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# Integral of [x2] with upper limit as 1.7 and lower limit as 0.where [ ] is the greatest integer function.

## 2 Answers

9 years ago

∫[x2]dx (x=0 to 1.7)=

∫[x2]dx (x=1 to 1.7)=

∫1*dx (x=1 to sqrt(2)) +∫2*dx (x=sqrt(2)to1.7)

=√2 -1+3.4-2√2=2.4-√2

9 years ago

Hi Gowri,

Split the interval from 0 to 1.7 as

(i) 0 to 1, where x^2 lies between 0 to 1, and hence [x^2]=0

(ii) 1 to √2, where x^2 lies between 1 to 2, so [x^2]=1

and (iii) √2 to 1.7, where x^2 lies between 2 to 3 and hence [x^2] = 2.

So intgral becomes

0.dx(in the limits 0 to 1) + 1.dx(in the limits 1 to root) + 2.dx(in the limits root to 1.7).

Hence the answer = 1*(√2-1)+2*(1.7-√2) = 2.4-√2

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).

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