Riemann integral/x^2 of 0 to 1/Explicit via step integrals/Example

We consider the function

$f\colon [0,1]\longrightarrow \mathbb {R} ,t\longmapsto t^{2},$

which is strictly increasing in this interval. Hence, for a subinterval ${}[a,b]\subseteq [0,1]$, the value ${}f(a)$ is the minimum, and ${}f(b)$ is the maximum of the function on this subinterval. Let ${}n$ be a positive natural number. We partition the interval ${}[0,1]$ into the ${}n$ subintervals ${}\left[i{\frac {1}{n}},(i+1){\frac {1}{n}}\right]$, ${}i=0,\ldots ,n-1$, of length ${}{\frac {1}{n}}$. The step integral for the corresponding lower step function is

${}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left(i{\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}i^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}-{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}-{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,$

(see exercise for the formula for the sum of the squares). Since the sequences ${}{\left(1/2n\right)}_{n\in \mathbb {N} }$ and ${}{\left(1/6n^{2}\right)}_{n\in \mathbb {N} }$ converge to ${}0$, the limit for ${}n\rightarrow \infty$ of these step integrals equals ${}{\frac {1}{3}}$. The step integral for the corresponding step function from above is

${}\sum _{i=0}^{n-1}{\frac {1}{n}}{\left((i+1){\frac {1}{n}}\right)}^{2}={\frac {1}{n^{3}}}\sum _{i=0}^{n-1}(i+1)^{2}={\frac {1}{n^{3}}}\sum _{j=1}^{n}j^{2}={\frac {1}{n^{3}}}{\left({\frac {1}{3}}n^{3}+{\frac {1}{2}}n^{2}+{\frac {1}{6}}n\right)}={\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\,.$

The limit of this sequence is again ${}{\frac {1}{3}}$. By fact, the upper integral and the lower integral coincide, hence the function is Riemann-integrable, and for the definite integral we get

${}\int _{0}^{1}t^{2}\,dt={\frac {1}{3}}\,.$