MATH 132 FINAL EXAM SOLUTIONS 1 Use substitution to evaluate 0 E F G H I J B B F13M132 4sol 1 B K L I N solution B B B B l I 2 Using integration by parts B B B 0 B G where 0 B E B B F B B G B B B H B B B I B B B J B B B K B B B L B B B M B B B N B B B solution B B B B B B B B B B B B G K 3x 1 3x 1 B B 3 Using partial fractions find a solution to B B B B A 68 B B B 3 68 B C 6 68 B D 68 B E 3 68 B B F 6 68 B G 68 B B L 3 68 B B M 6 68 B B N B B solution 3x 1 3x 1 B 3x 1 3x 1 B 68 B 68 B B 68 B G G 4 Find what becomes of the integral B B when you make the substitution 1 B 8 9 A 38 B 9 C 38 D E F G 38 H 9 I 8 J solution B 8 B 81 4 8 9 L 132F1 4 96 5 Find the area of the region enclosed by the curve C B and the line B C C B E F G H I J K L M N solution Intersection points and C B is always top and C B always bottom E B B B B B l H 6 Find the volume of the solid obtained by rotating the region enclosed by the curve C B and the lines C 2 and x 0 about the x axis E 1 F 1 G 1 H 1 I 1 J 9 1 K 1 L 1 I 1 N 1 solution C B and C 2 intersect at and Using the washer method we get Z 1 B B 1 B B l 1 K 7 Find the arc length of the curve B C C to 2 decimal places E 1 57 F 1 94 C 2 27 H 2 84 I 3 45 J 3 92 K 4 46 L 4 84 M 5 17 J 5 68 C C C B C C P C C l solution B C G 1 8 Find the value of the improper integral 0 1 B B if it converges E 0 B 2 C 4 D 6 I 8 J K L M N 3 1 1 solution lim 68 B lim 68 3 1 N y 9 Find the solution to the differential equation C t with initial condition C 68 E C F C G C H C 68 68 I C 68 J C 68 K C L C M C N C solution C C G C G Initial condition is C 68 so 68 G G and we get C C 68 I 132F1 4 96 8 1 Find the sum of the infinite series E 3 F 4 G H I J solution 8 8 8 8 8 2 K L 8 M 4 N H 1 Using the fewest terms possible approximate the sum 8 8 8 8 with an error of less than E F G H I F K L M N 8 solution 8 8 8 and so approximation will be the first 3 terms K 12 Find the complete interval of convergence either absolute or conditional of the power series 8 1 8 B 8 8 8 A B F B G B I B J B K B M N 8 8 Now check endpoints B H B L B l B l lB l 8 8 solution Ratio test 8 8 l B l8 8 lB l lB l B B V B 8 1 8 8 8 8 8 8 1 8 8 8 8 1 8 8 8 1 8 8 98 1 3 1 B K 13 If 8 B 8 is the Taylor series of 0 B ln 1 x centered at 8 E then F G H solution 0 w B w 0 w w B I J K L w M N B 0 w w B B 0 w w B B w 0 w w x I 132F1 4 96 14 Find the Taylor polynomial of order 6 for B sin 2x at B E B B B F B B B G B B B H B B B I B B B J B B B K B B B L B B B M B B B N B B B solution 38 B B B x B x 38 B B xB xB B 38 B B xB xB W9 X B B B M Using the Maclaurin series for 38 B approximate 38 B B with an error 0 002 E 0 109375 F 0 210432 G 0 3 H 0 465732 I 0 543876 J 0 623149 K 0 743987 L 0 841762 M 0 954129 N 1 097658 solution 38 B B B x B x B x so 38 B B B x B x B x B B 38 B B B B x B x B x B B B l 8 so approximation is G 9 B B 8 8 2 B 8 1 x 8 0 8 8 B 8 x 8 16 Find the Maclaurin series for the function E 8 B 8 2 8 2 x I 8 B 8 8x 8 0 M F 8 B 8 4 8 2 x G 8 0 8 B 8 J K 8 x 8 8 0 8 8 8 8 B B N 8 x 8 x 8 8 B B x x B B x x solution 9 B 9 B 8 8 8 B 8 x B x L D 8 B 8 4 8 3 x L 8 B 8 8 x 8 0 8 B B B B x x x x 9 B B x B x B x B 132f1 4 96 Find the Taylor series for 0 B B B B centered at E B B B solution 0 F B B B 0 w B B B 0 w G B B B 0 w w B B H B B B 0 w w x www www I B B B 0 B 0 J B B B x K 11 B B B X C69 W 3 B L 6 B B B B B M B B B G N B B B 8 If B 8 B 1 8 is the Taylor series for the function B 8 centered at 1 then find the coefficient A F G H I J K L M N solution 0 w B B 8 B 0 w w B B 8 B B 0 w w 1 0 w w 1 M x 19 Using the Binomial …
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