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I'm struggling with calculating $$V(e^U+e^{1-U})\qquad\text{while}\qquad U\sim U(0,1)$$ My problem is that I don't know how to approach exercises where there is a random variable in the exponents.
I tried to play with the expectation and with the definition but i can't advance.
Any clues guys?

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Let $Z=e^U+e^{1-U}$. Then $\text{Var}(Z)$ is computed as follows: \begin{align*} \text{Var}(Z)&=\mathbb{E}(Z^2)-\mathbb{E}(Z)^2\\ &=\int_0^1z^2\mathrm{d}u-\left(\int_0^1z\mathrm{d}u\right)^2\\ &=\int_0^1e^{2u}+2e+e^{2-2u}\mathrm{d}u-\left(\int_0^1e^u+e^{1-u}\mathrm{d}u\right)^2 \end{align*} Can you compute these integrals?

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    Thank you for your reply, but i have a question for you. I know that the expectation of a continuous variable is (with $U$ as noted)$$E(x)=\int_0^1xf_x(x)dx$$ So how do you know what is the density function of this $Z$ variable, also i don't see it in the integral. what am i missing? – 2017-01-11
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    In your example, $\mathbb{E}[g(U)]=\int_0^1g(u)f(u)\mathrm{d}u$, where $f(u)=1$. In the case where $g(u)=e^{u}+e^{1-u}$, you get my result above. – 2017-01-11
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    So tell me if i got it right, $e^u$ is uniformly distributed because $u$ is uniformly distributed so i can conclude that its density function is 1 because the domain is (0,1). And so is for the other function $e^{1-u}$? – 2017-01-11
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    Unfortunately, $e^u$ is not uniformly distributed. If you wanted to determine the probability density function for $e^u$, you have two options (at least): You can compute a cumulative distribution function $\mathbb{P}(e^U\leq u)$ and differentiate the result, or you can use a formula (see https://onlinecourses.science.psu.edu/stat414/node/157 ). – 2017-01-11