# find the value of $\int_{-a}^{a} \frac{f(x)} {1+e^x} dx$?

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

• Nope. $(1 + e^x)$ is not even... If instead you had $(1 + e^{|x|})$ well then... – David G. Stork Nov 13 '18 at 1:23

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