Exam 1 Solutions (Form A) Fall 2009.pdf

BYCPD0D0 BEBCBCBL C5CPD8CW BDBHBE BXDCCPD1 C1 CECTD6D7CXD3D2 BT CBD3D0D9D8CXD3D2D7 BDBA BV C4CTD8 u = 2 +x4BA CCCWCTD2 du = 4x3dxBA CBD9CQD7D8CXB9 D8D9D8CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 DDCXCTD0CSD7 1 4 A1 u1/2 du = 1 6u 3/2 + C = (2 + x 4)3/2 6 + C BEBA BW C4CTD8 u = 1 + exBA CCCWCTD2 du = ex dxBA CBD9CQB9 D7D8CXD8D9D8CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 DDCXCTD0CSD7 A1 u10 du = 1 11u 11 + C = (1 + e x)11 11 + C BFBA BT C4CTD8 u = 2 + tan?BA CCCWCTD2 du = sec2 ?d?BA CFCWCTD2 ? = 0, u = 2B8 CPD2CS DBCWCTD2 ? = pi 4, u =3 BA CBD9CQD7D8CXD8D9D8CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 DDCXCTD0CSD7 A2 3 2 du u = ln|u|| 3 2 = ln 3 ? ln 2 = ln parenleftbigg3 2 parenrightbigg BA BGBA BT BTD4D4D0DDCXD2CV D8CWCT CUD3D6D1D9D0CPB8 favg = 1 b? 0 A2 b 0 (2 + 6x)dx = 1b parenleftbig2x + 3x2vextendsinglevextendsingleb0. = 2b + 3b 2 b = 2 + 3b = 3BA CBD3D0DACXD2CV D8CWCXD7 CTD5D9CPD8CXD3D2 CUD3D6 b DDCXCTD0CSD7 b = 13BA BHBA BX C1D2D8CTCVD6CPD8CT CQDD D4CPD6D8D7B8 DBCXD8CW u = x, dv = e ?2x. CCCWCTD2 du = dx, v = ?12e?2xBA A1 xe?2x dx = ?12xe?2x ? A1 ?12e?2x dx = ?12xe?2x? 14e?2x+C = ?xe ?2x 2 ? e?2x 4 +C BA BIBA BW CDD7CXD2CV CPD2 CXCSCTD2D8CXD8DD CUD3D6 sin 2 x BM A1 2sin2 ?d? = A1 2 · 1 2(1 ? cos 2?)d? = ? ? 1 2 sin 2? + C = ? ? sin 2?2 + C BJBA BU BYD6D3D1 D8CWCT CVD6CPD4CW CQCTD0D3DBB8 A = A2 e 2 e ln x dxBA C1D2D8CTCVD6CPD8CT CQDD D4CPD6D8D7 DBCXD8CW u = ln x, dv = dxBA CCCWCTD2 du = dx x , v = xBA A = xln x| e2 e ? A2 e 2 e x· dxx = 2e2 ?e? (e2 ?e) = e2. 0 2 4 6 8 10?2.5 ?2 ?1.5 ?1 ?0.5 0 0.5 1 1.5 2 2.5 REGION Graph of y=ln(x) BKBA BX CCCPCZCT CP D8CWCXD2 D7CWCTCTD8 D3CU DBCPD8CTD6 DBCXD8CW D8CWCXCRCZD2CTD7D7 dx CPD8 CP CSCXD7D8CPD2CRCT x D1CTD8CTD6D7 CUD6D3D1 D8CWCT D8D3D4 D3CU D8CWCT D8CPD2CZBA CCCWCT DBCTCXCVCWD8 D3CU D8CWCT D7CWCTCTD8 B4CSCTD2D7CXD8DD D8CXD1CTD7 DAD3D0D9D1CTB5 CXD7 (10 3)(9.8)(2)(1)(dx) N BA CCCWCT CSCXD7D8CPD2CRCT D8CWCXD7 D7CWCTCTD8 D8D6CPDACTD0D7 CQCTCUD3D6CT D0CTCPDACXD2CV D8CWCT D8CPD2CZ CXD7 x D1CTD8CTD6D7 BA CCCWCT D8D3D8CPD0 DBD3D6CZ D6CTD5D9CXD6CTCS D8D3 D4D9D1D4 CWCPD0CU D8CWCT DBCPD8CTD6 D3D9D8 CXD7 A2 1/2 0 (19.6)(103)xdx = (9.8)(103)x2vextendsinglevextendsingle1/20 = 2.45 × 103J BLBA BW CBD0CXCRCT CXD2D8D3 CRD6D3D7D7B9D7CTCRD8CXD3D2D7 D4CTD6D4CTD2CSCXCRD9D0CPD6 D8D3 D8CWCT yB9CPDCCXD7 DBCXD8CW D8CWCXCRCZD2CTD7D7 dyBA CCCWCT DAD3D0D9D1CT D3CU CP D7D0CXCRCT CXD7 1 2pi parenleftbigg1 2x parenrightbigg2 dyBA CBCXD2CRCT y = ?x, x = y2B8 D7D3 D8CWCT D8D3D8CPD0 DAD3D0D9D1CT CXD7 CVCXDACTD2 CQDD A2 1 0 pi 8y 4 dy = pi 40y 5 vextendsinglevextendsingle vextendsingle 1 0 = pi40BA BDBCBA BV C4CTD8 u = 2x ? 3BA CCCWCTD2 du = 2dxBA CBD9CQD7D8CXB9 D8D9D8CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 CVCXDACTD7 D9D7 A1 g(x)dx = A1 f(2x ? 3)dx = 1 2 A1 f(u)du = F(u) 2 + C =F(2x? 3) 2 + CBA BDBDBA A1 tan3 x sec3 x dx = A1 tan2 x sec2 x (sec xtan x dx) = A1(sec2 x ? 1)sec2 x(sec xtan x dx) BA C4CTD8 u = sec xBA CCCWCTD2 du = sec xtan xdxBA CBD9CQD7D8CXD8D9D8CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 DDCXCTD0CSD7 A1(u2 ? 1)u2 du = A1(u4 ? u2)du = u5 5 ? u3 3 + C = sec5 x 5 ? sec3 x 3 + CBA BD BDBEBA C4CTD8 u = ?x. CCCWCTD2 du = dx2?xBA CBD9CQD7D8CXD8D9D8B9 CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 DDCXCTD0CSD7 2 A1 sin3 udu = 2A1 sin2 u(sin udu) = 2A1(1?cos2 u)(sin udu)BA C4CTD8 w = cos uBA CCCWCTD2 dw = ?sin u duBA CBD9CQD7D8CXD8D9D8CXD2CV CXD2D8D3 D8CWCT CXD2D8CTCVD6CPD0 DDCXCTD0CSD7 ?2 A1(1? w2)dw = ?2(w ? w 3 3 ) + C = ?2(cos u ? cos3 u 3 ) + C = ?2(cos ?x? cos3 ?x 3 ) + CBA BDBFBA CCCWCT D4D0D3D8 CXD7 CQCTD0D3DBBA CCCWCT CRD9D6DACTD7 CXD2D8CTD6D7CTCRD8 DBCWCTD2 1 ? x 2 = 2x ? 2 BA CBD3D0DACXD2CV CUD3D6 x DDCXCTD0CSD7 x2 + 2x ? 3 = 0, (x + 3)(x ? 1) = 0B8 D3D6 x = ?3, 1BA BYD6D3D1 D8CWCT CVD6CPD4CWB8 D8CWCT CPD6CTCP CXD7 A2 1 ?3 ((1 ? x2) ? (2x? 2))dx = A2 1 ?3 (?x2 ? 2x + 3)dxBA = ?x 3 3 ?x 2 + 3x vextendsinglevextendsingle vextendsinglevextendsingle 1 ?3 = parenleftbigg ?13 ? 1 + 3 parenrightbigg ? (9 ? 9 ? 9) = 323 ?4 ?3 ?2 ?1 0 1 2?12 ?10 ?8 ?6 ?4 ?2 0 2 4 Plot of f(x) and g(x) f(x) g(x) BDBGBA CCCWCT CVD6CPD4CW D3CU D8CWCT D6CTCVCXD3D2B8 DBCXD8CW CTD5D9CPD8CXD3D2D7 D3CU D8CWCT CPD4D4D6D3D4D6CXCPD8CT D0CXD2CTD7 B4CPD7 CUD9D2CRD8CXD3D2D7 D3CU yB5B8 CXD7 D7CWD3DBD2 CQCTD0D3DBBA CCCWCT DAD3D0D9D1CT D3CQD8CPCXD2CTCS CQDD D6D3D8CPD8CXD2CV D8CWCXD7 D6CTCVCXD3D2 CPCQD3D9D8 D8CWCT yB9CPDCCXD7 CXD7 V = pi A2 1 0 ((3y)2 ? (2y)2)dy = pi A2 1 0 5y2 dy = pi 53y3 vextendsinglevextendsingle vextendsinglevextendsingle 1 0 = 5pi3 BA ?0.5 0 0.5 1 1.5 2 2.5 3 3.5?0.5 0 0.5 1 1.5 x=2y x=3y BDBHBA CCCWCT CVD6CPD4CW D3CU D8CWCT D6CTCVCXD3D2 CXD7 D7CWD3DBD2 CQCTD0D3DBB8 CPD0D3D2CV DBCXD8CW CPD2 CPD6CQCXD8D6CPD6DD D7D8D6CXD4 DBCWCXCRCWB8 DBCWCTD2 D6D3D8CPD8CTCS CPCQD3D9D8 x = 1B8 D4D6D3CSD9CRCTD7 CP CRDDD0CXD2CSD6CXCRCPD0 D7CWCTD0D0BA CCCWCT DAD3D0D9D1CT D3CU D8CWCPD8 D7CWCTD0D0 CXD7 2pi(1 ? x)(x3)dxB8 D7D3 D8CWCT DAD3D0D9D1CT D3CU D8CWCT D7D3D0CXCS CXD7 V = A2 1 0 2pi(x3 ?x4)dx = 2pi parenleftbiggx4 4 ? x5 5 vextendsinglevextendsingle vextendsinglevextendsingle 1 0 = 2pi parenleftbigg1 4 ? 1 5 parenrightbigg = pi10BA 0 0.2 0.4 0.6 0.8 1 1.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 r=1?x Graph of y=x3 BDBIBA C1D2D8CTCVD6CPD8CT CQDD D4CPD6D8D7B8 DBCXD8CW u = sin x, dv = f??(x)dxBA CCCWCTD2 du = cos x, v = f?(x)BA CCCWCTD2 A2 pi 0 f??(x)sin x dx = f?(x)sin x|pi0 ? A2 pi 0 f?(x)cos x dxBA C1D2D8CTCVD6CPD8CT CQDD D4CPD6D8D7 CPCVCPCXD2B8 DBCXD8CW u = cos x,dv = BE f?(x)dxBA CCCWCTD2 du = ?sin xdx, v = f(x)BA CCCWCT CXD2D8CTCVD6CPD0 CQCTCRD3D1CTD7 f ?(x)sin x|pi 0 ? (f(x)cos x|pi0 ? A1 pi0 ?f(x)sin x dx)BA = f?(pi)sinpi ? f?(0)sin0 ? (f(pi)cospi ? f(0)cos0)? A2 pi 0 f(x)sin x dx = 0 ? 0 ? ((?1)(?1)? (1)(1)) ? 3 = ?3BA BF