早教吧作业答案频道 -->其他-->
设函数f(x)连续,且f(0)≠0,求极限limx→0∫x0(x−t)f(t)dtx∫x0f(x−t)dt.
题目详情
设函数f(x)连续,且f(0)≠0,求极限
.
lim |
x→0 |
| ||
x
|
▼优质解答
答案和解析
令x-t=u;
则:dt=d(-u)=-du;
f(x−t)dt=
f(u)d(−u)=
f(u)du.
因此:
=
=
(洛必达法则)
=
=
(洛必达法则)
=
=
.
则:dt=d(-u)=-du;
∫ | x 0 |
∫ | 0 x |
∫ | x 0 |
因此:
lim |
x→0 |
| ||
|
lim |
x→0 |
| ||||
|
=
lim |
x→0 |
| ||
x
|
=
lim |
x→0 |
| ||
|
=
lim |
x→0 |
f(x) |
f(x)+f(x)+xf′(x) |
=
f(0) |
f(0)+f(0)+0 |
=
1 |
2 |
看了 设函数f(x)连续,且f(0...的网友还看了以下:
设f(x)=x^n•sin(1/x)(x≠0),且f(0)=0,则f(x)在x=0处()设f(x) 2020-05-20 …
设在区间[0,1]上f''(x)>0,则f'(0)f'(1)和f(1)-f(0)的大小顺序是设在区 2020-06-08 …
1.设f(x)在[0,1]上连续,且f(0)=f(1),证明:存在x0∈[0,1],使得f(x0) 2020-06-18 …
设f(x)在闭区间[0,1]连续,在(0,1)内可导且f(0)=0,f(1)=1/3求证:彐ξ设f 2020-06-23 …
设f(x)=ax2bxc,满足f(a1)=0,f(a2)=0,f(a3)=1,(其中:a1,a2, 2020-07-09 …
设f(x)=ax2+bx-c,满足f(a1)=0,f(a2)=0,f(a3)=1,(其中:a1,a 2020-07-09 …
导数的连续性设f(x)可导,且f(0)=0,f(x)在0点的导数不为0,求w=lim(x→0){x 2020-07-16 …
设函数f(x)在x=0处可导,且f(0)=0,求下列极限,其中a不等于0,为常数limx→0[f( 2020-07-16 …
设函数f(x)具有二阶导数,g(x)=f(0)(1-x)+f(1)x,则在[0,1]上()A.当f 2020-07-20 …
微积分设函数f(x)在[0,1]上连续,在(0,1)内可导,有f(0)=f(1)=0.证明:至少微 2020-07-31 …