211 13 079
DESCRIPTION
211 13 079TRANSCRIPT
NAMA : RAIMUNDO SUHERDINNIM : 211 13 079JURUSAN : TEKNIK SIPILTUGAS : MATEMATIKA 4
UNIVERSITAS KATOLIK WIDYA MANDIRA KUPANG
2015
= 13𝑒−3𝑥
Pembuktian dari : ʃ x2 e-3x dx
mis : u = x2
du = 2x dxdv = e-3x
v = ʃ du = ʃ e-3x dx
Soal Wajib No. 19
න𝑢 .𝑑𝑢 = 𝑢𝑣−න𝑣𝑑𝑢 න𝑥2 .𝑒−3𝑥𝑑𝑥= 𝑥2൬−13𝑒−3𝑥൰− න13𝑒−3𝑥2𝑥
= −13𝑥2𝑒−3𝑥 − 23൬−13𝑥𝑒−3𝑥 + 13න𝑒−3𝑥൰𝑑𝑥
= −13𝑒−3𝑥൬𝑥2 + 23+ 29൰+ 𝑐
න𝑎𝑟𝑐cos2𝑥 𝑑𝑥 𝑝𝑒𝑛𝑦𝑒𝑙𝑒𝑠𝑎𝑖𝑎𝑛
න𝑢 𝑑𝑣 = 𝑢𝑣− න𝑣 𝑑𝑢 𝑢 = 𝑎𝑟𝑐cos2𝑥
𝑑𝑢 = − 2ξ1− 4𝑥2 𝑑𝑥 𝑑𝑣 = 𝑑𝑥
𝑣 = 𝑥
Soal Pilihan No. 14
න𝑎𝑟𝑐cos2𝑥 𝑑𝑥 = 𝑥 𝑎𝑟𝑐cos2𝑥−න𝑥 .− 2ξ1− 4𝑥2 𝑑𝑥
= 𝑥 𝑎𝑟𝑐cos2𝑥+ 2න 𝑥ξ1− 4𝑥2 𝑑𝑥 = 𝑥 𝑎𝑟𝑐cos2𝑥− 14න −8𝑥ξ1− 4𝑥2 𝑑𝑥 = 𝑥 𝑎𝑟𝑐cos2𝑥− 14(2 ඥ1− 4𝑥2) + 𝑐
= 𝑥 𝑎𝑟𝑐cos2𝑥− 12ඥ1− 4𝑥2 + 𝑐
න𝑠𝑖𝑛2 𝑥 𝑑𝑥= 12 න(1− cos2𝑥) 𝑑𝑥 = 12 න𝑑𝑥− 12නcos2𝑥 𝑑𝑥
= 12𝑥− 12 .12sin2𝑥+ 𝑐 = 12𝑥− 14sin2𝑥+ 𝑐
Soal Pilihan. No. 20
SEKIAN DAN
TERIMA KASIH