Integration to get expected mean and variance

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I have the following question:

The times in minutes needed to collect the tolls from motorists crossing a toll bridge has the probability density function $$f(x) = 2 exp(−2x), 0 ≤ x < ∞$$ A motorist approaches the bridge and counts 50 vehicles waiting in a queue to pay the toll. Only one toll booth is in operation. Use the central limit theorem to find the approximate probability that a motorist will have to wait more than 25 minutes before reaching the front of the queue.

In my answer book it says it uses integration by parts to get $$E[X]=1/2$$ and $$Var[X]=1/4$$

Can anyone explain how and why they did this? Surely integrating gets $$-exp(-2x)$$? Thanks

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Hint: In this case,

$$E[X] = \int_0^\infty xf(x)dx$$

$$Var(X) = \left(\int_0^\infty x^2f(x)dx\right) - (E[X])^2 = E[X^2]-(E[X])^2$$

Do you understand why the book says, use integration by parts?