kak soal integral.......

Materi integral substitusi <<<
Perbaikan nomor 1:
1/18 (4^6 - 1)
= 227,5

InTeGraL

No. 1
Misal ;
u = 3x + 1
du/dx = 3
dx = 1/3 du

∫(3x + 1)^5 dx [1...0]
= ∫u^5 . (1/3) du
= (1/3) ∫ u^5 du
= 1/3 . 1/6 u^6
= (1/18)(3x + 1)^6
= (1/18)(3.1 + 1)^6 - (1/8)(3.0 + 1)^6
= (1/18)(4^6 - 1)
= 4095/18
= 227,5

No. 2
Misal :
u = 2x² + 1
du/dx = 4x
x dx = 1/4 du

∫x√(2x² + 1) dx [2...0]
= ∫u^1/2 . 1/4 du
= (1/4)(1/(1/2 + 1)) u^(1/2 + 1)
= 1/4 . 2/3 u^3/2
= 1/6 u^3/2
= 1/6 (2x² + 1)^3/2
= 1/6 (2.2² + 1)^3/2 - 1/6 (2.0² + 1)^3/2
= 1/6 . 9^3/2 - 1/6 . 1
= 1/6 . 3^3 - 1/6
= 1/6 (3^3 - 1)
= 1/6 . 26
= 13/3
= 4,333