已知抛物线 =p(x)^2+qx (其中 lt 0,qgt 0) 在第一象限内与直线 x+y=5 相切,-|||-且此抛物线与x轴所围成的图形的面积为A.问p和q为何值时,A达到最大值,并求-|||-出此最大值.

15.求下列已知曲线所围成的图形,按指定的轴旋转所产生的旋转体的体积:-|||-(1) =(x)^2 ,=(y)^2 ,绕y轴;

求：int dfrac (1-ln x)({(x-ln x))^2}dx

2.按例4的方法求下列极限:-|||-(1) lim _(xarrow 0)dfrac ({e)^xsin x-x(1+x)}({x)^3} ;-|||-(2) lim _(xarrow infty )[ x-(x)^2ln (1+dfrac (1)(x))] ;-|||-(3) lim _(xarrow 0)dfrac (1)(x)(dfrac (1)(x)-cot x) .

33.已知 =((dfrac {1)(x))}^x 则 '= __

6.求下列矩阵的特征值和特征向量.-|||-3 -1-|||-(1) -1 3 ;-|||--1 1 0-|||-(2) -4 3 0 ;-|||-1 0 2-|||-1 2 3-|||-(3) 2 1 3 ;-|||-3 3 6-|||-0 0 0 1-|||-0 0 1 0-|||-(4) 0 1 0 0-|||-1 0 0 0-|||-2 2 -2-|||-(5) 2 5 -4-|||--2 -4 5-|||-a a a-|||-(6) ,其中 neq 0.-|||-a a a times n

设矩阵A= (3,-2) ,B= 2 4 https:/img.zuoyebang.cc/zyb_7142fe6d884b795b75323081826723ec.jpg,C= 1 -2 3-|||-1 3-|||--4 5 -6，则下列矩阵运算中有意义的是A.CBB.ABCC.BACD.CBA

.计算下列积分:-|||-(12)f(x)dx,其中 f(x)= {x)^2,xgt 1 .

1.判定下列各反常积分的收敛性,如果收敛,计算反常积分的值:-|||-(1) (int )_(1)^+infty dfrac (dx)({x)^4} =-|||-(2) (int )_(1)^+infty dfrac (dx)(sqrt {x)}-|||-(3) (int )_(0)^+infty (e)^-axdx(agt 0) =-|||-(4) (int )_(0)^+infty dfrac (dx)((1+x)(1+{x)^2)}-|||-(5) (int )_(0)^+infty (e)^-msin omega tdt(pgt 0,omega gt 0) =-|||-(6) (int )_(-infty )^+infty dfrac (dx)({x)^2+2x+2} ;-|||-(7) (int )_(0)^1dfrac (xdx)(sqrt {1-{x)^2}}-|||-(8) (int )_(0)^2dfrac (dx)({(1-x))^2} =-|||-(9) (int )_(1)^2dfrac (xdx)(sqrt {x-1)} =-|||-(10) (int )_(1)^edfrac (dx)(xsqrt {1-{(ln x))^2}}

设 b1 C1-|||-D= .2 b2 2 =6-|||-3a3 b3 C3，则 b1 C1-|||-D= .2 b2 2 =6-|||-3a3 b3 C3=（ ）A.-2B.3C.-4D.6

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