1 3 y3 , we thus see that I = 4 + 8 − 8 3 . Consequently, I = 28 3 . 020 10.0 points What value of x maximizes F when F (x) = ∫ x+4 x 1 t2 + 4 dt ? Correct answer: −2. Explanation: To apply the Fundamental Theorem of Cal- culus in the form d dx ( ∫ x a f(t) dt ) = f(x) , write F (x) = ∫ x+4 x 1 t2 + 4 dt = ∫ x+4 a 1 t2 + 4 dt − ∫ x a 1 t2 + 4 dt . Then F ′(x) = 1 (x + 4)2 + 4 − 1 x2 + 4 = x2 − (x + 4)2 ((x + 4)2 + 4)(x2 + 4) , which after simplification becomes F ′(x) = − 8(x + 2) ((x + 4)2 + 4)(x2 + 4) . Thus the only critical point of F occurs at x = −2. On the other hand, the sign chart + − −∞ ∞−2 shows that F ′(x) changes from positive to negative as x passes through −2. Conse- quently, F has a maximum at x = −2 .