Question

求1+y分之x的二重积分，积分区域是由y＝x分之1 x＝y x＝2所围成的。

Answer 1

y=1/x 与y=x的交点是(1,1)

∴

∫(1->2)dx ∫(1/x ->x) x/(1+y) dy

=∫(1->2)xdx ∫(1/x ->x) dy/(1+y)

=∫(1->2)xdx ∫(1/x ->x) d(y+1)/(1+y)

=∫(1->2)xdx  ln(y+1)|(1/x ->x)

=∫(1->2)xdx [ln(x+1) -ln(1/x +1)]

=∫(1->2)xdx [ln(x+1) -ln((x+1)/x)]

=∫(1->2)xdx [ln(x+1) -ln((x+1) + lnx]

=∫(1->2)xlnxdx

=1/2∫(1->2) lnx dx^2

=1/2 x^2 lnx |(1->2) -1/2∫(1->2) x^2 *1/x *dx

=1/2 (4ln2 -1ln1) -1/2∫(1->2) xdx

=2ln2 -1/4* x^2 |(1->2)

=2ln2 -1/4 (4-1)

=2ln2 -1 +1/4

=2ln2 -3/4