The enclosing area is s = ∫ (- 1) dx ∫ (x 2,2-x 2) dy = 2 ∫ (-1,1) (1)

2、y' - 1/x *y =lnx

Multiply both sides by1/x.

y'/x- 1/x^2*y=(lnx)/x

Namely: (y/x)'=(lnx)/x

Bilateral integral: y/x = ∫ (lnx)/xdx = ∫ lnxd (lnx) =1/2 (lnx) 2+C.

y= 1/2*x*(lnx)^2+Cx

3、y=x^4-2x^2+5

y'=4x^3-4x=4x(x- 1)(x+ 1)

When y'=0, x=0 or x=- 1 or x= 1, then the extreme points are these three points.

Calculate y (0) = 5 y (-1) = 4 y (1) = 4 y (-2) =13.

So the maximum value is 13 and the minimum value is 4.

4.∫ (0， 1)√(2x+ 1)/(x+ 1)dx √( 2x+ 1)= t x =(T2- 1)/2dx。

=∫( 1，√3) 2t/(t^2+ 1)*tdt

=∫( 1，√3)2( 1- 1/(t^2+ 1))dt

=2 (t arctangent) (1, √3)

=2(√3- 1)-π/6