Re: [積分&極限]後醫系94年考古題
2
x sin(xt)
If F(x) =∫--------dt, find F'(x) where x≠0
0 t
Recall The Leibniz's Rule For Differentiation Of Integrals :
Suppose f:[a,b]x[c,d]->R is continuous and f_x exists and is continuous
on [a,b]x[c,d].
Let u(x) and v(x) be continuously differentiable functions from
[c,d] into [a,b],and put
v(x)
F(x)=∫ f(t,x)dt,then :
u(x)
v(x)
F'(x)= f(v(x),x)v'(x) -f(u(x),x)u'(x) +∫f_x(t,x)dt
u(x)
--------
In this problem : sin(xt) 2
Obviously , f(t,x)=--------- and f_x(t,x)=cos(xt) are both continuous on R
t
2
;u(x)=0 and v(x)=x are continuously differentiable on R so we can just apply
this theorem to find F'(x) .
2
x
F'(x)= f(v(x),x)v'(x) -f(u(x),x)u'(x) +∫f_x(t,x)dt
0
2 2 3 3
sin(x*x ) x 2sin(x ) 1 x
=----------*2x- 0 + ∫cos(xt)dt =---------- + -∫cos(xt)d(xt)
2 0 x x 0
x
3
2sin(x ) 1 3
=--------- + -sin(x )
x x
3
3sin(x )
=-------
x
--------- sin(xt) 2
check: f(t,x)=--------- is continuous on R .
t
pf:It suffices to show that f(t,x) is continuous at (0,0) .
The key point is : when |x| and |t| are small enough ,
|sin(xt)| <|xt|.
sin(xt) sin(xt) xt
||(t,x)-(0,0)||<d , |---------- -0| =|-------| <|-----|=|x|<d .
t t t
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◆ From: 140.113.192.95
※ 編輯: nomorethings 來自: 140.113.192.95 (07/28 03:51)
※ 編輯: nomorethings 來自: 140.113.192.95 (07/28 03:53)
推
07/28 08:20, , 1F
07/28 08:20, 1F
→
07/28 08:21, , 2F
07/28 08:21, 2F
→
07/28 08:21, , 3F
07/28 08:21, 3F
※ 編輯: nomorethings 來自: 218.162.211.204 (07/29 17:34)
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