3.7.5 Pengamiran, SPM Praktis (Kertas 1) January 24, 2022April 10, 2021 by Soalan 13: Diberi y= x 2 2x−1 , tunjukkan dy dx = 2x( x−1 ) ( 2x−1 ) 2 . Seterusnya, nilaikan ∫ −2 2 x( x−1 ) 4 ( 2x−1 ) 2 dx . Penyelesaian: y= x 2 2x−1 dy dx = ( 2x−1 )( 2x )−x( 2 ) ( 2x−1 ) 2 = 4 x 2 −2x−2 x 2 ( 2x−1 ) 2 = 2 x 2 −2x ( 2x−1 ) 2 = 2x( x−1 ) ( 2x−1 ) 2 ( tertunjuk ) ∫ −2 2 2x( x−1 ) ( 2x−1 ) 2 dx = [ x 2 2x−1 ] −2 2 1 8 ∫ −2 2 2x( x−1 ) ( 2x−1 ) 2 dx = 1 8 [ x 2 2x−1 ] −2 2 1 4 ∫ −2 2 x( x−1 ) ( 2x−1 ) 2 dx = 1 8 [ ( 2 2 2( 2 )−1 )−( ( −2 ) 2 2( −2 )−1 ) ] = 1 8 [ ( 4 3 )−( 4 −5 ) ] = 1 8 ( 32 15 ) = 4 15