Show that $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ [duplicate]

I don't really know where to start with this one. Can you just ignore the $f(..)$ and deal exclusively with what's inside the brackets?

marked as duplicate by N3buchadnezzar, Surb, Daniel Fischer♦, Aditya Hase, HakimDec 12 '14 at 13:39

• Make an appropriate substitution in the RHS integral. – David Mitra Dec 12 '14 at 13:12
• let $y=a-x$ then usual substitution – Santosh Linkha Dec 12 '14 at 13:13

Hint Substitute $y = a-x$ and remember that $\int_{c}^dg(x)dx = -\int_{d}^c g(x)dx$.
Use the $u$-substitution $u=a-x$, which gives $\mathrm{d}u=\mathrm{d}x$, so we get $$\displaystyle\int_{x=0}^{x=a} f(x) \, \mathrm{d}x = \displaystyle\int_{u=a}^{u=0} -f(u) \, \mathrm{d}u = \displaystyle\int_{a}^{0} -f(a-x') \, \mathrm{d}x' = \displaystyle\int_{0}^{a} f(a-x) \, \mathrm{d}x.$$
Let $a-x=t$ and $dx=-dt$ so we have $$\int_{0}^{a}f(a-x)dx=-\int_{a}^{0}f(t)dt=\int_{0}^{a}f(t)dt$$
$$\int_0^a f(x)dx = \int_0^a f(a-y) dy$$
From the rhs, the substitution $a-y = x$ gives $$\int_{a}^0 f(x) (-dx)$$ which is equal to the lhs.