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Let $a$ be a postive real number. If $f$ is a continious and even function defined on the interval $[-a,a]$, then find the value of $$\int_{-a}^{a} \frac{f(x)} {1+e^x} dx. $$

My answer is :
$$2 \int_{0}^{a} \frac{f(x)} {1+e^x} dx $$ because $\int_{-a}^{a} = 2\int_{0}^{a}$.

Is it correct??

any hints/solution will be appreciated

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    $\begingroup$ Nope. $(1 + e^x)$ is not even... If instead you had $(1 + e^{|x|})$ well then... $\endgroup$ Commented Nov 13, 2018 at 1:23

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$$ I = \int_0^a + \int_{-a}^0 \frac {f(x)}{1+\mathrm e^x} \,\mathrm dx = \int_0^a \frac {f(x)\,\mathrm dx}{1+\mathrm e^x} + \int_0^a \frac {f(x)\, \mathrm dx}{1 + \mathrm e^{-x}} = \int_0^a \frac {f(x)(1 + \mathrm e^x)}{\mathrm e^x + 1}\,\mathrm dx = \int_0^a f. $$

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  • $\begingroup$ thanks u ..@Xbh $\endgroup$
    – jasmine
    Commented Nov 13, 2018 at 1:39
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    $\begingroup$ You are welcome, glad to help. $\endgroup$
    – xbh
    Commented Nov 13, 2018 at 1:43

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