Let $[t]$ denote the largest integer less than or equal to $t$. If $$\int_\limits0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) \mathrm{d} x=\mathrm{a}+\mathrm{b} \sqrt{2}-\sqrt{3}-\sqrt{5}+\mathrm{c} \sqrt{6}-\sqrt{7}$$, where $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{Z}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is equal to __________.
Answer (integer)
23
Solution
<p>$$\int_\limits0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) d x \quad=\int_\limits0^1 0 d x+\int_1^{\sqrt{2}} 1 d x+\int_\limits{\sqrt{2}}^{\sqrt{3}} 3 d x+$$</p>
<p>$$
\begin{aligned}
& \int_\limits{\sqrt{3}}^2 4 d x+\int_\limits2^{\sqrt{5}} 6 d x+\int_\limits{\sqrt{5}}^{\sqrt{6}} 7 d x+\int_\limits{\sqrt{6}}^{\sqrt{7}} 9 d x+\int_\limits{\sqrt{7}}^{\sqrt{8}} 10 d x+\int_\limits{\sqrt{8}}^3 12 d x \\
& =31-6 \sqrt{2}-\sqrt{3}-\sqrt{5}-\sqrt{7}-2 \sqrt{6} \\
& \Rightarrow a=31, b=-6, c=-2 \\
& \Rightarrow a+b+c=23
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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