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Integral Calculus> definite...

question mark

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

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

aryan arora , 14 Years ago

Grade 12th Pass

1 Answers

Jitender Singh

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

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