We consider the function
-
which is
strictly increasing
in this interval. Hence, for a subinterval
,
the value
is the
minimum,
and
is the
maximum
of the function on this subinterval. Let
be a positive natural number. We partition the interval
into the
subintervals
,
,
of length
. The
step integral
for the corresponding
lower step function
is
-
![{\displaystyle {}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left(i{\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}i^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}-{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}-{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a3fed70d4badcc255803fe3671b386c7277cdc5)
(see
exercise
for the formula for the sum of the squares).
Since the
sequences
and
converge
to
, the
limit
for
of these step integrals equals
. The
step integral
for the corresponding
step function from above
is
-
![{\displaystyle {}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left((i+1){\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}(i+1)^{2}={\frac {1}{n^{3}}}\sum _{j=1}^{n}j^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}+{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/965b39d6f191970c2af7646e38cfeac2e5677b9a)
The limit of this sequence is again
. By
fact,
the
upper integral
and the
lower integral
coincide, hence the function is
Riemann-integrable,
and for the
definite integral
we get
-
![{\displaystyle {}\int _{0}^{1}t^{2}\,dt={\frac {1}{3}}\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d34e58f7b0b9bc82e30d2a423feb54b200e1d83)