HW 8 - Portillo, Tom Homework 8 Due: Oct 24 2006, 3:00 am...

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Portillo, Tom – Homework 8 – Due: Oct 24 2006, 3:00 am – Inst: David Benzvi 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – fnd 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 0 7 4 + x 2 dx. 1. I 10 . 4794 2. I 10 . 2794 3. I 9 . 87944 4. I 10 . 6794 5. I 10 . 0794 correct Explanation: When the interval [0 , 4] is partitioned into 4 equal subintervals the trapezoidal rule esti- mates the integral I = Z 4 0 f ( x ) dx as I 1 2 f (0)+2 f (1)+2 f (2)+2 f (3)+ f (4) · . When f ( x ) = 7 4 + x 2 , thereFore, I 7 2 1 2 + 2 4 + 1 + 2 4 + 4 + 2 4 + 9 + 1 4 + 16 · . Consequently, I 10 . 0794 . 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 frst 5 seconds oF a race as shown in the table t (secs) vel (m/sec) 0 0 0.5 4 . 6 1.0 7 . 6 1.5 8 . 9 2.0 9 . 1 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 . 49 meters 2. dist 44 . 53 meters 3. dist 44 . 51 meters 4. dist 44 . 55 meters correct 5. dist 44 . 47 meters Explanation: The distance covered during those 5 sec- onds is given by the integral I = Z 5 0 v ( t ) dt where v ( t ) is the velocity oF the runner at time
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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 of the values oF v ( t ) From the table we thus see that I 44 . 55 meters . keywords: de±nite integral, table, Simpson’s rule, velocity, distance 003 (part 1 oF 1) 10 points ²ind the total area under the graph oF y = 9 x 4 For x 1 . 1. Area = 3 correct 2. Area = 5 3 3. Area = 7 3 4. Area = 5. Area = 8 3 6. Area = 2 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 4 dx. Now Z t 1 9 x - 4 dx = - 3 h x - 3 i t 1 . Consequently, Area = lim t →∞ 3 h 1 - t - 3 i = 3 . keywords: improper integral, area, limit 004 (part 1 oF 1) 10 points Determine iF the improper integral I = Z 0 -∞ 4 2 x - 5 dx converges, and iF it does, ±nd its value. 1. I = 5 4 2. I = 2 3. I = 4 5 4. I = 1 2 5. integral is not convergent correct Explanation: The integral is improper because oF the in- ±nite interval oF integration. To test For con- vergence, thereFore, we have to check iF the limit lim t →-∞ Z 0 t 4 2 x - 5 dx exists. But Z 0 t 4 2 x - 5 dx = h 2 ln | 2 x - 5 | i 0 t = 2 ln 5 - 2 ln | 2 t -
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HW 8 - Portillo, Tom Homework 8 Due: Oct 24 2006, 3:00 am...

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