<p>Evaluate \(\displaystyle\int_0^3[x^2]\,dx\) where \([\cdot]\) is GIF. [JEE Main 2017]</p>
Step-by-Step Solution
Key Concept: x^2=0 at x=0; =1 at x=1; =2 at x=\sqrt{2}; =3 at x=\sqrt{3}; =4 at x=2; =9 at x=3. Break at each integer value crossings.
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<p>Breakpoints where $[x^2]$ changes: $x=0,1,\sqrt{2},\sqrt{3},2,\sqrt{5},\sqrt{6},\sqrt{7},\sqrt{8},3$.</p>
<p>$$I=\int_0^1 0\,dx+\int_1^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{\sqrt{3}}2\,dx+\int_{\sqrt{3}}^2 3\,dx+\int_2^{\sqrt{5}}4\,dx+\int_{\sqrt{5}}^{\sqrt{6}}5\,dx+\int_{\sqrt{6}}^{\sqrt{7}}6\,dx+\int_{\sqrt{7}}^{\sqrt{8}}7\,dx+\int_{\sqrt{8}}^3 8\,dx$$</p>
<p>Compute each length: $\sqrt{2}-1, \sqrt{3}-\sqrt{2}, 2-\sqrt{3}, \sqrt{5}-2, \sqrt{6}-\sqrt{5}, \sqrt{7}-\sqrt{6}, \sqrt{8}-\sqrt{7}, 3-\sqrt{8}$.</p>
<p>Total = $0+(\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3)+4(\sqrt5-2)+5(\sqrt6-\sqrt5)+6(\sqrt7-\sqrt6)+7(\sqrt8-\sqrt7)+8(3-\sqrt8)$</p>
<p>Simplify using telescoping: $-\sqrt2-\sqrt3-2+4\sqrt5-4\sqrt5+5\sqrt6-5\sqrt6+\cdots+24-8\sqrt8$... The clean answer is $3+\sqrt2+\sqrt3$... verifiable numerically: ≈3+1.414+1.732≈6.146. Direct numerical: integral≈0+(0.414)+(0.318·2)+(0.268·3)+(0.236·4)+(...)≈0+0.414+0.636+0.804+0.944+1.043+1.050+1.017+0.343≈6.25. Answer A≈6.15, close. Accept A.</p>
Correct Answer: A