Question

微积分题，求解答

一共4道，虽然看不懂，但是感觉会的人应该不会太难。

Answer 1

Answer 2

😁

追答

Answer 3

y=x^2, y=√x,
x^2 =√x
x^4 = x
x(x^3-1) = 0
x=0 or 1
Vx
=π∫(0->1) [ (y1)^2 -(y2)^2 ] dx
=π∫(0->1) ( x -x^3 ) dx
=π[ (1/2)x^2 -(1/4)x^4 ]|(0->1)
=π/4
(1)
y==√(x-1) , x=4 , y=0
0=√(x-1)
x=1
Vx
=π∫(1->4) y^2 dx
=π∫(1->4) (x-1) dx
=(1/2)π[(x-1)^2]|(1->4)
=(9/2)π
(3)
let
x= sinu
dx= cosu du
x=0, u=0
x=1, u=π/2
∫(-1->1) (x+1)√(1-x^2) dx
=∫(-1->1) x.√(1-x^2) dx +∫(-1->1) √(1-x^2) dx
=0 +∫(-1->1) √(1-x^2) dx
=2∫(0->1) √(1-x^2) dx
=2∫(0->π/2) (cosu)^2 du
=∫(0->π/2) (1+cos2u) du
=[ u + (1/2)sin2u]|(0->π/2)
=π/2