bantu jawab nomer 5 aja
Jawab:
integral tentu
₁ᵃ∫ (1 + x ) = a
[ x + 1/2 x²]ᵃ₁= a
(a- 1) + 1/2(a² - 1) = a . . . kalikan 2
2(a- 1) + (a² - 1) = 2a
2a - 2+ a² - 1 = 2a
a² - 3= 0
a = ± √3
