$\displaystyle Var(X) = E(X^2) - \left[ E(X) \right]^2$

Since:

$\displaystyle \int_0^1 ce^{2x} = 1$

$\displaystyle {c{e^{2} - 1} \over 2} = 1$

$\displaystyle c = {2 \over {e^2 - 1}}$

$\displaystyle E(X) = \int cxe^{2x}$

$\displaystyle = {{c} \over {2}}xe^{2x} - {{c} \over {2}} \int e^{2x}dx$

$\displaystyle = {{c} \over {2}} e^{2x}x - {{c} \over {4}} e^{2x} + C$

$\displaystyle = ce^{2x} \left( {x \over 2} - {1 \over 4} \right) + C$

$\displaystyle E(X^2) = \int cx^2e^{2x}$

$\displaystyle = {c \over 2} e^{2x}x^2 - c \int e^{2x}xdx$

$\displaystyle = {c \over 2} e^{2x}x^2 - {c \over 2} e^{2x}x + {c \over 2} int e^{2x} dx$

$\displaystyle = {c \over 2} e^{2x}x^2 - {c \over 2} e^{2x}x + {c \over 4} e^{2x} + C$

$\displaystyle = {ce^{2x} \over 4} \left( 2x^2 - 2x + 1 \right) + C$

Now we need to integrate from zero to one.

$\displaystyle E(X) = \int_0^1 cxe^{2x}$

$\displaystyle = c \left | e^{2x} \left( {x \over 2}-{1 \over 4} \right) \right | _0^1$

$\displaystyle = c \left[ e^2 \left( {1 \over 2} - {1 \over 4} \right) - e^0 \left( {0 \over 2} - {1 \over 4} \right) \right]$

$\displaystyle = c \left({{e^2 + 1} \over 4} \right)$

$\displaystyle E(X^2) = \int_0^1 cx^2e^{2x}$

$\displaystyle = c \left | {{e^{2x}} \over 4} ( 2x^2 - 2x + 1 ) \right |_0^1$

$\displaystyle = c \left [ {e^2 \over 4} (2 - 2 + 1) - {e^0 \over 4} \right ]$

$\displaystyle = c \left ( {{e^2 - 1} \over 4} \right )$

Plugging in we get:

$\displaystyle Var(X) = c \left ( {{e^2-1} \over 4} \right) - \left [ c \left ( {{e^2 + 1} \over 4} \right ) \right ]^2$

$\displaystyle = c \left({{e^2 - 1} \over 4} \right) - {{c^2(e^4 + 2e^2 + 1)} \over 16}$

$\displaystyle = c \left({{e^2 - 1} \over 4} \right) - {{c^2(e^2 + 1)^2} \over 16}$

Now plug in the value of c:

$\displaystyle = {2 \over e^2 - 1} \left( {e^2 - 1 \over 4} \right) - {\left( {2 \over e^2 - 1} \right)^2(e^2 + 1)^2 \over 16}$

$\displaystyle = {1 \over 4} - {e^2 + 1 \over 8}$

$\displaystyle = {1 - e^2 \over 8}$

Therefore:

$\displaystyle Var(X) = {1 - e^2 \over 8}$

Is any of this right?

NOTE: I had to translate from Open office's formula expressions to the latex ones used here. I don't think I made any error's but some of the equations may make no sense and then go back to what it would have been had it made sense. I hope that makes sense