3.4 Kamiran Tentu (Bahagian 1)


3.4 Kamiran Tentu (Bahagian 1)
   a b f( x )dx=F( b )F( a )  
Contoh:
Nilaikan yang berikut.
(a)  1 0 ( 3 x 2 2x+5 )dx (b)  0 2 ( 2x+1 ) 3 dx


Penyelesaian:
(a)  1 0 ( 3 x 2 2x+5 )dx = [ 3 x 3 3 2 x 2 2 +5x ] 1 0 = [ x 3 x 2 +5x ] 1 0 =0[ ( 1 ) 3 ( 1 ) 2 +5( 1 ) ] =0( 115 ) =7

(b)  0 2 ( 2x+1 ) 3 dx = [ ( 2x+1 ) 4 4( 2 ) ] 0 2 = [ ( 2x+1 ) 4 8 ] 0 2 =[ ( 2( 2 )+1 ) 4 8 ][ ( 2( 0 )+1 ) 4 8 ] = 625 8 1 8 =78


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