## Trending News

# 實變重積分問題

Suppose that f is L(R),

∫∫_(R^2)f(4x)f(x+y)dxdy=1,calculate ∫_(R)f(x)dx

### 1 Answer

- SamLv 610 years agoFavorite Answer
圖片參考：http://imgcld.yimg.com/8/n/AC06918685/o/1511050508...

Fromgeometric viewpoint, for any a in R, the graph of y=f(x+a) is justthe graph of y=f(x) shifting by a constant - a, so the integral of f(x+a) on Ris equivalent to the integral of f(x) on R, i.e.∫[R]f(x+a)dx = ∫[R]f(x)dx for all a in R.Using substitution u=x+a, we can alsoobtain the same result. Since let u=x+a, dx=du and the region of integral isalso R, we have : (*) ∫[R]f(x+a)dx= ∫[R]f(u)du for all a in R.Hence,∫∫[R^2]f(4x)f(x+y)dxdy{ by Fubini theorem }= ∫[R]{∫[R]f(4x)f(x+y)dy}dx{ since f(4x) is independent to y, soit is a constant for y. } = ∫[R] f(4x){∫[R] f(x+y)dy}dx{ by (*) }= ∫[R] f(4x){∫[R] f(u)du}dx { ∫[R] f(u)du is aconstant }=∫[R] f(u)du*∫[R] f(4x)dx=1/4*∫[R] f(u)du*∫[R] f(4x)d4x{ let v=4x }=1/4*∫[R] f(u)du*∫[R] f(v)dv=1/4*∫[R] f(x)dx*∫[R] f(x)dx{ u and v are dumb variables, so can replaceby x. }=1/4*[∫[R] f(x)dx]^2=1 {assumption}= > ∫[R] f(x)dx = 2.[[Done]]

2011-05-06 03:49:15 補充：

If you consider the following examples, you will see the key point of the problem.

ex01

let

f(x)=1 0<=x<=2,

f(x)=0 otherwise;

then calculate the double integral.

2011-05-06 03:49:26 補充：

ex02

let

f(x)=x 0<=x<=2,

f(x)=0 otherwise;

then calculate the double integral.

ex03

draw any graph above the x-axis, and assume that the area between it and

the x-axis is 2, then calculate the double integral.