A Case of Pre-FTC Definite Integral

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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}.

Pumpkin Pi


Fig. 1

Among many images of carved pumpkin, I like the one above (see Fig. 1) the most. It shows Leibniz’s formula for calculating the value of \pi. Namely,


To derive this formula, we begin with finding the derivative of \arctan{x}:

Let y = \arctan{x}, we have x=tan(y), and

\frac{d}{dx}x=\frac{d}{dx}\tan{y}=\frac{d}{dy}\tan{y}\frac{dy}{dx}=\sec^{2}{y}\frac{dy}{dx} = (1+tan^{2}{y})\frac{dy}{dx}

= (1+x^{2})\frac{dy}{dx}.

Since \frac{d}{dx}x=1,


It follows that by (1) and the Fundamental Theorem of Calculus,

\int\limits_{0}^{1}\frac{1}{1+x^2}dx = \arctan{x}\bigg|_{0}^{1}=\frac{\pi}{4}


\frac{\pi}{4} = \int\limits_{0}^{1}\frac{1}{1+x^2} dx\quad\quad\quad\quad\quad\quad(2)

From carrying out polynomial long division, we observe

\frac{1}{1+x^2} = 1 + \frac{-x^2}{1+x^2},

\frac{1}{1+x^2} = 1 - x^2 + \frac{x^4}{1+x^2},

\frac{1}{1+x^2} = 1 - x^2 + x^4  + \frac{-x^6}{1+x^2},

\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +  \frac{x^8}{1+x^2}.

It seems that

\frac{1}{1+x^2} = \sum\limits_{k=1}^{n}{(-1)^{k+1}x^{2k-2}} + \frac{(-1)^{n} x^{2n}}{1+x^2}\quad\quad\quad\quad\quad\quad(3)

Assuming (3) is true, we integrate it with respect to x from 0 to 1,

 \int\limits_{0}^{1}\frac{1}{1+x^2}dx=\int\limits_{0}^{1}\sum\limits_{k=1}^{n}(-1)^{k+1}x^{2k-2} dx + \int\limits_{0}^{1}\frac{(-1)^{n} x^{2n}}{1+x^2}dx

= \sum\limits_{k=1}^{n}(-1)^{k+1}\int\limits_{0}^{1}x^{2k-2}dx +(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx

= \sum\limits_{k=1}^{n}(-1)^{k+1}\frac{x^{2k-1}}{2k-1}\bigg|_{0}^{1}+(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx .

As a result of integration,  (2) becomes

\frac{\pi}{4} = \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1} + (-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx,


\frac{\pi}{4} - \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1} =(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx.


|\frac{\pi}{4} -  \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}|=|(-1)^{n}\int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx | = \int\limits_{0}^{1}\frac{x^{2n}}{1+x^2}dx < \int\limits_{0}^{1}x^{2n}dx

=\frac{x^{2n+1}}{2n+1}\bigg|_{0}^{1}= \frac{1}{2n+1}.

Moreover, \forall \epsilon > 0, we obtain n > \frac{1}{2}(\frac{1}{\epsilon}-1) through solving \frac{1}{2n+1} < \epsilon. It means that \forall \epsilon > 0, \exists n^*=\frac{1}{2}(\frac{1}{\epsilon}-1) such that  for all n > n^*, |\frac{\pi}{4} -  \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}|<\epsilon, i.e.,

\lim\limits_{n\to\infty}{ \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}}=\frac{\pi}{4}.


\pi = 4 \sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}\quad\quad\quad(4)

The numerical value of \pi is therefore approximated according to (4) by the partial sum

 4 \sum\limits_{k=1}^{n}\frac{(-1)^{k+1}}{2k-1}=4(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dots+(-1)^{n+1}\frac{1}{2n-1})\quad\quad\quad\quad(5)

Its value converges to \pi as n increases.

However, (5) is by no means a practical way of finding the value of \pi, since its convergence is so slow that many terms must be summed up before a reasonably accurate result emerges (see Fig. 2)

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Fig. 2

I doubt Leibniz has ever used his own formula to obtain the value of \pi !

Let me leave you with an exercise: Prove (3)