Compute This Standard Deviation More Efficiently 
The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of $250$. In the event that the automobile is damaged, repair costs can be modeled by a uniform random variable on the interval $(0, 1500)$.  Determine the standard deviation of the insurance payment in the event that the automobile is damaged.

Let $X$ denote the damage, so it is the uniform random variable on $(0, 1500)$.  Then, we set $Y$ as the payment:
$$Y \ \ = \ \ \begin{cases}
\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \text{when } & X < 250 \\
\ X - 250 & 250 \le X \le 1500
\end{cases}$$
The computation for $\text{var } Y = E(Y^2) - (EY)^2$ is straight forward, but ugly:
\begin{align}
E(Y^2) \ \ & = \ \ \int_{250}^{1500} (x - 250)^2 \cdot \tfrac{1}{1500} \, dx \\
EY \ \ & = \ \ \int_{250}^{1500} (x - 250) \cdot \tfrac{1}{1500} \, dx
\end{align}
It will definitely take  me (and many others ...) several attempts and a lot of time to get to the correct answer.  This was supposed to be an exam question, and we need a better way to do this.  Any suggestion?
Computation:
Let me be more specific.  The integration is not difficult.  It is at the end when you have so many substitutions that I screw up:
\begin{align}
E(Y^2) \ \ & = \ \ \int_{250}^{ 1500} (x-250)^2 \cdot \tfrac{1}{1500} \, dx \\
& = \ \ \int_{250}^{ 1500} \tfrac{x^2 - 500x + 250^2}{1500} \, dx \\
& = \ \ \frac{1}{1500} \left[ \, \frac{x^3}{3} - 250x^2 + 250^2x \, \right]_{\, 250}^{\, 1500} \ \ = \ \ ???
\end{align}
The change of variable helps.  I think too often I just dive straight into the question and don't see that I'm doing things stupidly.  You're right; until I clean up my calculations, I will never get far.
 A: Define the Bernoulli random variable $$L = \begin{cases} 1, & X \ge 250, \\ 0, & X < 250, \end{cases}$$ with $\Pr[L = 1] = \frac{1500-250}{1500} = \frac{5}{6}.$ Then we can write $Y = L(X - 250)$, and $$\begin{align*} \operatorname{Var}[Y] &= \operatorname{Var}[\operatorname{E}[Y \mid L]] + \operatorname{E}[\operatorname{Var}[Y \mid L]] \\
&= \operatorname{E}[Y \mid L]^2 \Pr[L = 1] (1 - \Pr[L = 1]) + \operatorname{Var}[Y \mid L] \Pr[L = 1],
\end{align*}$$ by the law of total variance.  The conditional expectation is $$\operatorname{E}[Y \mid L] = \begin{cases} 625, & L = 1 \\ 0, & L = 0, \end{cases}$$ because the midpoint of $[250, 1500]$ is $\operatorname{E}[X \mid L = 1] = 825$, then subtract off $250$ to get $625$.  The conditional variance is clearly $$\operatorname{Var}[Y \mid L] = \begin{cases} \operatorname{Var}[X \mid L] = \frac{390625}{3}, & L = 1 \\ 0, & L = 0, \end{cases}$$ since the variance is not affected by a location shift, and the variance of a $\operatorname{Uniform}(a,b)$ random variable is $(b-a)^2/12$.  Then we have $$\operatorname{Var}[Y] = 625^2 \cdot \frac{5}{6} \cdot \frac{1}{6} + \frac{390625}{3}\cdot \frac{5}{6} = \frac{1953125}{12}.$$
Whether this approach is easier than the direct integration is another question entirely.

It is worth noting that this approach works in a more general context:  suppose we have a ground-up loss random variable $X$ and the insurer's amount paid on the claim $Y = (X - d)_+ = \max\{X-d, 0\}$ subject to an ordinary deductible of $d$.  Then we have $$\operatorname{Var}[Y] = \left((\operatorname{E}[X \mid X > d] - d)^2 \Pr[X \le d] + \operatorname{Var}[X \mid X > d]\right)\Pr[X > d].$$  So if we can compute the moments of the conditional loss random variable in excess of the deductible, we can get the desired variance from the above.  For example, if $X \sim \operatorname{Exponential}(\mu)$, then the memorylessness property gives us $$\operatorname{E}[X \mid X > d] = d + \operatorname{E}[X] = \mu + d, \quad \operatorname{Var}[X \mid X > d] = \operatorname{Var}[X] = \mu^2,$$ and the formula simplifies to $$\operatorname{Var}[Y] = \mu^2 (2 - e^{-d/\mu}) e^{-d/\mu}.$$  Of course, this computation is simple because of memorylessness.

In regard to the original integration, which I still think is the simplest approach, we should note $$\int_{x=250}^{1500} \frac{(x-250)^k}{1500} \, dx = \int_{u=0}^{1250} \frac{u^k}{1500} \, du = \left[\frac{u^{k+1}}{1500(k+1)}\right]_{u=0}^{1250} = \frac{1250^{k+1}}{1500(k+1)},$$ for $k > 0$, using the simple substitution $u = x - 250$, $du = dx$.  Thus we simply have $$\operatorname{Var}[Y] = \frac{1250^3}{1500(3)} - \left(\frac{1250^2}{1500(2)}\right)^2.$$
