# Prove $\int _0^1\:e^xf\left(1-x\right)dx=\int _0^1\:e^x\left(f′\left(1-x\right)\right)dx$.

I'm having trouble with following problem, thinking about integration by parts but just getting circular answer:

Let $f$ by continuous on $[0,1]$ and differentiable on $(0,1)$, and also $f(0)=f(1)=0$. Prove $$\int _0^1\:e^xf\left(1-x\right)=\int _0^1\:e^x\left(f′\left(1-x\right)\right)$$

• Note that when you do $(f(1-x))' = -f'(1-x)$ where the minus sign is due to the derivative of $1-x$. When you integrate the LHS by parts, you have a term which is null because of the condition $f(0)=f(1)=0$, namely $e^x f(1-x) |_{0}^{1}$. Then the answer follows. – Alessio Ranallo Sep 22 '17 at 14:08
• Very polite title but I wonder if it can be shortened? :-) – Kevin Sep 22 '17 at 14:16

Let $f(1-x)=u$ and $e^xdx=dv$ then $f'(1-x)\times-1dx=du$ and $e^x=v$, with integration by parts
$$\int udv=uv-\int v du$$
Using integration by parts and exploiting of the fact $f(0)= f(1)=0$, it all boils down to $$\int_0^1 e^x f(1-x) \mathrm{d}x = -\int_0^1 e^x \Big(f(1-x)\Big)' \mathrm{d}x+ [e^x f(1-x)]_{0}^1 = \int_0^1 e^x f'(1-x) \mathrm{d}x$$ as desired, after some caution with the argument of the $f$ function and a chain rule application, as $(1-x)' = -1$.