Expected Value for variables $x_1Q) There are three-sample values $x_1, x_2$ and $x_3$ following P.D.F $f(x_i) = 2x_i, (0 < x_i<1)$ in $(0,1)$
Find the $E(X)$ for $X=x_2$ (Here the $x_1 < x_2 < x_3$ and $x_i$ are independent variables.)

Here is my attempt.
Since the $x_i$ are independent, $f(x_1, x_2, x_3) = 8x_1x_2x_3$ 
Say the $f(x_2)$ be P.D.F only for one variable $x_2(=X)$
Then $f(x_2) = \int_{x_2} ^1 \int_{0}^{x_2} f(x_1, x_2, x_3) dx_1 dx_3  = 2X^3(1-X^2)$
Hence, $E(X) = \int_0 ^1 Xf(X) dX = {4 \over 35}$ 
But the answer was $24 \over 35$
What the point do I have a mistake? I can't find what I've missed.
Thanks. 
 A: First, let's be careful with our notation. It's totally possible to define $x_1, x_2, x_3$ as independent random variables with the given density function. But if you do that, you can't enforce the condition that $x_1 < x_2 < x_3$, because this contradicts independence. Instead, the right idea is to define the order statistics, which are three other variables:
\begin{align*}
  x_{(1)} & := \min\{x_1, x_2, x_3\} \\
  x_{(3)} & := \max\{x_1, x_2, x_3\} \\
  x_{(2)} & := \text{(the other one)}
\end{align*}
(It's possible to write down a definition for $x_{(2)}$, of course, but it's not really worth the effort as long as the meaning is clear.) Keeping the notation straight is important; the $x_i$ variables have the given density function and are independent of one another. The $x_{(i)}$ variables, though, are very much dependent on each other, and they do not have the original density function anymore. Intuitively, it should be clear that $x_{(1)}$ is much more likely to be found in, say, the interval $[0, 0.1]$ than $x_{(3)}$ is. These new variables are called the order statistics, and there is a long body of literature on how to deal with these.
The line in your solution that I don't understand is the line where you say 

Then $f(x_2) = \int_{x_2} ^1 \int_{0}^{x_2} f(x_1, x_2, x_3) dx_1 dx_3  = 2X^3(1-X^2)$

and as far as I can tell, you may be making an error in conflating the difference between $x_2$ and $x_{(2)}$. You may have noticed that your answer was off by a factor of $6$; this is not an accident, and the factor of $3!$ comes from the ways to rearrange the three variables.
Hopefully I've said enough at this point to identify your mistake and point you on the right track; please let me know if not.
A: Define $A=\{ X_1 < X_2 <X_3 \}$
 so I think you want to find $E(X_2| X_1 < X_2 <X_3)$
$$E(X_2| X_1 < X_2 <X_3)=E(X_2|A)=\frac{E(X_2 1_{A})}{P(A)}=\frac{\frac{4}{35}}{\frac{1}{6}}$$ Conditional_expectation_with_respect_to_an_event
$$P(A)=P(X_1 < X_2 <X_3)=\frac{1}{6}$$ and
$$E(X_2 1_{A})=E(X_2 1_{\{X_1 < X_2 <X_3\}})=\iiint_{0<x_i<1} x_2 1_{\{X_1 < X_2 <X_3\}} f_{(X_1,X_2,X_3)}(x_1,x_2,x_3) dx_1 \, dx_2 \, dx_3$$
$$=
\iiint_{x_1<x_2<x_3,0<x_i<1} x_2 f_{(X_1,X_2,X_3)}(x_1,x_2,x_3) dx_1 \, dx_2 \, dx_3$$
$$=
\int_0^1   \int_{x_2}^1  \int_0^{x_2} x_2 f_{(X_1,X_2,X_3)}(x_1,x_2,x_3) dx_1 \, dx_2 \, dx_3$$
$$\overset{by \ your \ calculation}{=} \int_0^1    2x_2^4 (1-x_2^2) dx_2 $$
$$= \frac{4}{35}$$
$$P(A)=P(X_1<X_2<X_3)=\frac{1}{3!}=\frac{1}{6}$$
since all of following event are equal (since $X_i$ are $i.i.d$)
$$X_1<X_2<X_3$$
$$X_1<X_3<X_2$$
$$X_2<X_1<X_3$$
$$X_2<X_3<X_1$$
$$X_3<X_1<X_2$$
$$X_3<X_2<X_1$$
A: You are dealing with order statistics here and should work e.g. with notation $x_{(i)}$ where $\{x_1,x_2,x_3\}=\{x_{(1)},x_{(2)},x_{(3)}\}$ and $x_{(1)}<x_{(2)}<x_{(3)}$.
Also in this answer we make use of the general rule that $$\mathbb EX=\int_0^{\infty}P(X>0)dx$$for non-negative random variable $X$.
Let $N=\left|\left\{ i\in\left\{ 1,2,3\right\} \mid x_{i}\leq x\right\} \right|$ so that $\{x_{(2)}\leq x\}=\{N\geq2\}$.
Here $N$ has binomial distribution with parameters $n=3$ and: $$p=P\left(x_{1}\leq x\right)=\int_{0}^{x}2ydy=x^{2}$$ so that:
$$P\left(x_{\left(2\right)}\leq x\right)=P\left(N=2\right)+P\left(N=3\right)=x^{6}+3x^{4}\left(1-x^{2}\right)=3x^{4}-2x^{6}$$ 
and applying the mentioned rule we find: $$\mathbb{E}x_{\left(2\right)}=\int_{0}^{\infty}P\left(x_{\left(2\right)}>x\right)dx=\int_{0}^{1}1-3x^{4}+2x^{6}dx=\left[x-\frac{3}{5}x^{5}+\frac{2}{7}x^{7}\right]_{0}^{1}=\frac{24}{35}$$
