Tugas Kalkulus 7 Juni 2009
Posted by supardiyo in Semester I, Semester II.Tags: Tugas Kalkulus
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1. 2∫ 2 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
- 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
