Why the variance of x is 1/4 for a uniform distribution range in 0 to 1?

Question

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Why not Var$(x) = \frac{1}{12}$?

Answer 1

$$E(X)=\int_0^1 x dx = \frac12$$ $$E(X^2)=\int_0^1 x^2 dx = \frac13$$ $$Var(X)=E(X^2)-(E(X))^2=\frac13-(\frac12)^2=\frac{1}{12}$$