求由曲线y=lnx和直线x+y=1,y=1所围图形的面积

1、如图，阴影部分面积即为所求面积。这种形状用y作为积分变量比较方便一点

将两条曲线分别转变为y的函数，可得x=-y+1，x=e^y，积分变量为y从0→1

S阴影=∫(0→1)(x2-x1)dy

=∫(0→1)[e^y-(-y+1)]dy

=∫(0→1)e^ydy+∫(0→1)(y-1)dy 

=(e-1)+(0+1)

=e

2、用分步积分法：

∫(0→1)xarctanxdx

=1/2∫(0→1)arctanxdx^2

=1/2[(0→1)x^2arctanx-∫(0→1)x^2d(arctanx)]

=1/2[π/4-∫(0→1)x^2/(1+x^2)dx]

=1/2[π/4-∫(0→1)dx+∫(0→1)1/(1+x^2)dx]

=1/2[π/4-1+(0→1)arctanx]

=1/2[π/4-1+π/4]

=π/4-1/2

3、分步积分法：

∫(1→π)xlnxdx

=1/2∫(1→π)lnxd(x^2)

=1/2[(1→π)x^2lnx-∫(1→π)x^2dlnx]

=1/2[(1→π)x^2lnx-∫(1→π)xdx]

=1/2[(1→π)x^2lnx-(1→π)x^2/2]

=1/2[(π^2lnπ-0)-(π^2-1)/2]

=1/2π^2(lnπ-1)+1/4