Prove that:
Int (limits a,0) f(x) dx = Int (limits a,0) f(a-x) dx
a is the top limit, 0 is the bottom limit.
I know this is true but how is the proof written... thanks.
2007-12-23 07:13:26 · 3 answers · asked by ndavos 2 in Science & Mathematics ➔ Mathematics
For the RHS, set u=a-x. du=-dx. Lower limit is a-0=a and the upper limit is a-a=0. This gives -Int(limits 0,a)f(u)du. Change the sign by reversing the limits to get Int(limits a,0)f(u)du. Change your variable of integration from u back to x, as it is not important what you call it. Proven.
2007-12-23 07:27:15 · answer #1 · answered by jcsuperstar714 4 · 0⤊ 0⤋
lets see , set u = a-x , du = -dx,
at x = a , u = 0 ,
at x = 0 , u = a ,
then , integral ( 0,a) f(a-x)dx = int (0,a) f(u)du = int(0,a) f(x)dx
x here is a dummy variable
2007-12-23 07:39:29 · answer #2 · answered by Nur S 4 · 0⤊ 0⤋
call z= a-x so dz = -dx so you have the Int(0,a) -f(z)dz =
Int ((a,0)f(z) dz =-Int(a,0) f(a-x)dx
It seams that a (-) sign is lacking
2007-12-23 07:27:26 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋