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integral of [f(x)] between the limits 2 to 6 , where f(x)=[(x-1)^2] /2[x]+1 , where [.] represents greatest integer function .

integral of  [f(x)] between the limits 2 to 6 , where


f(x)=[(x-1)^2] /2[x]+1 , where [.] represents greatest integer function .

Grade:12th Pass

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
I = \int_{2}^{6}[\frac{[(x-1)^{2}]}{2[x]+1}]dx
2\leq x< 3,[\frac{[(x-1)^{2}]}{2[x]+1}]=0
3\leq x< \sqrt{7}+1,[\frac{[(x-1)^{2}]}{2[x]+1}]=0
\sqrt{7}+1\leq x< 4,[\frac{[(x-1)^{2}]}{2[x]+1}]=1
4\leq x< 5,[\frac{[(x-1)^{2}]}{2[x]+1}]=1
5\leq x< \sqrt{22}+1,[\frac{[(x-1)^{2}]}{2[x]+1}]=1
\sqrt{22}+1\leq x< 6, [\frac{[(x-1)^{2}]}{2[x]+1}]=2
I = \int_{2}^{\sqrt{7}+1}[f(x)]dx+\int_{\sqrt{7}+1}^{\sqrt{22}+1}[f(x)]dx+\int_{\sqrt{22}+1}^{6}[f(x)]dx
I = \int_{2}^{\sqrt{7}+1}0.dx+\int_{\sqrt{7}+1}^{\sqrt{22}+1}1.dx+\int_{\sqrt{22}+1}^{6}2.dx
I = (x)_{\sqrt{7}+1}^{\sqrt{22}+1}+(2x)_{\sqrt{22}+1}^{6}
I = \sqrt{22}-\sqrt{7}+10-2.\sqrt{22}
I = 10-\sqrt{22}-\sqrt{7}
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty

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