Exam 2 | Questions with Answer of Calculus with Analytic Geometry II | MAT 271, Exams of Analytical Geometry and Calculus

Download Exam 2 | Questions with Answer of Calculus with Analytic Geometry II | MAT 271 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Mastery test on Tuesday Jan 30 in PSA 21 Testing Center -Review- A. Review the following problems for Mastery Test: 1.  dxxx cos 2.  dxxx cos 2 3.  dxex x2 4.  dxxe x cos 5.  dxxln 6.  dxxx ln 7.   dxx1tan 8.  dxxx ln 2 9.   dxxx )1ln( 10.  dxe x 11.   dxx1cos 12.  dxxx ln 13.  dxx)2ln( 14.  dttt 2sin 15.  dxxx )5ln( 5 16.  dxx x 2 ln 17.  dxxx 3ln 18.   dxxxx )sin7ln( 3/2 19.  dxxe x55 20.    dx xx x 96 92 2 21.   dx x 9 1 22.    dx xx xx )1)(1( 835 2 2 23. dxxx  cos23sin 24.  dxex x323 25.   dx x x 3 2 92 26.  dtt e t 2 /1 2 27.  dxx2 28.   dx e e x x 2 29.   dx e e x x 5ln 30.    dx xe xe x x 2 42 31.    dy y y 1 1 32.  dxxx ))2cos(sin()2cos( 33.   dxxx 1 34a.  dxxx )(sec)tan( 2 34b.   dx u u 21 35.   dx x x 41 2 36.  dxx x)sin(ln 37.    dx x x )92sin(1 )92cos( 38.   dx e e t t21 39.    dx xx x 96 92 2 40.   dx x x 1 2 41.   dx xx e x 62 ln 42. dxxx 4/ 0 2 )(sec)tan(  43. dx x x  4 1 )4ln( 44. dxx   1 2 45. dxe x  1 1 2 46. dxx  2 2 3 )sin( 47. dxe x 16 1 48. dxxarc 2/1 0 )(cos MAT 271 SPRING 2007 49. dxxx 2/ 0 45 sincos  50. dt t t  3 1 arctan Answer key for Mastery test: You need to add a constant with each answer. 1. xxx sincos  2. xxxxx cos2sin2sin2  3. )22( 2xxe x  4. )sin(cos2/1 xxe x  5. )1(ln xx 6. )1ln2(4/1 2 xx 7. )1ln(2/1tan 21 xxx  8. )1ln3(9/1 3 xx 9. )1ln()1(4/32/14/1)1ln()1(2/1 22  xxxxxx 10. )1(2 xe x 11. 21 1cos xxx  12. )1ln2(8/1 2 xx 13. )12(ln xx 14. xxx 2cos2/12sin4/1  15. )15ln6(36/1 6 xx 16. )1(ln/1  xx 17. )1ln2(4/3 2 xx 18. xxxxx cos7ln5/3 3/5  19. )15(5/1 5 xe x 20. )3/(3)3ln(2  xx 21. 92 x 22. )1ln(5tan3 1  xx 23. 2/3)cos23(3/1 x 24. 3xe 25. 2/12/3 183/4  xx 26. te /12/1 27. 0,;0, 22  xxxx 28. )2ln( xe 29. )5ln(ln xe 30. )ln(2 2xe x  31. 14)1(3/2 2/3  yy 32. )2sin(sin2/1 x 33. 2/32/5 )1(3/2)1(5/2  xx 34a. 2/3(tan)3/2 34b. 21 u 35. )(tan 21 x 36. )cos(ln x 37. ))92sin(1ln(2/1  x 38. xx ee   39. )3ln(2)3/(3  xx 40. )1ln(2/2 xxx  41. )6ln( x 42. 3/2 43. 2.88 44. 3/2 45. 0 46. 0 47. 59.3276 4 e 48. 0.6576 49. 8/315 50. 36.13/32)2ln(2/   B. 1.  dxx2sin 2 Answer: 2.  xdx3cos 2 Answer: 3.  dxx )2ln( Answer: 4.    dy y y 1 1 Answer: 5.  dxx x)cos(ln Answer: Cxxx x x xx vduuv         21 2 1 1cos 1 cos C. Following trigonometric identities are useful: )2cos1(2/1sinsin21 )2cos1(2/1cos1cos2 sincos2cos 2sin2/1cossincossin22sin 22 22 22 xxx xxx xxx xxxxxx     )3coscos3(4/1coscos3cos43cos )3sinsin3(4/1sinsin4sin33sin 33 33 xxxxxx xxxxxx   Example 1 Evaluate    Cxxdxxxxdx )3sin3/1sin3(4/1)3coscos3(4/1cos 3 Or we could solve by substitution as      Cuuduuxdxxxdxxxdx 32223 3/1)1(cos)sin1(coscoscos where xdxduxu cossin  Thus we have   Cxxxdx 33 sin3/1sincos . Our results are equivalent. Example 2 Evaluate    Cuuuduuuuxdxx 9/7/25/)2(sincos 97586445 Where xdxduxu cossin  and 22224 )sin1()(coscos xxx  Example 3 Evaluate dxx 2/ 0 5cos     2/ 0 2/ 0 224 2/ 0 5 cos)sin1(coscoscos   xdxxxdxxdxx We have     duuuduuxdx )21()1(cos 42225 , where ux sin . Do the rest! Example 4.  3/ 6/ 3csc   xdx   7825.1csccotcotcsc(ln2/1csc cotcsclncsccot cotcsc )cot(csccsc csccotcsccsccotcsc2 csccsccsccot)1(csccsccsccot cotcsccsccotcsccsccsc 3/ 6/ 3/ 6/ 3 3 32 223                   xxxxdxx xxxx dx xx xxx xxdxxxdxx xdxdxxxdxxxx dxxxxxvduuvudvxdxxxdx where xvxdxdv xdxxduxu cotcsc cotcsccsc 2   Example 5.      Cuumduum dxmxmxdxmxmxdxmx )3/(/1)1( 1 )sin())(cos1()sin()(sin)(sin 32223 where dxmxmdumxu )sin()cos( 