Variance of continuous Random Variable negative? I am having trouble finding the Variance for this question.

The proportion of salt X left in the salt shakers at the end of the day at a crowded restaurant has a probability density function given by
f(x) { 2x for 0 < x < 1

   0 other wise 


There are 80 salt shakers in the restaurant with each one having a capacity of 3 ounces of salt and they are all filled and used independently of each other. Find the expectation and standard deviation of T = the total amount of salt needed to fill the 80 salt shakers at the end of the day?

Here is what I am getting -
E(X) = an integral 0 to 1 (2x^2) , which then equals 2/3.
I then take the opposite of the 2/3 being that is how much left in the salt shaker and get 1/3 which is how much is gone.
I then multiply 1/3(3) and see that one ounce needs to be refilled in each salt shaker.
Then I multiply 1(80) to account for the 80 salt shakers.
So my E(T) = 80
Now to find the Variance wouldn't I just find the integral from 0 to 1 of f(x)(x^2)? and then subtract E(t)^2 from that to fond the Variance?
When I do that I am getting a negative number and I know that it isn't right.
Any help?
 A: This seems like a textbook problem rather than a restaurant management problem,
so I will give an outline leading to the answer. Your answer is a good start.
Now, I hope you will turn it into a coherent mathematical demonstration with reasons for each step: Use $E(a + bX)= a + bE(X),\,$ $Var(a + bX) = b^2Var(X),\,$ and for independent
random variables $X_1, X_2,$ you have $Var(X_1 + X_2) = Var(X_1)+Var(X_2).$ 
The random variable $X$ has $E(X) = 2/3,$ which you have already found.
You can also show that $Var(X) = 1/18.$
The salt used in the $i$th shaker is $Y_i = 3(1-X_i),$ so 
$$E(Y_i) = E(3 - 3X_i) = 3 - E(3X_i) = 3 - 2 = 1.$$
Now you need to find $Var(Y_i),$ based on $Var(X).$ 
Finally, $T = \sum_{i=1}^{80}Y_i.$ Use that to find $E(T)$ and $Var(T).$
Here is a simulation in R statistical software, based on the fact that
$X \sim \mathsf{Beta}(2,1).$ With a million iterations (days) one can
expect results for $E(T)$ and $SD(T)$ accurate to about three significant digits,
perhaps better.
set.seed(1018)
t = replicate(10^6, sum(3*(1-rbeta(80,2,1))))
mean(t);  sd(t);  var(t)
[1] 80.00621  # aprx E(T) = 80
[1] 6.325178
[1] 40.00787
mean(t > 90)
[1] 0.059156  # aprx P(T > 90)

Here is a histogram of the simulated distribution of $T$ (in Halloween orange).
Because of the Central Limit Theorem, the distribution of $T$ is very nearly normal (normal density curve shown in blue). Can you find a normal approximation
of $P(T > 90)?$

Note: Perhaps you'll want to look at Wikipedia's article on beta distributions.
A: The answer to the question above was found by finding F(X^2) which = 1/2
1/2 - (2/3)^2 = .05
.05(80) = 4.44 (80 being the amount of salt shaker summed up.)
4.44(3)^2 = 40 
root(40) = 6.3 
