JEE Main 2024 (Online) 6th April Evening ShiftQuestion 1
Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_{0}^{3}\left([x^{2}]+\left[\frac{x^{2}}{2}\right]\right)dx=a+b\sqrt{2}-\sqrt{3}-\sqrt{5}+c\sqrt{6}-\sqrt{7}$, where $a,b,c\in\mathbf{Z}$, then $a+b+c$ is equal to $\_\_\_\_\_\_$ .
Definite Integration
Answer$23$
Solution
Concept. $[x^2]$ counts the integers $j\ge1$ with $x^2\ge j$; each such $j$ adds $3-\sqrt j$ to the integral on $[0,3]$. Formulas used. (1) $\int_0^3[x^2]\,dx=\sum_{j=1}^{8}\big(3-\sqrt j\big)$ and $\int_0^3\big[\frac{x^2}2\big]\,dx=\sum_{j=1}^{4}\big(3-\sqrt{2j}\big)$Step 1 - First part. By (1), $$24-\big(1+\sqrt2+\sqrt3+2+\sqrt5+\sqrt6+\sqrt7+2\sqrt2\big)=21-3\sqrt2-\sqrt3-\sqrt5-\sqrt6-\sqrt7.$$ Step 2 - Second part. By (1), $$12-\big(\sqrt2+2+\sqrt6+2\sqrt2\big)=10-3\sqrt2-\sqrt6.$$ Step 3 - Add. $$31-6\sqrt2-\sqrt3-\sqrt5-2\sqrt6-\sqrt7.$$ So $a=31$, $b=-6$, $c=-2$. Step 4 - Value. $a+b+c=23$. Answer: 23