A Case of Pre-FTC Definite Integral

Screen Shot 2017-08-29 at 9.23.39 PM.png

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}.


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}}.


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.


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


\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}.

2 thoughts on “A Case of Pre-FTC Definite Integral

  1. Pingback: Introducing log | Vroom

  2. Pingback: Evaluate a Definite Integral without FTC | Vroom

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