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

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