It is similar to:
Question:) How much is $500 \times 10^3$?
Answer1:) $500 \times 10^3 = 500 \times (9+1)^3 = 5(9^3 + 3\times 9^2 + 3\times 9 + 1) \dots = $
Answer2:) It is $500 \times 1000 = 1000 + 1000 + \dots = $
The point is the integral can be evaluated easily as
$\int_{0}^{1} (1-x)^3 \text{ d}x = -(1-x)^4/4 |_0^1 = -(1-1)^4/4 + (1-0)^4/4 = 1/4$
The joke is that most students did it the hard way by expanding $(1-x)^3$, while one student, computed the anti-derivative easily (giving some hope to the teacher), but then made it even harder by trying to expand that instead of just plugging in the values...