文档介绍：limx→-1x+1x2+1=()→0时,sin3x是2x的()(x)=x+1,x<0,在x=0处()x2,x≥(x)=xeπ2,则f'(x)=()A.(1+x)eπ2B.(12+x)eπ2C.(1+x2)eπ2D.(1+2x)(x)=x4-4x的单调增区间的是()A.(-∞,+∞)B.(-∞,0)C.(-1,1)D.(1,+∞)(x)在区间-3,3上连续,则-11f(3x)dx=()-33f(t)-11f(t)-33f(t)dt7.(x-2+sinx)dx=()A.-2x-1+cosx+cB.-2x-3+.-x-33-cosx+cD.–x-1-cosx+(x)=0x(t-1)dt,则f“(x)=()A.-=xy,则∂z∂x=()-+=cos(xy),∂2y∂x2=()(xy)(xy)C.-y2sin(xy)D.-y2cos(xy)→0sin1x=.→∞(1-2x)x3=.e-=ln(4x-x2),则y'(1)=.=x+sinx,则dy=.(1+cosx)=x32+e-x,则y”=.34x-12+e-(x)dx=cos(lnx)+C,则fx=.-sin(lnx)x17.-11xxdx=.(xlnx)=.xlnx+=x2,直线x=1及x轴所围成的平面有界图形的面积S=.=eyx,则∂z∂x|(1,1)=.-→1ex-elnxlimx→1ex-elnx=limx→1ex1x==cos(x2+1),求y'.y'=cos(x2+1)'=-sin(x2+1)∙(x2+1)'=-2xsin(x2+1)+x2dxx4+x2dx=1214+x2d(4+x2)=12ln(4+x2)+,其中fx=x,&x<111+x,&x≥104fxdx=01xdx+0111+xdx=x22|10+ln(1+x)|41=12+(x)是连续函数,且0xf(t)e-tdt=x,求01f(x),得f(x)e-x=1f(x)=ex01f(x)dx=01exdx=ex|10=e-(x)=lnx-x.(1)求f(x)的单调区间和极值;f(x)的定义域为(0,+∞),f'(x)=1x-'(x)=0得驻点x=<x<1时,f'(x)>0;当x>1时,f'(x)<(x)的单调增区间是(0,1),单调减区间是(1,+∞).f(x)在x=1处取得极大值f(1)=-1(2)判断曲线y=f(x)的凹凸性。因为f″(x)=-1x2<0,所以曲线y=f(x)(x,y)=x22-xy+