求曲线y=e^x与y=2及x=0围成的图形的面积A,和绕y轴旋转一周的体积、

联立y = e^x 和y = 2,可得二者的交点为(ln2,2)
x = 0为y轴,三者所围的图形的面积为f(x) = 2 - e^x在0和ln2之间的定积分
F(x) = ∫(2-e^x)dx = 2x - e^x + C
A = F(ln2) - F(0) = (2ln2 - 2) - (0 - 1) = 2ln2 -1
计算绕y轴旋转一周的体积时,用y作为自变量更简便.y = e^x,x = lny; 在y处的半径为lny,截面积为π(lny)²,积分区间为[1,2]
G(y) = π∫(lny)²dy = π(yln²y -2ylny + 2y) +C
G(2) = π(2ln²2 - 4ln2+ 4) +C
G(1) = 2π + C
V = G(2) - G(1) = 2π(ln²2 - 2ln2 + 1) = 2π(ln2 -1)²

1

A=∫(0,ln2) (2-e^x)dx=(2x-e^x)|(0,ln2)=2ln2-(2-1)=2ln2-1
V=∫(0,ln2) 2πx(2-e^x)dx=2π[(x^2)|(0,ln2)-∫(0,ln2) xe^x*dx]
=2π[(ln2)^2-(xe^x)|(0,ln2)+∫(0,ln2) e^x*dx]=2π[(ln2)^2-2ln2+1]=2π(1-ln2)^2=2π[ln(e/2)]^2

S = ∫ [0,ln2] (2 - e^x ) dx = 2 ln2 - e^x | [0,ln2]
= 2 ln2 - 1
V = π ∫ [1,2] (lny)² dy
= 2π ∫ [0,ln2] x * e^x dx
= 2π (x-1) e^x | [0,ln2]
= 2π (2 ln2 - 1)是绕...

2求由曲线y=x ，及直线x=1,x=2,y=0所围成的平面图形绕x轴旋转一周所成的旋转体的体积。 体积=π*(e^x)^2*dx 定积分，积分区间ln2→ln