Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) \mathrm{d} x=\mathrm{a}+\mathrm{b} \sqrt{2}-\sqrt{3}-\sqrt{5}+\mathrm{c} \sqrt{6}-\sqrt{7}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{Z}$, then…

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Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) \mathrm{d} x=\mathrm{a}+\mathrm{b} \sqrt{2}-\sqrt{3}-\sqrt{5}+\mathrm{c} \sqrt{6}-\sqrt{7}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{Z}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to_______

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^3\left[x^2ight] d x+\int_0^3\left[\frac{x^2}{2}ight] d x \ & =\int_0^1 0 d x+\int_1^{12} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x\end{aligned}$ $\begin{aligned} & +\int_{\sqrt{3}}^2 3 \mathrm{dx}+\int_2^{\sqrt{5}} 4 \mathrm{dx}+\int_{\sqrt{5}}^{\sqrt{6}} 5 \mathrm{dx} \ & +\int_{\sqrt{6}}^{\sqrt{7}} 6 \mathrm{dx}+\int_{\sqrt{7}}^{\sqrt{8}} 7 \mathrm{dx}+\int_{\sqrt{8}}^3 8 \mathrm{dx} \ & +\int_0^{\sqrt{2}} 0 \mathrm{dx}+\int_{\sqrt{2}}^2 1 \mathrm{dx} \ & +\int_2^{\sqrt{6}} 2 \mathrm{dx}+\int_{\sqrt{6}}^{\sqrt{8}} 3 \mathrm{dx}+\int_{\sqrt{8}}^3 4 \mathrm{dx}=31-6 \sqrt{2}-\sqrt{3}-\sqrt{5} \ & -2 \sqrt{6}-\sqrt{7} \ & \mathrm{a}=31 \quad \mathrm{~b}=-6 \quad \mathrm{c}=-2 \ & \mathrm{a}+\mathrm{b}+\mathrm{c}=31-6-2=23\end{aligned}$

Let $[t]$ denote the largest integer less than or equal to $t$. If…

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