# Let’s Evaluate a Definite Integral Without FTC!

For the readers of “A Case of Pre-FTC Definite Integral“, this post illustrates yet another way of evaluating the following definite integral without FTC:

$\int\limits_{a}^{b}x^k\;dx\quad\quad(0\le a \le b, k \in N^+)$

Let us divide $[0, b]$ into $n$ unequal subintervals (see Fig. 1) such that

$x_n = r^{n-1}b,\; ...,\; x_3=r^2b,\; x_2=rb,\; x_1=b$,

i.e.,

$x_i = r^{i-1}b, i = 1, 2, ..., n\quad\quad\quad(1)$

in general and

$r = 1-\frac{1}{n}\quad\quad\quad(2)$

so that the subintervals become finer as $n$ increases.

Fig. 1

We have, for the sum of the rectangles above the curve $y=x^k$,

$S_r=\sum\limits_{i=1}^{n}(x^{i}-x^{i+1})\cdot(x_i)^k$

$= \sum\limits_{i=1}^{n}(r^{i-1}b-r^i b)\cdot(r^{i-1}b)^k$

$=b^{k+1}\sum\limits_{i=1}^{n}(r^{i-1}-r^i)\cdot(r^{i-1})^k$

$=b^{k+1}((1-r) + (r-r^2)r^k + (r^2-r^3)r^{2k} + (r^3-r^4)r^{3k} + ... + (r^{n-1}-r^n)r^{(n-1)k})$

$=b^{k+1}((1-r) + (1-r)r^{k+1} + (1-r)r^{2k+2} + (1-r)r^{3k+3} + ... + (1-r)r^{(n-1)k+ (n-1)})$

$=b^{k+1}(1-r)(1+r^{k+1} + r^{2(k+1)} + r^{3(k+1)} + ... +r^{(n-1)(k+1)})$

$=b^{k+1}(1-r)(1+(r^{k+1})+(r^{k+1})^2 + (r^{k+1})^3+... + (r^{k+1})^{n-1})$

$\overset{s=r^{k+1}}{=}b^{k+1}(1-r)(1+s+s^2+s^3+ ... + s^{n-1})$

$= b^{k+1}(1-r)\frac{1-s^n}{1-s}\quad\quad\quad\quad($see “Beer Theorems and Their Proofs$)$

$=b^{k+1}(1-r)\frac{1-(r^{k+1})^n}{1-r^{k+1}}$

$=b^{k+1}(1-r)\frac{1-(r^{k+1})^n}{(1-r)(1+r+r^2+r^3+...+ r^k)}$

$=b^{k+1}\frac{1-(r^{k+1})^n}{1+r+r^2+r^3+ ... +r^k}.\quad\quad\quad(3)$

Since $0 as $n\rightarrow \infty, r \rightarrow 1, (r^{k+1})^n\rightarrow 0$. Consequently,

$(3) \rightarrow b^{k+1}\frac{1- 0}{1 + \underbrace{1^1 + 1^2+ 1^3 + ... + 1^k}_{k\;1's}}=\frac{b^{k+1}}{1+k}$.

And so,

$\int\limits_{0}^{b}x^k \;dx= \frac{b^{k+1}}{k+1}.$

It follows that

$\int\limits_{a}^{b}x^k\;dx = \int\limits_{0}^{b}x^k\;dx - \int\limits_{0}^{a}x^k\;dx =\frac{b^{k+1}}{k+1}-\frac{a^{k+1}}{k+1},$

i.e.,

For $k \in N^+, \int\limits_{a}^{b}x^k\;dx = \frac{b^{k+1}-a^{k+1}}{k+1}.$