# Ngô Quốc Anh

## September 26, 2009

### How to calculate limit by using definition of definite integral?

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 14:23

Let $f : [a, b] \to \mathbb R$ be a continuous function, not necessarily nonnegative. Partition $[a, b]$ into $n$ consecutive sub-intervals $[x_{i-1}, x_i]$ ($i = 1, 2, ..., n$) each of length $\Delta x = \frac{b-a}{n}$, where we set $a=x_0$, $b=x_n$ and $x_1, x_2,...,x_{n-1}$ to be successive points between $a$ and $b$ with $x_k-x_{k-1}=\Delta x$. Let $c_k$ be any intermediate point in the sub-interval $[x_{k-1},x_k]$. Then the sum

$\displaystyle\sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x}$

is called a Riemann sum for $f$ on $[a, b]$.

Suppose we let the number of partition in $P$ tends to infinity.

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)\Delta x} = I.$

We call $I$ the Riemann integral (or definite integral) of $f$ over $[a, b]$ and we write

$\displaystyle I = \int_a^b {f(x)dx} .$

In other words,

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {{c_k}} \right)}$

if the limit on the right side exists.

If we put $c_k=x_{k-1}$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + (k - 1)\frac{{b - a}}{n}} \right)} .$

Example 1. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right).$

Solution. Clearly

$\displaystyle\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}} = \sum\limits_{k = 1}^n {\frac{1}{{n + (k - 1)}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + \frac{{k - 1}}{n}}}} .$

Then if we choose $a=0$, $b=1$ we then get

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \int_0^1 {\frac{{dx}}{{1 + x}}} .$

With this it is easy to see that

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{n + 1}} + \cdots + \frac{1}{{2n - 2}} + \frac{1}{{2n - 1}}} \right) = \ln 2$

since

$\displaystyle\int_0^1 {\frac{{dx}}{{1 + x}}} = \ln 2.$

If we put $c_k=x_k$ we the obtain

$\displaystyle\int_a^b {f(x)dx} = \frac{{b - a}}{n}\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {f\left( {a + k\frac{{b - a}}{n}} \right)} .$

Example 2. Find

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {{(n - 1)}^2}}} + \frac{n}{{{n^2} + {n^2}}}} \right).$

Solution. Clearly,

$\displaystyle\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}} = \sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{\left( {\frac{k}{n}} \right)}^2}}}}$

which yields

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\frac{n}{{{n^2} + {1^2}}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}} \right) = \int_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{\pi }{4}.$

Remark. It is worth mentioning that in general it is not true that

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right).$

For example, we all know that for each fixed $k$

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}} = 0$

but

$\displaystyle\mathop {\lim }\limits_{n \to \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{n + k}}} } \right) = \ln 2 \ne 0 = \sum\limits_{k = 1}^n {\left( {\mathop {\lim }\limits_{n \to \infty } \frac{1}{{n + k}}} \right)} .$

The point is

$\displaystyle\lim \left( {summation} \right) = summation\left( {\lim } \right)$

holds true only for finite summation.