bantu jawab gan sis sebisanya

Limit
1a) lim(x->a)  (x² - a²)/(x - a)
lim(x->a) (x -a)(x + a)/(x- a)
lim(x-> a) (x+a)
limit = a+a = 2a

1b) lim(x->a)  (x³ - a³) /(x-a)
lim(x->a) (x -a)(x² + 2ax + a²) /(x-a)
lim(x->a) (x² + 2ax + a²)
x= 0 ,limit = a² + 2a²+ a² = 4 a²

3) Persmaan lingkaran jarijari r = 3
x²+ y² = r²
y²= r² - x²
y = √(r²-x²)
luas  y  = 4 ₀³∫√(9 - x²)  dx
misal x = 3 sin u ---> dx = 3 cos u du
untuk x= 0 --> sin u= 0 --> u  =0
untuk x = 3 --> sin u  = 1 --> u = π/2
9 - x² = 9 - 9sin² u

L= 4 (0...π/2) ∫√(9 - 9 sin² u). (3 cos u ) du
L = 4 (0..π/2) ∫ √9(1-sin²u)(3)(cos u) du
L = 4 (0..π/2) ∫ (3cos x)(3)(cos u) du
L = 4 (0..π/2) ∫ 9 cos² du
L = 4 (0..π/2)  ∫ (9/2) cos 2u  +9/2) du
L = 4  (9/4 sin 2u  +  9/2 u )] (π/2 ,....0)
L= 9  sin 2u + 18 u]π/2 ..0
L=9 (sin π- sin 0)  + 18(π/2 - 0)
L = 0 + 9π
L= 9π