integral 0 sampai 2 dari 2x(8-x^2) gimana kak

jawab

₀²∫ 2x (8-x²) dx
substitusi
u = 8 - x²
du =2x dx

batas integral
u = 8 - x²
x = 0, u = 8
x = 2, u = 4

₄⁸∫ u du = [ 1/2 u²]⁸₄
= 1/2 (64 - 16)
= 1/2 (48)
= 24