Let \(f(x) = x^2\) ; \(0\leq x < 1\) and \(f(x) = \sqrt{x}\) ; \(1\leq x \leq 2\)

find \(\displaystyle\int_0^2 \ f(x)\ dx\)

answer is \(\dfrac{a\sqrt{b}-1}{3}\) find \(a+b\)

\( a \) and \(b\) are positive integers, and \(b\) is not divisible by the square of any prime.

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