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Unformatted text preview: Cheung, Anthony – Homework 8 – Due: Oct 24 2006, 3:00 am – Inst: David Benzvi 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points After partitioning the interval [0 , 4] into 4 equal subintervals, use the trapezoidal rule to estimate the integral I = Z 4 8 √ 1 + x 2 dx. 1. I ≈ 16 . 9345 2. I ≈ 17 . 3345 3. I ≈ 17 . 1345 4. I ≈ 16 . 7345 correct 5. I ≈ 16 . 5345 Explanation: When the interval [0 , 4] is partitioned into 4 equal subintervals the trapezoidal rule esti mates the integral I = Z 4 f ( x ) dx as I ≈ 1 2 ‡ f (0)+2 f (1)+2 f (2)+2 f (3)+ f (4) · . When f ( x ) = 8 √ 1 + x 2 , therefore, I ≈ 4 ‡ 1 + 2 √ 1 + 1 + 2 √ 1 + 4 + 2 √ 1 + 9 + 1 √ 1 + 16 · . Consequently, I ≈ 16 . 7345 . keywords: partition, trapezoidal rule, inte gral, estimate 002 (part 1 of 1) 10 points A radar gun was used to record the speed of a runner during the first 5 seconds of a race as shown in the table t (secs) vel (m/sec) 0.5 4 . 8 1.0 7 . 1 1.5 8 . 2 2.0 9 2.5 10 . 1 3.0 10 . 3 3.5 10 . 7 4.0 10 . 8 4.5 10 . 88 5.0 10 . 99 Use Simpson’s Rule and all the given data to estimate the distance the runner covered during those 5 seconds. 1. dist ≈ 44 . 03 meters 2. dist ≈ 43 . 97 meters 3. dist ≈ 43 . 99 meters 4. dist ≈ 43 . 95 meters 5. dist ≈ 44 . 01 meters correct Explanation: The distance covered during those 5 sec onds is given by the integral I = Z 5 v ( t ) dt where v ( t ) is the velocity of the runner at time Cheung, Anthony – Homework 8 – Due: Oct 24 2006, 3:00 am – Inst: David Benzvi 2 t . Simpson’s Rule estimates this integral as I ≈ 1 6 n v (0) + 4 v ‡ 1 2 · + 2 v (1) + 4 v ‡ 3 2 · + 2 v (2) + 4 v ‡ 5 2 · + 2 v (3) + 4 v ‡ 7 2 · + 2 v (4) + 4 v ‡ 9 2 · + v (5) o . Reading off the values of v ( t ) from the table we thus see that I ≈ 44 . 01 meters . keywords: definite integral, table, Simpson’s rule, velocity, distance 003 (part 1 of 1) 10 points Find the total area under the graph of y = 9 x 3 for x ≥ 1 . 1. Area = 5 2 2. Area = 7 2 3. Area = 9 2 correct 4. Area = 3 5. Area = 4 6. Area = ∞ Explanation: The total area under the graph for x ≥ 1 is an improper integral whose value is the limit lim t →∞ Z t 1 9 x 3 dx. Now Z t 1 9 x 3 dx = 9 2 h x 2 i t 1 . Consequently, Area = lim t →∞ 9 2 h 1 t 2 i = 9 2 . keywords: improper integral, area, limit 004 (part 1 of 1) 10 points Determine if the improper integral I = Z∞ 4 2 x 7 dx converges, and if it does, find its value. 1. integral is not convergent correct 2. I = 2 3. I = 4 7 4. I = 7 4 5. I = 1 2 Explanation: The integral is improper because of the in finite interval of integration. To test for con vergence, therefore, we have to check if the limit lim t →∞ Z t 4 2 x 7 dx exists. But Z t 4 2 x 7 dx = h 2 ln  2 x 7  i t = ‡ 2 ln 7 2 ln  2 t...
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This note was uploaded on 01/27/2010 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas.
 Spring '08
 RAdin
 Calculus

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