Math Definite integral Partition of an interval

Partition of an interval



The more partitions you set with the upper or lower sum, the more accurate the result becomes.

If you now let the number of partitions $n$ go against infinity, you will get the surface you are looking for. For this one forms the limit of the upper and lower sums for $n\to\infty$.



That's just the derivation of the definite integral.


$f(x)=x^2$ in the interval $[0; 1]$

  1. Calculate lower sum

    The width is $\frac1n$ and the height is $x^2$.
    $s=\frac1n\cdot(0^2+(\frac1n)^2+(\frac2n)^2+...$ $+(\frac{n-1}{n})^2)$

    Remove brackets by squaring the numerator and denominator.
    $s=\frac1n\cdot(0^2+\frac{1^2}{n^2}+\frac{2^2}{n^2}+...$ $+\frac{(n-1)^2}{n^2})$

    Each denominator contains $n^2$, so you can expand brackets

    There is a trick to rewrite the parenthesis: All squares up to $m$ ($1^2+2^2+...+m^2$) make up:
    Applied to our example, this means:
    $s=\frac{1}{n^3}\cdot\frac16\cdot\color{red}{(n-1)}\cdot$ $(\color{red}{n-1}+1)\cdot(2\color{red}{(n-1)}+1)$
    $s=\frac16\cdot\frac{1}{n^3}\cdot(n-1)\cdot n\cdot(2n-1)$
  2. Calculate upper sum

    The same again for the upper sum.
    $S=\frac1n\cdot((\frac1n)^2+(\frac2n)^2+...$ $+(\frac{n-1}{n})^2+1^2)$
    $S=\frac{1}{n^3}\cdot(1^2+2^2+...$ $+(n-1)^2+n^2)$
    $S=\frac16\cdot\frac{1}{n^3}\cdot n\cdot (n+1)\cdot(2n+1)$
  3. Find limit

    Now the limit must be found.
    $s=\frac16\cdot\frac{1}{n^3}\cdot(n-1)\cdot n\cdot(2n-1)$ $=\frac16\cdot\frac{n-1}{n}\cdot \frac{n}{n}\cdot \frac{2n-1}{n}$

    $\lim\limits_{n \to \infty}{s}=$ $\lim\limits_{n \to \infty}{(\frac16\cdot\frac{n-1}{n}\cdot \frac{n}{n}\cdot \frac{2n-1}{n})}$ $=\frac16\cdot1\cdot1\cdot2=\frac13$

    $S=\frac16\cdot\frac{1}{n^3}\cdot n\cdot (n+1)\cdot(2n+1)$ $=\frac16\cdot\frac{n}{n}\cdot\frac{n+1}{n}\cdot\frac{2n+1}{n}$

    $\lim\limits_{n \to \infty}{S}=$ $\lim\limits_{n \to \infty}{(\frac16\cdot\frac{n}{n}\cdot\frac{n+1}{n}\cdot\frac{2n+1}{n})}$ $=\frac16\cdot1\cdot1\cdot2=\frac13$
  4. Determine area

    For the area applies:
    $s \le A \le S$

    $\frac13 \le A \le \frac13$

    The area you are looking for is thus: $A=\frac13$