Question

Ini soal matematika. aku butuh bantuan...

Answer 1

1.₀ᵃ∫ (x+1) dx  = 4 dengan a > 0
1/2 x² + x]ᵃ₀ = 4
1/2 a² + a = 4
a² + 2a - 8 = 0
(a+ 4)(a - 2) = 0
a = - 4 ,atau a = 2
a> 0 maka HP a = 2

3. ∫(x+3)/(x² +6x -3)³ dx
= ∫(x+3)(x²+6x-3)⁻³ dx
= (x+3)/(2x+6) (-1/2)(x² +6x - 3)⁻² + c
= (1/2)(-1/2)(1/(x²+6x-3)² ) + c

= - 1/4. (1/ x²+6x -3)² + c
= - 1/{4(x²+6x-3)²} + c
.
6. x + 2y = 2
y =  0 , x = 2
x + 2y = 2
2y = x - 2
y = 1/2 x -1
.

V = π  ₀²∫ (1/2 x - 1)² dx
V = π ₀²∫ 1/4 x²- x + 1 dx
V = π [1/12 x³ - 1/2 x² + x]²₀
V = π [8/12 - 1/2(4) +  2]
V = π [8/12]
V = 3/4 π