JEE Main 2024 — 6 April, Shift 2 — Question 24

$\int_{0}^{3}\left[x^{2}\right] d x+\int_{0}^{3}\left[\frac{x^{2}}{2}\right] d x$

$=\int_{0}^{1} 0 d x+\int_{1}^{12} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x$

$+\int_{\sqrt{3}}^{2} 3 \mathrm{dx}+\int_{2}^{\sqrt{5}} 4 \mathrm{dx}+\int_{\sqrt{5}}^{\sqrt{6}} 5 \mathrm{dx}$ $+\int_{\sqrt{6}}^{\sqrt{7}} 6 d x+\int_{\sqrt{7}}^{\sqrt{8}} 7 d x+\int_{\sqrt{8}}^{3} 8 d x$ $+\int_{0}^{\sqrt{2}} 0 d x+\int_{\sqrt{2}}^{2} 1 d x$

$+\int_{2}^{\sqrt{6}} 2 d x+\int_{\sqrt{6}}^{\sqrt{8}} 3 d x+\int_{\sqrt{8}}^{3} 4 d x=31-6 \sqrt{2}-\sqrt{3}-\sqrt{5}$

$-2 \sqrt{6}-\sqrt{7}$

$a=31 \quad b=-6 \quad c=-2$

$a+b+c=31-6-2=23$