₀²∫ (4x) √(4 – x²) dx Tolong bantuannya

Jawab

₀²∫(4x) √(4-x²) dx

u =  4 – x²
du = -2x dx
4x dx = – 2 du

batas integral
u = 4- x²
x= 0 → u = 4
x = 2 → u= 0

₀²∫(4x)√(4-x²) dx = ₄⁰∫ (-2) U^(1/2) du
= -2(2/3) u√u ]⁰₄
= -4/3 ( 0 – 4√4)
= – 4/3 (0- 8)
=  32/3