定积分的计算.doc

定积分计算牛顿-莱布尼茨公式abfxdx=Fb-F(a)换元法假设函数f(x)在区间[a,b]上连续,函数x=φ(t)满足条件:(1)φα=a,φβ=b,(2)φ(t)在α,β上具有连续导数,且其值域Rφ=[a,b]则有abfxdx=αβfφtφ'tdt分部积分法abudv=[uv]ab-abvdu利用奇偶性、周期性-aafxdx=20afxdxfx=f(-x)0fx=-f(x)aa+Tfxdx=0Tfxdx利用已有公式0π2sinnxdx=0π2cosnxdx=n-1n∙n-3n-2⋯12∙π2为偶数n-1n∙n-3n-2⋯23∙1n为大于1的奇数历年真题1、121x3e1xdx(2007,数一,4分)【解析】121x3e1xdx=-121xe1xd1x=-121xde1x=-1xe1x12+12e1xd1x=-12e12+e+e1x12=e22、0π2xcosxdx(2010,数一,4分)【解析】令x=t,则x=t2,dx=2tdt0π2xcosxdx=0π2t2costdt=0π2t2dsint=2t2sint0π-0π4tsintdt=4tcost0π-0π4costdt=-4π3、02x2x-x2dx(2012,数一,4分)【解析】02x2x-x2dx=02x1-(x-1)2dx令x-1=sint,则dx=costdt02x2x-x2dx=-π2π21+sintcos2tdt=20π2cos2tdt=2∙12∙π2=π24、设fx=1xln⁡(1+t)tdt,计算01f(x)xdx(2000,数二,5分)【解析】因为fx=1xln⁡(1+t)tdt所以f'x=ln⁡(1+x)x且f1=001f(x)xdx=201f(x)dx=2f(x)x01-201xf'xdx=-201ln⁡(x+1)xdx=-4xln⁡(x+1)01+401xx+1dx=-4ln2+401xx+1dx令u=x,则01xx+1dx=201u2u2+1du=2(u-arctanu)01=2-π2所以01f(x)xdx=8-2π-4ln25、-π2π2(sinx1+cosx+x)dx(2015,数一,4分)【解析】-π2π2(s