文档介绍：第十九章含参变量的积分
§1 含参变量的正常积分
1. 求下列极限:
R √
(1) lim 1 x2 + a2dx;
a→0 −1
R
(2) lim 2 x2 cos ax dx;
a→0 0
R 1+a dx
(3) lim 2 2 .
a→0 a 1+x +a
0(x),其中:
R 2
x −xy2
(1) F (x) = x e dy;
R √
cos x x 1−y2
(2) F (x) = sin x e dy;
R b+x sin(xy)
(3) F (x) = a+x y dy;
h i
R x R x2
(4) 0 t2 f(t, s)ds dt.
(x)为连续函数,
Z ·Z ¸
1 x x
F (x) = 2 f(x + ξ+ η)dη dξ
h 0 0
求F 00(x).
.研究函数
4 Z
1 yf(x)
F (y) = 2 2 dx
0 x + y
的连续性,其中f(x)是[0,1]上连续且为正的函数.
:
R π
2 2 2
(1) 0 ln(a − sin x)dx (a > 1);
1
R
π 2
(2) 0 ln(1 − 2a cos x + a )dx(|a| < 1);
R π
2 2 2 2 2
(3) 0 ln(a sin x + b cos x)dx (a, b 6= 0);
R π
2 arctan(a tan x)
(4) 0 tan x dx(|a| < 1).
:
R 1 xb−xa
(1) 0 ln x dx (a > 0, b > 0);
R 1 ¡ 1 ¢ xb−xa
(2) 0 sin ln x ln x dx (a > 0, b > 0).
,试求下列函数的二阶导数:
R x
(1) F (x) = 0 (x + y)f(y)dy;
R b
(2) F (x) = a f(y)|x − y|dy (a < b);
R 1 R 1 x2−y2 R 1 R 1 x2−y2
: 0 dx 0 (x2+y2)2 dy 6= 0 dy 0 (x2+y2)2 dx.
R p
1 2 2
(y) = 0 ln x + y dx,问是否成立
Z 1 p
0 ∂ 2 2
F (0) = ln x + y |y=0dx.
0 ∂y
Z 2π
F (x) = ex cos θ cos(x sin θ)dθ
0
求证F (x) ≡ 2π.
(x)为两次可微函数,ϕ(x)为可微函数,证明函数
Z
1 1 x+at
u(x, t) = [f(x − at) + f(x + at)] + ϕ(z)dz
2 2a x−at
满足弦振动方程
∂2u ∂2u
= a2
∂t2 ∂x2
及初始条件
u(x, 0) = f(x), ut(x, 0) = ϕ(x).
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