题目有问题,应当是二者和轴所围的区域.
S = ∫₀¹(x + 1 - 2√x)dx
= (x²/2 + x - 2*(2/3)x√x)|₀¹
= 1/2 + 1 - 4/3
= 1/6
V = ∫₀¹π[(x + 1)² - (2√x)²]dx
= π∫₀¹(x² - 2x + 1)dx
= π(x³/3 - x² + x)₀¹
= π(1/3 - 1 + 1)
= π/3
题目有问题,应当是二者和轴所围的区域.
S = ∫₀¹(x + 1 - 2√x)dx
= (x²/2 + x - 2*(2/3)x√x)|₀¹
= 1/2 + 1 - 4/3
= 1/6
V = ∫₀¹π[(x + 1)² - (2√x)²]dx
= π∫₀¹(x² - 2x + 1)dx
= π(x³/3 - x² + x)₀¹
= π(1/3 - 1 + 1)
= π/3