Math 132 Exam I Spring 2008 This exam contains 15 multiple choice questions and 2 hand graded questions The multiple choice questions are worth 5 points each and the hand graded questions are worth a total of 25 points The latter questions will be evaluated not only for having the correct solutions but also for clarity Points may be taken for confusing and disorganized writing even when the answer is correct 1 Evaluate the sum 30 X 2j j 2 j 2 A 10374 B 10375 C 10376 D 10377 E 10378 F 10379 G 10380 H 10381 I 10382 J 10383 30 X j 2 2j j 2 3 30 X 2j j 2 j 1 3 2 30 X j 1 3 2 j 30 X j 1 30 31 2 3 930 9455 10382 1 j2 30 31 61 6 2 Find a formula for the Riemann sum RN for the function f x x2 over the interval 1 2 A RN 1 N PN 1 B RN 1 N PN j 1 2 N C RN 2 N PN j 1 2 N D RN 1 N PN j 1 2 N E RN 1 N PN 1 2 1 j 1 N F RN 2 N PN 1 1 j 2 N G RN 2 N PN 1 1 j 1 2 N H RN 1 N PN 1 2 j 1 N I RN 1 N PN 1 1 j 2 N J RN 1 N PN 1 1 j 1 2 N j 0 2 j 1 2 j 1 1 j 1 1 j 0 j 0 j 0 j 0 j 0 j 0 x 1 N RN x j 1 2 N aj 1 N 1 X j 0 j N 2 N 1 1 X j 1 1 f aj 1 N j 0 N 2 3 The limit 4 N 3 X 3j lim 2 N N N j 1 represents a definite integral What is this integral R3 A 0 x4 dx R5 B 2 x4 dx R3 C 0 2 3x 4 dx R3 D 0 2 3x N 4 dx R5 E 2 x N 4 dx R5 F 2 2 x 4 dx R5 G 2 2 3x 4 dx RN H 1 2 3x N 4 dx R1 I 0 2 x 4 dx R2 J 0 2 x 4 dx The Riemann sum can be written in the form x N X f aj j 1 where f x x4 and the set of points aj 2 3j N j 1 30 approximates the interval 2 5 Notice that x 5 2 N 3 N This corresponds to the integral Z 5 x4 dx 2 3 4 Evaluate the sum of integrals Z 2 Z f x dx 0 6 f x dx 0 where f x is the function shown in the graph 3 2 1 0 1 2 0 1 2 3 4 5 6 A B 4 C 4 D 2 E 2 F G 2 H 3 I 3 2 J 4 Notice that Z 2 Z 0 6 Z f x dx f x dx 0 6 f x dx 2 is the area of the larger half disc of radius 2 which is 2 4 5 Calculate the integral Z 5 3f x 5g x dx 0 assuming that R5 0 f x dx 7 and R5 0 g x dx 5 A 3 B 3 C 4 D 4 E 7 F 7 G 2 H 2 I 5 J 5 Z 5 Z 3f x 5g x dx 3 0 5 Z f x dx 5 0 3 7 5 5 4 5 5 g x dx 0 6 Calculate the integral 2 Z x2 1 dx 0 A 4 B 3 C 2 D 1 E 0 F 1 G 2 H 3 I 4 J 5 We have x2 1 0 over the interval 1 1 and positive for x 1 and x 1 Thus Z 2 Z 1 Z 2 2 2 x 1 dx 1 x dx x2 1 dx 0 0 1 x x3 3 10 x3 3 x 21 1 1 3 8 3 2 1 3 1 2 6 7 Calculate the integral Z 0 3x 2ex dx 2 A 4e2 2 B 2e 2 8 C e 2 8 D e 2 8 E 2e 2 8 F 2e 2 4 G 2e2 8 H 2e2 4 I 2e2 8 J 4e 2 4 Z 0 3x 2ex dx 3x2 2 2ex 0 2 2 6 2e 2 2e 2 8 2 7 8 Calculate the integral Z a 3a dt t A ln 2 a B ln 3 a C a ln 3 D a ln 2 E ln 3 ln a F ln 3a ln a G ln 2a H ln 3 a I ln 3a J ln 3 Z a 3a 3a dt ln t 3a ln 3 a ln 3a ln a ln t a 8 9 A function G s is defined by the integral Z cos s u4 3u du G s 6 Find G0 s A G0 s cos s cos4 s 3 cos s B G0 s cos s cos4 s 3 cos s 6 C G0 s sin s cos4 s 3 cos s D G0 s sin s cos4 s 3 cos s 6 E G0 s sin s cos4 s 3 cos s 4 cos3 s sin s 3 sin s F G0 s sin s cos4 s 3 cos s 64 18 G G0 s cos s cos4 s 3 cos s 64 18 H G0 s sin s cos4 s 3 cos s 64 18 I G0 s cos s sin4 s 3 sin s J G0 s sin s cos4 s 3 cos s 6 Rx Writing F x 6 u4 3u du and g s cos s we have G s F g s The chain rule gives G0 s F 0 g s g 0 s Therefore G0 s cos4 s 3 cos s sin s 9 10 Calculate the derivative d dx Z x2 tan tdt x A 2x sec2 x2 sec2 x 2 x B sec2 x2 x2 sec2 x x C tan x2 2x tan x x D sec2 x2 x2 tan x x E tan x2 x2 tan x x F x2 tan x2 tan x x G 2x tan x2 tan x 2 x H 2x tan x2 tan x x I x tan x2 tan x 2 x J x tan x2 tan x 2 x d dx Z x2 x Z x d tan tdt tan tdt dx 0 0 2x tan x2 tan x 2 x d tan tdt dx Z x2 10 11 Water flows into an empty reservoir at a rate of 600 5t gallons per hour What is the quantity of water in the reservoir after 2 hours A 1207 B 1208 C 1209 D 1210 E 1211 F 1212 G 1213 H 1214 I 1215 J 1216 This …
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