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Integral of [x2] with upper limit as 1.7 and lower limit as 0.
where [ ] is the greatest integer function.
∫[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
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]) + 2.dx(in the limits root[2] 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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