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Let $f:[-1,1]\to \mathbb{R}$ diferentiable, with $f'$ integrable, such that $$\frac{\int_{-1}^{1}e^xf(x)dx}{f(1)-f(-1)}=2(e+e^{-1})$$ Prove that there exists $c\in (-1,1)$ such that $$\int_{-1}^{1}\frac{f(x)}{e^x}dx=4ef'(c)$$

My progress so far:

I integrated $\int_{-1}^{1}e^x f(x) dx$ by parts.

I got $(e-e^{-1})-\int_{-1}^{1} f'(x)e^x dx =2(e+e^{-1})$.

By Mean Value Theorem for integrals, $$\frac{e-e^{-1}}{e+e^{-1}}=f'(c)e^c$$ for $c \in (-1,1)$.

But how to proceed now?

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  • $\begingroup$ Your integration by parts is erroneous. You should get $e f(1)-e^{-1}f(-1)-$ $\int_0^1e^x f'(x)\;dx.$ $\endgroup$ – DanielWainfleet Apr 26 '16 at 6:29
  • $\begingroup$ sorry, yes it was erroneous $\endgroup$ – Legend Killer Apr 26 '16 at 6:36
  • $\begingroup$ It is confusing, since, in the context, you have $e^x$ in the denominator, which is different from in the title. $\endgroup$ – xpaul Apr 27 '16 at 19:10
  • $\begingroup$ The MVT should give you $e^tf'(t)=\frac{e^{-1}-e}{2}$, for $t\in (-1,1)$ not the result you have written. $\endgroup$ – MathematicianByMistake May 4 '16 at 10:40
  • $\begingroup$ I did it faultily from the beginning , my bad...still will the correct MVT yield positive results? $\endgroup$ – Legend Killer May 4 '16 at 11:00
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Here is one version, $$A=\int_{-1}^1 e^xf(x)dx$$ $$B=\int_{-1}^1 e^{-x}f(x)dx$$

So B is what we need to estimate.

Using integration by parts. $$A=f(1)e-f(-1)e^{-1}-\int_{-1}^1 e^xf'(x)dx$$ $$B=-f(1)e^{-1}+f(-1)e+\int_{-1}^1 e^{-x}f'(x)dx$$

equation (1) $$A-B=(f(1)-f(-1))(e+e^{-1})-\int_{-1}^1 (e^x+e^{-x})f'(x)dx$$

from the condition, $$A=2(f(1)-f(-1))(e+e^{-1})$$

So with mean value, $$B=(f(1)-f(-1))(e+e^{-1})+\int_{-1}^1 (e^x+e^{-x})f'(x)dx$$ $$B=2\frac {f(1)-f(-1)}2(e+e^{-1})+2(e^{c}+e^{-c})f'(c)$$

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  • $\begingroup$ Any ideas how to bring $4e$ in the RHS? $\endgroup$ – Legend Killer May 5 '16 at 2:38
  • $\begingroup$ I guess $$\frac{\int_{-1}^1e^{-x}f(x)dx}{f(1)-f(-1)}=2e$$ $\endgroup$ – user115350 May 5 '16 at 22:03
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Am giving a try. Did a bit progress in my way and so want to share it. $$I = \int_{-1}^{1} e^x \cdot (f(x) + f(-x))dx$$ $$ = 0 - \int_{-1}^1 e^x \cdot (f'(x) - f'(-x))dx$$ $$ = 2f'(c)(e-\frac{1}{e})$$

Now, $$I = \int_{-1}^1 e^x f(x) + \int_{-1}^{1} e^{-x} f(x)$$ $$ \implies \int_{-1}^{1} e^{-x} f(x)= 2(e-\frac{1}{e})\cdot(f'(c) - f(1) + f(-1))$$

And am stuck out here only

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