Trapezium Rule for Definite Integrals

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Theorem

Let $f$ be a real function which is integrable on the closed interval $\closedint a b$.

Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ form a normal subdivision of $\closedint a b$:

$\forall i \in \set {1, 2, \ldots, n}: x_i - x_{i - 1} = \dfrac {b - a} n$


Then the definite integral of $f$ with respect to $x$ from $a$ to $b$ can be approximated as:

$\ds \int_a^b \map f x \rd x \approx \dfrac h 2 \paren {\map f {x_0} + \map f {x_n} + \sum_{i \mathop = 1}^{n - 1} 2 \map f {x_i} }$

where $h = \dfrac {b - a} n$.


Error Term

The error can be quantified as:

$\dfrac {\paren {b - a}^3 \map {f' '} \xi} {12 n^2}$

where $\xi \in \closedint a b$.


Proof

The geometric interpretation of a definite integral states that the area between the $4$ lines $x = a$, $x = b$, $y = 0$ and $y = \map f x$ is equal to $\ds \int_a^b \map f x \rd x$.


We approximate this area by dividing it into trapezia:


Trapezium-rule.png


Consider the trapezium $T_i$ whose vertices are $\tuple {x_i, \map f {x_i}, x_{i + 1}, \map f {x_{i + 1} } }$ for some $i \in \set {0, 1, \ldots, n}$.

The area of $T_i$ is given by Area of Trapezium as:

$\map \Area {T_i} = \dfrac {\map f {x_i} + \map f {x_{i + 1} } } 2 h$


Now let us consider the summation of all such areas:

\(\ds \sum_{i \mathop = 0}^{n - 1} \map \Area {T_i}\) \(=\) \(\ds \sum_{i \mathop = 0}^{n - 1} \dfrac {\map f {x_i} + \map f {x_{i + 1} } } 2 h\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 0}^{n - 1} \dfrac h 2 \paren {\map f {x_i} + \map f {x_{i + 1} } }\)
\(\ds \) \(=\) \(\ds \dfrac h 2 \paren {\map f {x_0} + \map f {x_n} + \sum_{i \mathop = 1}^{n - 1} 2 \map f {x_i} }\)

Hence the result.

$\blacksquare$


Also known as

US sources refer to this rule as:

the trapezoidal formula
the trapezoidal rule
the trapezoid rule

as a result of the fact that, in the US, the terms trapezoid and trapezium have the opposite definitions.


Also see


Sources

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