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.

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