MATH 132 FINAL EXAM solutions 132F12 4sol 1 1 4 1 Use substitution to evaluate 0 tan B B B E F G H I J K L I N solution 8 B B B l G e 2 Use integration by parts to evaluate 1 B 68 B B E F G H I J K L M N solution 68 B B B B B B B 68 B B B J B 68 B B l B 3 Using partial fractions find a solution to 2 B B B B 2 2 A 8 2B 1 B B B C 68 B B D 68 B 68 B E 68 B 68 B 68 B B B B F 68 B G 68 x 1 B L 68 B B M B N 68 B B solution B B B partial fractions 68 B 68 B 68 B I 4 Find what becomes of the integral 1 1 B B B when you make the substitution B 38 9 A 38 B 9 C 38 D E F G 38 H 9 I 38 J 9 solution B 38 B 9 38 38 9 38 9 9 38 K 5 Find the area of the region enclosed by the curve C B and the line C 2 2 B E F G H I J K N L M solution B B B B B 8 B intersection points E B B B B B B B B l H 132F12 4sol 2 6 Find the volume of the solid obtained by rotating the region enclosed by the curve B C the lines C C and the C axis about the y axis E 1 F 1 G 1 H 1 I 1 J 9 1 K 1 L 1 M 1 N 1 C solution Z 1 C C 1 C 1 C l 1 M 7 Find the arc length of the curve B C 2 C 4 8 14 E F G 3 H I 2 F K L M 20 J B C C C l solution B C C C C P G B 8 Find the value of the improper integral 0 B if it converges B E 0 B 2 C 4 D 6 I 8 J K L I N 3 1 solution B B B l K 9 Find the solution to the initial value differential equation C C C E C F C G C H C I C 68 J C 68 K C L C M C N C solution C C 68 C G C O O G C O O C G 1 Find the sum of the infinite series E 3 F 4 G solution 8 H 8 8 I 8 J 8 2 8 K 8 8 L M 4 N G f sol 3 1 Approximate the sum 8 8 8x with an error of less than 2 x 10 3 E 367 F 371 G 384 H 396 I 408 J 414 K 429 L 437 M 4458 N 452 solution x x x x x x Approx E x x x x x 12 Find the complete interval of convergence either absolute or conditional of the power series A B J B 8 F K solution Ratio Test lB l x 8 8 8 8 8 8 B 8 8 8 B B G B H B I B L B M B N 8 lB l8 8 8 8 8 8 lB l8 B 8 8 lB l p lB l Endpoints B 8 8 8 8 8 8 8 8 8 converges Therefore interval is 8 diverges B B 13 Find the Taylor polynomial of order for 0 B B at B E B B F B B G B B H B B I B B J B B K B B L B B M B B N B B l solution 0 0 w B B 0 w 0 w w B 0 w w x T B B B F B 14 If the Maclaurin series for B B is 8 B8 then find E F G H I J 8 K L M B B B x x x B B B B x x x B B B B B c N solution eB B e B B B B B B B H 132f12 4sol 4 approximate 68 B B Using the Maclaurin series for 68 1 B with an error 0 002 E 0 109375 F 0 210432 G 0 342871 H 0 465732 I 0 543876 J 0 623149 K 0 743987 L 0 841762 M 0 954129 N 1 097658 B solution 68 B B B B B B B B B 68 B B l 68 B B E B 16 Find the Maclaurin series for the function B B 8 B E 8 0 I 8 0 M 8 8 B 8 2 8 2 F 8 B 8 8 8 B 8 8 8 0 J 8 8 B B 8 B B G 8 0 8 B 8 8 N solution 8 B B 8 B 8 4 8 2 K 8 8 B 8 2 8 1 8 B 8 8 H 8 0 8 B 8 4 8 3 L 8 8 B 8 8 8 B 8 8 B B B B B B 8 B B 8 0 8 B 8 4 8 3 B B B H Find the Taylor series for 0 B B B centered at E B B G B B I B B K B B M B B F B B H B B J B B L B B N B B solution 0 0 w B B 0 w 0 w w B x 0 8 B for all 8 8 for all 8 Taylor series B B I 132f12 4sol 5 8 If 38 B 8 B 1 8 is the Taylor series for the function 38 B 8 1 then find the coeficient c of B 1 2 centered at A L F M G H I J K N 0 w w 1 x 0 B 38 B 38 1 solution 0 w B 9 B 0 w w B 38 B 0 w w 1 E 19 From the formula of the Binomial Series we get that B 8 B8 with A 0 B 8 G H I J K L M N solution 1 x B B B B B B B B B B I 20 Using the Alternating Series Estimation Theorem and the Maclauren Series for eB estimate B B with an error less than or equal to 1 x 10 E 0 B 2765 C 3876 D 0 5762 E 0 F 0 K 0 L 0 M 523 N B B B B x x x x e B B B x B x B x B x B B B B B B B B B B B B B l so approximation is solution eB B L
View Full Document
Unlocking...