Prove that $ \int \limits_a^b f(x) dx$ = $ \int \limits_a^b f(a+b-x) dx$

Question

Hi everyone I have been trying to prove that that $ \int \limits_a^b f(x) dx$ = $ \int \limits_a^b f(a+b-x) dx$ . Heres my attempt:

LS: $ \int \limits_a^b f(x) dx$ = $ \int \limits f(b) - \int \limits f(a) $

RS: $ \int \limits_a^b f(a+b-x) dx$ = $ \int \limits f(a+b-b) - \int \limits f(a+b-a) $ = $ \int \limits f(a) - \int \limits f(b) $

as you can see I have really not been able to make the 2 sides equal each other. Any help on this i appreciated! Thank you

Change of variable. – Martín-Blas Pérez Pinilla Feb 09 '14 at 21:39 — Martín-Blas Pérez Pinilla, Feb 09 '14 at 21:39

score 2 · Answer 1 · answered Feb 09 '14 at 21:39

2

Hint: Make the change of variables $ y=a+b-x $.

answered Feb 09 '14 at 21:39

Mhenni Benghorbal

score 2 · Accepted Answer · edited Feb 09 '14 at 21:55

2

Take $y=a+b-x$. Then $dy/dx=-1$ and thus $∫_a^bf(a+b-x)dx= ∫_b^a-f(y)dy = ∫_a^bf(x)dx$

edited Feb 09 '14 at 21:55

bodo

answered Feb 09 '14 at 21:41

Tubai

Those are some unusual unicode characters you have there. How did they come into your possession? – bodo Feb 09 '14 at 21:48
i wrote first on word and then made copy-paste – Tubai Feb 09 '14 at 21:49
I took the liberty of unburdening you from them. – bodo Feb 09 '14 at 21:57

2 Answers2