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(A) We need to evaluate the integral $I = \int_1^2 [x^2] dx$. Let $t = x^2$, then $dt = 2x dx$, which implies $dx = \frac{dt}{2\sqrt{t}}$. When $x=1$,

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$5-\sqrt{2}-\sqrt{3}$

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(A) We need to evaluate the integral $I = \int_1^2 [x^2] dx$. Let $t = x^2$, then $dt = 2x dx$, which implies $dx = \frac{dt}{2\sqrt{t}}$. When $x=1$, $t=1$. When $x=2$, $t=4$. So, $I = \int_1^4 [t] \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int_1^4 \frac{[t]}{\sqrt{t}} dt$. We split the integral based on the intervals where $[t]$ is constant: $I = \frac{1}{2} \left( \int_1^2 \frac{1}{\sqrt{t}} dt + \int_2^3 \frac{2}{\sqrt{t}} dt + \int_3^4 \frac{3}{\sqrt{t}} dt \right)$. Evaluating each integral: $\int_1^2 t^{-1/2} dt = [2\sqrt{t}]_1^2 = 2(\sqrt{2}-1)$. $\int_2^3 2t^{-1/2} dt = [4\sqrt{t}]_2^3 = 4(\sqrt{3}-\sqrt{2})$. $\int_3^4 3t^{-1/2} dt = [6\sqrt{t}]_3^4 = 6(2-\sqrt{3}) = 12-6\sqrt{3}$. Summing these up: $I = \frac{1}{2} [2\sqrt{2}-2 + 4\sqrt{3}-4\sqrt{2} + 12-6\sqrt{3}] = \frac{1}{2} [10 - 2\sqrt{2} - 2\sqrt{3}] = 5 - \sqrt{2} - \sqrt{3}$.

If $[\cdot]$ denotes the greatest integer function, then $\int_1^2 [x^2] dx =$

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