0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Alessio Ranallo Sep 22 '17 at 14:08
  • $\begingroup$ Very polite title but I wonder if it can be shortened? :-) $\endgroup$ – Kevin Sep 22 '17 at 14:16
0
$\begingroup$

Let $f(1-x)=u$ and $e^xdx=dv$ then $f'(1-x)\times-1dx=du$ and $e^x=v$, with integration by parts

$$\int udv=uv-\int v du$$

we have

\begin{align} \int _0^1\:e^xf\left(1-x\right)dx &= f(1-x)e^x|_0^1-\int _0^1e^xf'(1-x)\times-1dx \\ &= f(0)e^1-f(1)e^0+\int _0^1e^xf'(1-x)dx \\ &= \int _0^1e^xf'(1-x)dx \end{align}

$\endgroup$
1
$\begingroup$

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 $(1-x)' = -1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.