求由曲线y=lnx和直线x+y=1,y=1所围图形的面积 曲线y=cosx直线y=3π/2-x和y轴围成图形的面积
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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
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绛旓細鎵鍥村浘褰㈤潰绉负(b-a)銆傝В锛氭牴鎹鎰忓彲寰楁墍鍥村浘褰㈤潰绉彲鐢ㄥ畾绉垎琛ㄧず锛屽嵆闈㈢Н=鈭(lna,lnb)xdy锛屽張y=lnx锛岄偅涔坸=e^y銆傚洜姝も埆(lna,lnb)xdy=鈭(lna,lnb)e^ydy =e^y(lna,lnb)=e^lnb-e^lna=b-a銆傚嵆闈㈢Н涓篵-a銆
