How to show $$ [x^2] $$ is Riemann Integrable in [0,2] ?
I will explain how I proceeded my doubt is with greatest integer function part,
after splitting into 3 intervals
$$[0,1],[{1,\sqrt 2 }], [{\sqrt 2,\sqrt 3 }], [{\sqrt 3,2] }$$
In the interval $$[0,1]$$ Supremum of function is 1 and Infimum is 0.
In the interval $$[{1,\sqrt 2 }]$$ Supremum of function is 2 and Infimum is 1.
In the interval $$[{\sqrt 2,\sqrt 3 }]$$ Supremum of function is 3 and Infimum is 2.
Now U(P,f) = $$ 1(1-0)+2(\sqrt 2-1)+ 3(\sqrt 3-\sqrt 2)+4(2-\sqrt 3) $$
and L(P,f) = $$ 0(1-0)+1(\sqrt 2-1)+ 2(\sqrt 3-\sqrt 2)+3(2-\sqrt 3) $$
and they are different values. Please help.
Best Answer
Since no one is helping with the answer, I figured out an answer after referring through various questions in Riemann Integrability.
Since the function $ [x^2] $ is discontinuous at $1,\sqrt 2, \sqrt 3 $ in the interval [0,2], we partition the interval as follows:
x0($=0$), x1 , x2 .... xi, y0($=1$), y1 , y2.... yi,z0 (=$\sqrt 2$),z1 , z2.... zi, k0 (=$\sqrt 3$),k1 , k2.... ki,m0$(=2)$
Now U(P,f)= 0*$\Sigma$xi + 1*(y0-xi) +1*$\Sigma$yi + 2*(z0-yi) + 2*$\Sigma$zi +3*(k0-zi)+ 3*$\Sigma$ki + 4*(m0-ki)
=$0+1*(\sqrt2-1)+2*(\sqrt3-\sqrt2)+3*(2-\sqrt3)+$1*(y0-xi) +2*(z0-yi) +3*(k0-zi) + 4*(m0-ki)
Take
$\epsilon$ = 1*(y0-xi) +2*(z0-yi) +3*(k0-zi) + 4*(m0-ki)
Now L(P,f)= 0*$\Sigma$xi + 0*(y0-xi) +1*$\Sigma$yi + 1*(z0-yi) + 2*$\Sigma$zi +2*(k0-zi)+ 3*$\Sigma$ki + 3*(m0-ki)
=$0+1*(\sqrt2-1)+2*(\sqrt3-\sqrt2)+3*(2-\sqrt3)+$ 1*(z0-yi) + +2*(k0-zi)+ 3*(m0-ki)
Now taking U(P,f) - L(P,f) < $\epsilon$
Hence $ [x^2] $ is Riemann Integrable in [0,2]