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Tugas Kalkulus 7 Juni 2009

Posted by supardiyo in Semester I, Semester II.
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1. 22 12x3 –x2 +3x + 1 dx

=3x4 – ⅓x3 +3/2x2 +x  -2 2

=[ 3(2) 4– ⅓(2) 3+3/2(2) 2+ (2) ] – [ -3(-2) 4 – ⅓ (-2) 3+3/2(-2) 2+ (-2)]

=[48 – 8/3 + 8 ] – [ 48 + 8/3 – 8 ]

=[ 144/3 – 8/3 + 4/3 ] – [ 144/3 + 8/3 – 24/3 ]

= 160/3 – [ 128/3 ] = 32/3

  1. 0 4 x √x – 3x dx

= x 1 . x1/2 – 3x dx

= x3/2 – 3x dx

= 1/3/2 + 1 x3/2 – 3/2 x2 0 4

= 2/5 x 5/2 – 3/2 x2 0 4

= [2/5 (4) 5/2 – 3/2 (4) 2] – 0

=[2/5 (42 . 41/2) – 3/2 .16 ] – 0

= [2/5 (16 . 2 ) – 3/2 .16 ]

=  [2/5 ( 32 ) – 3/2 .16 ]

= 64/5 – 48/2

= 64/5 – 24

= 64/5 – 120/5 = -56/5

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