[x] represents the greatest integer function. | vedclass

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(D) We evaluate the integral $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx$ by splitting the interval based on the values where $[x]$ changes: $\int_{\sqrt{3}

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$2$

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(D) We evaluate the integral $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx$ by splitting the interval based on the values where $[x]$ changes: $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = \int_{\sqrt{3}}^{2}[x] \, dx + \int_{2}^{3}[x] \, dx + \int_{3}^{4}[x] \, dx + \int_{4}^{\sqrt{18}}[x] \, dx$ Since $\sqrt{3} \approx 1.732$ and $\sqrt{18} = 3\sqrt{2} \approx 4.242$,we have: $\int_{\sqrt{3}}^{2} 1 \, dx + \int_{2}^{3} 2 \, dx + \int_{3}^{4} 3 \, dx + \int_{4}^{3\sqrt{2}} 4 \, dx$ $= (2 - \sqrt{3}) + 2(3 - 2) + 3(4 - 3) + 4(3\sqrt{2} - 4)$ $= 2 - \sqrt{3} + 2 + 3 + 12\sqrt{2} - 16$ $= (2 + 2 + 3 - 16) + 12\sqrt{2} - \sqrt{3}$ $= -9 + 12\sqrt{2} - \sqrt{3}$ Comparing this with $a + b\sqrt{2} + c\sqrt{3}$,we get $a = -9$,$b = 12$,and $c = -1$. Therefore,$a + b + c = -9 + 12 - 1 = 2$.

$[x]$ represents the greatest integer function. If $\int_{\sqrt{3}}^{\sqrt{18}}[x] \, dx = a + b\sqrt{2} + c\sqrt{3}$,then $a + b + c =$

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