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[.] represents a greatest integer function.If $\int_{\sqrt{3}}^{\sqrt{18}}[x] d x=a+b \sqrt{2}+c \sqrt{3}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}=$
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$2$
$\int_{\sqrt{3}}^{\sqrt{18}}[x] d x=$ $\int_{\sqrt{3}}^{\sqrt{4}}[x] d x+\int_2^3[x] d x$ $+\int_3^4[x] d x+\int_4^{\sqrt{18}}[x] d x$
$=\int_{\sqrt{3}}^{\sqrt{4}} 1 d x+\int_2^3 2 d x+$ $\int_3^4 3 d x+\int_4^{\sqrt{18}} 4 d x$
$=(2-\sqrt{3})+(2)+(3)+4(\sqrt{18}-4)$
$=2-\sqrt{3}+2+3+4 \sqrt{18}-16$
$=-9-\sqrt{3}+12 \sqrt{2}$
$\begin{aligned} & =12 \sqrt{2}-\sqrt{3}-9 \\ & =b \sqrt{2}+c \sqrt{3}+9\end{aligned}$
Where $a=-9, b=12, c=-1$
$a+b+c=2$
$=\int_{\sqrt{3}}^{\sqrt{4}} 1 d x+\int_2^3 2 d x+$ $\int_3^4 3 d x+\int_4^{\sqrt{18}} 4 d x$
$=(2-\sqrt{3})+(2)+(3)+4(\sqrt{18}-4)$
$=2-\sqrt{3}+2+3+4 \sqrt{18}-16$
$=-9-\sqrt{3}+12 \sqrt{2}$
$\begin{aligned} & =12 \sqrt{2}-\sqrt{3}-9 \\ & =b \sqrt{2}+c \sqrt{3}+9\end{aligned}$
Where $a=-9, b=12, c=-1$
$a+b+c=2$
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