**Unformatted text preview:**

9 30 Evaluationof Definiteintegrals rule and 2 methods to comple At anti derivatives this point you have one Power Rule integration by u substitution integration by parts so far all practice has been indefinite integrals Let s apply the Fundamental Theorem of calculus 1 dx Si u du t it So 2x x 2 1 let u du 2xdx III Fa it f 9 Utu du 2x dx du xdx dx a 1 4 a du 343121 161312 o 64 a Si 4d u du dx v 21 4 du 1 41dX 2 1 471525 1 4 dx 5 2 1 4 2151 5 4 o 2 376 4 15 5 4132 43 0 4 3 1013 30 8 4 3 27 36 since from 2 02 a 5 2 2 5 dx So a du 4 2 5 a51 F 0 du 2d 2 4 5 5 431 5 a f x iKx 3 5dx 1 311 3 a f 2xe dx 2x U 2dx du e dx v ex ze axe i affdx zxe li flie 21n He de z z ext 4 11 2 dx 41 su du 2 4 441 5 54 14 625 1 624 3120 IS e du eu fe ex fire dx Edu xd it 1 8 at du 3Rd Idu Rdx it I 3 Ex nxdx U lnx dx du du dx v xinx IS xl ldx Inx eine ell lie 11mi e I o dtl D e 1 ext Find the area of the shaded region CM meaiierns f total area flxldx is flxldx Fx dx let u 9 R 2 d dx du du if 4 3 9 1312 9 9 0 I 9 02 9 total area 9 9 IS u du u 0 9 27 9

View Full Document