Calculus/Integration techniques/Infinite Sums

The most basic, and arguably the most difficult, type of evaluation is to use the formal definition of a Riemann integral.

Exact Integrals as Limits of Sums

Using the definition of an integral, we can evaluate the limit as $n$ goes to infinity. This technique requires a fairly high degree of familiarity with summation identities. This technique is often referred to as evaluation "by definition," and can be used to find definite integrals, as long as the integrands are fairly simple. We start with definition of the integral:

$\int \limits _{a}^{b}f(x)dx$	$=\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]$	Then picking $x_{k}^{*}$ to be $x_{k}=a+k{\frac {b-a}{n}}$ we get,
	$=\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f{\big (}a+k{\tfrac {b-a}{n}}{\big )}\right]$

In some simple cases, this expression can be reduced to a real number, which can be interpreted as the area under the curve if $f(x)$ is positive on $[a,b]$ .

Example 1

Find $\int \limits _{0}^{2}x^{2}dx$ by writing the integral as a limit of Riemann sums.

$\int \limits _{0}^{2}x^{2}dx$	$=\lim _{n\to \infty }\left[{\frac {b-a}{n}}\sum _{k=1}^{n}f(x_{k}^{*})\right]$
	$=\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {2k}{n}}{\big )}\right]$
	$=\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}\left({\frac {2k}{n}}\right)^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {2}{n}}\sum _{k=1}^{n}{\frac {4k^{2}}{n^{2}}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {8}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {4}{3}}\cdot {\frac {2n^{2}+3n+1}{n^{2}}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {8}{3}}+{\frac {4}{n}}+{\frac {4}{3n^{2}}}\right]$
	$={\frac {8}{3}}$

In other cases, it is even possible to evaluate indefinite integrals using the formal definition. We can define the indefinite integral as follows:

$\int f(x)dx$	$=\int \limits _{0}^{x}f(t)dt=\lim _{n\to \infty }\left[{\frac {x-0}{n}}\sum _{k=1}^{n}f(t_{k}^{*})\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}0+{\tfrac {k(x-0)}{n}}{\big )}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]$

Example 2

Suppose $f(x)=x^{2}$ , then we can evaluate the indefinite integral as follows.

$\int \limits _{0}^{x}f(t)dt$	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}f{\big (}{\tfrac {kx}{n}}{\big )}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}\left({\frac {kx}{n}}\right)^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x}{n}}\sum _{k=1}^{n}{\frac {k^{2}\cdot x^{2}}{n^{2}}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\sum _{k=1}^{n}k^{2}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\right]$
	$=\lim _{n\to \infty }\left[{\frac {x^{3}}{n^{3}}}\cdot {\frac {2n^{3}+3n^{2}+n}{6}}\right]$
	$=x^{3}\cdot \lim _{n\to \infty }\left[{\frac {2n^{3}}{6n^{3}}}+{\frac {3n^{2}}{6n^{3}}}+{\frac {n}{6n^{3}}}\right]$
	$=x^{3}\cdot \lim _{n\to \infty }\left[{\frac {1}{3}}+{\frac {1}{2n}}+{\frac {1}{6n^{2}}}\right]$
	$=x^{3}\cdot \left({\frac {1}{3}}\right)$
	$={\frac {x^{3}}{3}}$

This article is issued from Wikibooks. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.