THE INTERNATIONAL UNIVERSITY VIETNAM NATIONAL UNIVERSITY HCMC FINAL EXAMINATION January 14th 2013 Duration 120 minutes Chair of Department of Mathematics SUBJECT CALCULUS I Lecturers Tran Thai Duong Pham Huu Anh Ngoc Nguyen Minh Quan Signatures Signature Full name Professor Phan Quoc Khanh INSTRUCTIONS Each student is allowed a scienti c calculator and a maximum of two double sided sheets of reference material size A4 or similar stapled together and marked with their name and ID All other documents and electronic devices are forbidden Students are not allowed to borrow anything especially calculators and reference material from each other 1 15 points A farmer has 1500 feet of fencing and wants to fence o a rectangular eld that borders a straight river He needs no fence along the river What are the dimensions of the eld that has the largest area See the below picture 2 Suppose the percentage of households in Ho Chi Minh city with at least one video cassette recorder has been modeled by the function 75 f t 1 74e 0 6t where t is in years and t 0 represents the year 1985 a 10 points Show that f t is strictly increasing in 0 b 10 points Find the time T at which the number of video cassette recorders was increas ing most rapidly i e f cid 48 T is the maximum increasing rate T 0 3 10 points Find the maximum and minimum values of the function f x xe x x 0 Please turn over 1 4 15 points Find the antiderivative cid 90 18dx x 3 x2 9 5 a 10 points Evaluate x ln xdx 1 cid 82 0 cid 82 0 xdx x2 1 b 10 points Determine whether the improper integral is convergent or divergent no point will be given for any plain statement without a proof a 10 points Find the area of the region enclosed by the two curves y x3 and 6 y x b 10 points Find the volume of the solid obtained by rotating the region bounded by the given two curves y x3 and y x about the x axis END 2 Answers for Calculus I Final Examination January 2013 15 points Let x and y be the depth and width of the rectangle in feet We want to 1 maximize where It follows that A xy 2x y 1500 x y 0 A xy x 1500 2x 1500x 2x2 A x 0 x 750 Note that A cid 48 x 0 if and only if x 375 Furthermore we have A 0 0 A 375 281250 A 750 0 Thus the maximum value of A is A 375 281250 f t2 Therefore A attains the maximum value at x 375 ft and y 750 ft 2 a 10 points We have f cid 48 t 75 0 6 74 e 0 6t 1 74e 0 6t 2 3330e 0 6t 1 74e 0 6t 2 0 Therefore f is increasing strictly b 10 points We want to maximize the function g t f cid 48 t t 0 Note that g cid 48 t f cid 48 cid 48 t 3330 0 6 e 0 6t 1 74e 0 6t 2 3330 2 74 0 6 e 0 6t 2 1 74e 0 6t 3 Thus the critical number is de ned by the equation 1 74e 0 6t 2 74 e 0 6t 0 or 1 74e 0 6t 0 This implies the unique critical number T ln 74 0 x T Therefore g t attains its maximum at time 0 6 Moreover g cid 48 x 0 x 0 T and g cid 48 x T ln 74 0 6 cid 39 7 17 years 10 points Since f cid 48 x e x 1 x the unique critical number is at x 1 3 On the other hand f cid 48 x 0 x 0 1 and f cid 48 x 0 x 1 Moreover we have f 1 e 1 f 0 0 and f x 0 x 0 Therefore by the rst derivative test the maximum value of f is f 1 e 1 and the minimum value of f is f 0 0 3 4 15 points Solve for A B C in the equation or 18 A Bx C x 3 x2 9 x 3 x2 9 x2 9 18 A x2 9 Bx C x 3 Choose x 3 and get 18 18A or A 1 Choose x 0 and get 18 9 3C or C 3 Choose x 1 and get 18 10 B 3 4 or B 1 Hence cid 90 cid 90 cid 90 18dx x 3 x2 9 ln x 3 1 2 dx x 3 xdx x2 9 3dx x2 9 ln x2 9 arctan C cid 90 cid 17 cid 16 x 3 5 a 10 points By the method of integration by parts 1 cid 90 t 1 cid 90 0 x ln xdx lim t 0 x ln xdx 1 cid 90 cid 26 du dx t x v x2 2 0 1 cid 90 cid 26 u ln x cid 12 cid 12 cid 12 cid 12 1 x2 2 ln x dv xdx t cid 18 1 cid 90 t cid 19 x ln xdx lim t 0 t2 2 ln t t2 4 1 4 1 4 cid 90 0 t cid 90 0 xdx x2 1 lim t xdx x2 1 t cid 90 0 xdx x2 1 1 2 du u 1 2 ln cid 0 t2 1 cid 1 t2 1 cid 90 1 4 x ln xdx 1 2 xdx t2 2 ln t t2 4 1 4 Thus by using L Hospital s rule we obtain b 10 points Solution 1 By the de nition of improper integral By a change of variable u x2 1 du 2xdx we obtain cid 82 1 cid 82 1 cid 82 0 6 de ned by Therefore xdx x2 1 is divergent Solution 2 Note that x x2 1 cid 62 1 2x cid 62 0 x cid 62 1 Moreover dx x is divergent Thus by the comparison test xdx x2 1 is divergent It implies that xdx x2 1 is also divergent a 10 points The points of intersection are the origin and 1 1 The desired area is b 10 points The volume is cid 0 x x3 cid 1 dx A 2 3 x3 2 x4 4 5 12 cid 0 x x6 cid 1 dx cid 20 x2 2 V x7 7 5 14 1 cid 90 0 1 cid 90 0 cid 82 1 cid 12 cid 12 cid 12 cid 12 1 cid 21 1 0 0 5