2. Incorrect. The answer is true, not false

Applying the definition of expectation for a continuous random variable:

E(Y) = ∫0 yf(y)dy

E(Y) = ∫0 y(2e-2y) dy

E(Y) = 2 [ ye-2y/(-2)-2 e-2y/4] 0

E(Y) = 1/2(equivalent to 500 barrels of oil)

Regarding E(Y2), we have that:

E(Y2) = ∫0 y2f(y)dy

E(Y2) = 2∫0 y2e-2ydy

E(Y2) = 2 [y2e-2y/(-2)-2y e-2y/(-8)]0

E(Y2) = 1/2