Definite Integral as Limit of a Sum

DEFINITE INTEGRAL AS LIMIT OF A SUM

An alternative way of describing \int\limits_{a}^{b}{{}}f(x)dx\ is that the definite integral \int\limits_{a}^{b}{{}}f(x)dx\is a limiting case of the summation of an infinite series, provided f(x) is continuous on [a, b] i.e., $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{n\to \infty }{\mathrm{limh}}\sum _{r=0}^{n-1}f(a+rh)whereh=\frac{b-a}{n}$. The converse is also true i.e., if we have an infinite series of the above form, it can be expressed as a definite integral.

Method to express the infinite series as definite integral:

(i) Express the given series in the form $\sum \frac{1}{n}f\left(\frac{r}{n}\right)$

(ii) Then the limit is its sum when $n\to \infty$, i.e. $\underset{n\to \infty }{\mathrm{Lim}}\sum \frac{1}{n}f\left(\frac{r}{n}\right)$

(iii) Replace $\frac{r}{n}$ by x and $\frac{1}{n}$ by dx and $\underset{n\to \infty }{\mathrm{Lim}}\sum$by the sign of $\int$.

(iv) The lower and the upper limit of integration are the limiting values of $\frac{r}{n}$ for the first and the last term of r respectively.

Some particular cases of the above are

(a) $\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=1}^n\frac{1}{n}f\left(\frac{r}{n}\right)$ $or\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=0}^{n-1}\frac{1}{n}f\left(\frac{r}{n}\right)=\underset{0}{\overset{1}{\int }}f(x)dx$

(b) $\underset{n\to \infty }{\mathrm{Lim}}\sum _{r=1}^{pn}\frac{1}{n}f\left(\frac{r}{n}\right)=\underset{\alpha }{\overset{\beta }{\int }}f(x)dx$

where $\alpha =\underset{n\to \infty }{\mathrm{Lim}}\frac{r}{n}=0$ (as r = 1) and $\beta =\underset{n\to \infty }{\mathrm{Lim}}\frac{r}{n}=p$ (as r = pn)

Illustration : Show that $\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{n+n}\right)$= ln2.

Solution: Let I = $\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{n+n}\right)$

= $\underset{n\to \infty }{\lim }\left(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{n+n}\right)$ = $\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n\frac{1}{1+\frac{r}{n}}$

Now $\alpha$ = $\underset{n\to \infty }{\lim }\frac{1}{n}=0$ ( as r = 1 )

and $\beta$ = $\underset{n\to \infty }{\lim }\frac{r}{n}=1$ (as r = n)

$\Rightarrow$ $I=\underset{0}{\overset{1}{\int }}\frac{1}{1+x}dx={\left[\ln \left(1+x\right)\right]}_0^1$ $\Rightarrow$ I = ln2.