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$f:[0,1]\rightarrow \mathbb{R}$ be continuous, then $ \int_{0}^{1}f(x)e^{-x}dx=?$

I understand I need to do integration by parts, taking $f(x)$ in 2nd function, so that I can apply mean value theorem of Integral calculus $\int_{0}^{1}f(x)=(1-0)f(c)$ for some $c\in(0,1)$, but I am not able to get the final answer

will the answer b from the following

  1. $f'(0)-f'(1)e^{-1}$

  2. $f(c)(1-e^{-1})$

  3. $e^{-c}\int_{0}^{1}f(x)dx$

  • 4
    Could you bit a little more precise in asking your question? I can't tell what you want here.2012-07-28

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You do not need integration by parts. Option 1 is a red herring: there is no way for the integral to be independent of the values of $f$ in the interior of the interval. This question tests your knowledge of the Mean value theorem for integrals (to the extent that a multiple-choice question can do such a thing). Both 2 and 3 look like an application of this theorem, but one is legitimate and one is not.

Stop and think before reading further.

To understand which, you should recall that the functions that remains in the integral must have constant sign (it's considered to be a "weight", most often nonnegative). So we can apply MVT by leaving $e^{-x}$ inside. We cannot do the same with $f$, since there is no information about its sign.