# A Case of Pre-FTC Definite Integral

Fig. 1

By recalling the general formula for power summation derived in “Little Bird and a Recursive Generator“, namely

$\sum\limits_{i=1}^{n}{i^{p}} = \frac{n(n+1)^{p}- \sum\limits_{j=2}^{p}\binom{p}{j}\sum\limits_{i=1}^{n}i^{p-j+1}}{p+1}\quad\quad\quad(1)$

We can obtain the result

$\int\limits_{a}^{b}x^{p} dx = {{b^{p+1}-a^{p+1} }\over {p+1}}$

where $b\ge a\ge 0$ and $p \in Z_{0}^{+}$, without the Fundamental Theorem of Calculus.

To this end, we divide the interval $[0, b]$ into $n$ sub-intervals of equal length as shown in Fig. 1. Let $s_{n}$ denotes the sum of the areas of the rectangles. The value $s_{n}$ tends to as $n$ increases, is the definite integral of $x^{p}$ from $x=0$ to $x=b$, i.e.,

$\int\limits_{0}^{b} x^{p} dx = \lim\limits_{n \to \infty} s_{n}$.

Since

$s_{n} = \sum\limits_{i=1}^{n-1}\frac{b}{n}(i\frac{b}{n})^p= \frac{b^{p+1}}{n^{p+1}}\sum\limits_{i=1}^{n-1}i^p$

It follows from (1) that

$s_{n} = \frac{b^{p+1}}{n^{p+1}}\cdot\frac{(n-1)n^{p}- \sum\limits_{j=2}^{p}\binom{p}{j}\sum\limits_{i=1}^{n-1}i^{p-j+1} }{p+1}$

$= \frac{b^{p+1}}{p+1}(\frac{(n-1)n^{p}}{n^{p+1}}-\frac{\sum\limits_{j=2}^{p}\binom{p}{j}\sum\limits_{i=1}^{n-1}i^{p-j+1}}{n^{p+1}})$

$= \frac{b^{p+1}}{p+1}((1-\frac{1}{n})-\sum\limits_{j=2}^{p}\binom{p}{j}\frac{\sum\limits_{i=1}^{n-1}i^{p-j+1}}{n^{p+1}})$.

For  $0 \leq p < 2, s_{n}$ reduces to

$s_{n} = \frac{b^{p+1}(1-\frac{1}{n})}{p+1}$,

and the result immediately follows:

$\int\limits_{0}^{b}x^{p} dx = \lim\limits_{n \to \infty} s_{n}=\lim \limits_{n \to \infty}\frac{b^{p+1}(1-\frac{1}{n})}{p+1}=\frac{b^{p+1}}{p+1}$.

For $p \ge 2$, we let

$t_{n} = \frac {\sum\limits_{i=1}^{n-1}i^{p-j+1}}{n^{p+1}}$.

Clearly,

$0 < t_{n} < \frac{(n-1) n^{p-j+1}}{n^{p+1}}=\frac{n-1}{n^{j}}$.

Since $j \ge 2$,

$\lim\limits_{n \to \infty}\frac{n-1}{n^j}=0$.

i.e.,

$\lim\limits_{n \to \infty} t_{n} = 0$.

Therefore,

$\int\limits_{0}^{b}x^{p} dx = \lim\limits_{n \to \infty} s_{n}=\lim\limits_{n \to \infty} \frac{b^{p+1}}{p+1}((1-\frac{1}{n})-\sum\limits_{j=2}^{p}\binom{p}{j}t_{n})$

$=\frac{b^{p+1}}{p+1}(\lim\limits_{n \to \infty}{(1-\frac{1}{n})}-\sum\limits_{j=2}^{p}\binom{p}{j}\lim\limits_{n \to \infty}{t_{n}})=\frac{b^{p+1}}{p+1}$.

Applying this result to the area from $0$ to $a$ we have

$\int\limits_{0}^{a}x^{p} dx = \frac {a^{p+1}}{p+1}$,

and by subtraction of the areas,

$\int\limits_{a}^{b}x^{p} dx = \int\limits_{0}^{b}x^{p} dx -\int\limits_{0}^{a}x^{p} dx = \frac {b^{p+1}-a^{p+1}}{p+1}$.

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