cuma satu ,tentang integral

integral ( 4 - x^2) dx
= 4x - (1/3)x^3 + c
= 4x - (1/3)x^3 ( batas atas -2, bawah -3)
= { (4.-2 - (1/3).(-2)^3 } - { (4.-3-(1/3).(-3)^3 }
= { -8+8/3)-{ -12+9}
= (-24/3 + 8/3)-(-3)
= (-16/3)+(3)
= -16/3 + 9/3
= -7/3 (berada di bawah sumbu x)
= 7/3

InTegraL

∫(4 - x²) dx [3_-2]
= 4x - 1/3 x³
= 4(3 - (-2)) - 1/3 (3³ - (-2)³)
= 4 . 5 - 1/3 . 35
= 20 - 35/3
= 20 - 11 2/3
= 8 1/3