(30)窗体上有两个文本框Text1、Text2以及一个命令按钮Commandl,编写下列程序: Dim y As Integer Private Sub Commandl_Click() Dim x As Integerx=2 Text1.Text=p2(pl(x),y) Text2.Text=p2(x) End Sub Private Function p2(x As Integer,y As Integer) x=x+y:y=x+y pl=x+y End Function Private Function p2(x As Integer,y As Integer)As Integer p2=2*x+y End Function 当单击1次和单击2次命令按钮后,文本框Text1和Text2内的值分别为( )。A.2 4 B.2 4 C.4 4 D.10 10 2 4 4 8 8 8 58 58
A、P1X1P2X2=I
B、P1X1P2X2=I
C、X1=–(P2/P1)X2I/P1
D、X2=–(P2/P1)X1I/P2
分析下面的程序 main() { int*p1,*p2,*p; int x=4,y=6; p1=&x;p2=&y; if(x<y) {p=p1;p1=p2;p2=p; } printf("%d,%d,",*p1,*p2); printf("%d,%d\n",x,y); } 程序的输出结果为_______。
A.6,4,4,6
B.4,6,6,4
C.4,6,4,6
D.6,4,6,4
已知集合P={x|0 ≤x ≤5,x∈Z},Q={y|y=|x2-1|,x∈P},则P∩Q中元素的个数是( ).
(A)3.
(B)6.
(C)8.
(D)9.
设X的分布列为,概率P(2≤X<5)=( )。
A.p2+p3+p4+p5
B.p2+p3+p4
C.P(X<5)-P(X<2)
D.1-P(X<2)-P(X>4)
E.P(X≤4)-P(X<2)
第一章 一元函数积分学1.1 有理函数积分Pm(x) Qn(x)dx第一步将假分式用长除法化为真分式R(x) + S(x)Q(x)dx第二步将分母化为因式相乘的形式S(x)R(x) + dxq1(x)q2(x) qk(x)第三步用待定系数法或者留数法将真分式化为部分分式R(x) +s1(x)q1(x)+s2(x)q2(x)+ +sk(x) qk(x)dx第四步部分分式有以下四种形式1 Adx = Aln(x a) xa2 A 1kk dx = A(x a) (xa) 1k3 x2+px+q dxBx+Cp p Bx + C dx = B(x +2 ) + C 2dxx2 + px + qp p2(x + 2 )2 + q 4p p= pB(x + 2 ) ) + d(x + d(x + p)C 2p p2 p p22 (x +2 )2 + q (x +2 )2 + q 4 4pp2d(x + 2 + q p d(x + p)2 )B2 ) 4+ (C =p p2 p p22(x + 2 )2 + q (x +2 )2 + q 4 4ln 4 (x + q)2 + q 2 2+ qB p2p C x +p p= arctan2 2p2 p2q q 4 44 Bx+C(x2+px+q)k dx 有递推公式p pBx + C B(x + 2 ) + C dx = 2h 4 ik dx(x2 + px + q)kp p2(x + 2 )2 + q 取 t = x + p2 , m = C 2 , n = q pp2原式化简为4Bt + mB + d(t2 + n) m 2 dt = dt(t2 + n)k (t2 + n)k (t2 + n)k1.1 有理函数积分例题 1.1 求积分 解x+3x2+2x+4 dxx + 3 dx = x2 + 2x + 4x + 32 dx(x + 2)设 x+3(x+2)2 =因此Ax+2 +B(x+2)2解得A = 1x + 3 1B = 1 +2 =x + 2(x + 2) x + 3 dx = x + 32 dxx2 + 2x + 4 (x + 2)12(x + 2)= 1x + 2+12 dx(x + 2)= ln (x + 2) 1x + 2+ C例题 1.2 求积分 解97xx2+12x+38 dx9 7x 7 (x + 6) + 51 dx =d (x + 6)x + 12x + 38 22 + 2(x + 6)令 t=x+6, 则有原式 = 7t + 51t2 + 2dt例题 1.3 求积分 方法 1: 三角换元= 7 dt + 51 t2 + 2 t2 + 2t 1dtd t2 + 2 2 + d 72 1 51 1 tt2 + 2 2= 22 2t+ 12 arctan 21.1 有理函数积分方法 2: 分母次数高, 用分部积分降低次数解x2dx = (a2 + x2)2x xdx(a2 + x2)21=1= d x2 + a2x(a2 + x2)21xdx2 + a2例题 1.4 求积分 1(a2+x2)2 dx, a 0解 a2 2 dx =1 1(a2 + x2)1x= 2x2 + a21x= 2x2 + a2a22 dx(a2 + x2)+ 12 1x2 + a2dxarctan a + C2a1 x=a12 a2 + x2 x22 dx(a2 + x2)=a2 11a2 + x2 (a2 + x2)2a2 1 x2dx dxa2 aarctan a 2 + + Carctan 1 x 1 1 x 1 xa3 x2 + a2 2a 1 x 1 x arctan + + C2a3 a 2a2 x2 + a2从这一题来看, 这种题目都可以归结为求上一道例题的积分, 当然, 最终选择用三角函数换元, 还是分部积分降次便是见仁见智了.例题 1.5 求积分 11+x3 dx解 dx = 1 1dx1 + x3 (1 + x) (1 x + x2)3 1 + x 1 x + x1 2 x= + 2 dxln x2 x + 11 1= ln (1 + x) 3 21.1.1 组合积分法 (积木法 by 魏念辉)+ 3 arctan 3 1.1 有理函数积分再改变一下分母的形式 1 dx = 1dxx6 + 1 (1 + x2) (1 x2 + x4)这时分子改成 1 x2 + x4 也很好积分2注意到 12x+x4 dx 也是一个比较好积分的函数1x1 x dx = 1 21x2dx1 + kx2 + x4 1x2 + k + x2= x dxx 1 x x 12 2 + k= 11 x dx 2 x 2 + k1x 因此分子可改成 1 + x2 2综上所述, 现有 x6 + 1, x5, x2, 1 x1 x2 + x4, 1 x4, 1 + 2x2 + x4 六块积木现在用这六块积木组成分子2 x2 + 2 + x4 + 411 = 1 x 1 x2 1 + 2x2 + x4 2 x2 + 411 = 1 x. . .则有2 x2 + 2 + x4 + 41 1 x 1 x1 dx = dxx + 1 x + 1 23 11 + x2d x + x11 1 2 + dx dx3=x + x2 1 + (x3)143 ln x2 + 3x + 1! x2 1 1 1 3x + 1= arctan x arctan x3 +6 23#+ C例题 1.7 求积分 sin x+ cos x sin x+ cos x dx解 选取积木 sin x + cos x, cos x sin x 则有dx = 2 + 2 1.1 有理函数积分 1 dx = 14 + tan2 x sec2 xdx4 + tan2 x 4 + tan2 x3dtan x3 1 = dx 1 + 2 22tan x1 1 tan x=x + Carctan 2 1.1 有理函数积分2 1 + x2 + 211 x 1 dx = dxx4 + kx2 + 1 x4 + kx2 + 12 dx1.1 有理函数积分解1 + x41 + x6dx = 1 + x43 dx1 + (x2)= = = = 1 + x4dxdx(1 + x2) (1 x2 + x4)1 + x4 x2 + x2(1 + x2) (1 x2 + x4)1 x2dx + dx1 + x2 (1 + x2) (1 x2 + x4)1 dx + x2dx1 + x2 1 + x6= 1dx +1 + x21dx321 + (x3)= arctan x +13arctan x3 + C例题 1.15 求积分 解1x(x3+27) dx1 dx =x(x + 27) 3x2dxx3(x3 + 27)=11dx3x3(x3 + 27)