Midpoint rule

Midpoint rule

Tangent trapezoidal rule

The center point rule (also: rectangle rule or tangent trapezoid rule ) is a numerical method for the approximate calculation of integrals ( numerical quadrature ). It is based on the continuous summation of closely spaced mean values ​​of the function to be integrated.

description

Boxing rule

With the left-hand (left box rule) or right-hand box rule (right box rule) , the interval length is multiplied by the function value of the function to be integrated at the left or right edge point: ${\displaystyle (b-a)}$

${\displaystyle \int \limits _{a}^{b}f(x)\,dx\approx (b-a)\cdot f(a)\quad {\text{resp.}}\quad \int \limits _{a}^{b}f(x)\,dx\approx (b-a)\cdot f(b)}$.

The box rule plays an important role in the derivation of the Riemann integral . The left-hand boxing rule corresponds to the lower sums and the right-hand box rule corresponds to the upper sums. It is also comparable to the one-sided difference quotient .

The box rule is exactly for polynomial functions of degree at most 0 (i.e. for constant functions) and thus of order 1.

Midpoint rule

Take the midpoint of the interval and multiply the width of the interval by the function value of the integrand at this point in order to obtain an approximate value of the integral: ${\displaystyle (a+b)/2}$ ${\displaystyle [a,b]}$ ${\displaystyle (b-a)}$

${\displaystyle \int \limits _{a}^{b}f(x)\,dx\approx (b-a)\cdot f\!\left({\frac {a+b}{2}}\right)}$.

If you turn the horizontal straight line at the point counterclockwise in the above picture of the center point rule, you get the tangent for the point . The picture below of the tangent trapezoidal rule results. Since the trapezoid thus obtained has the same area as the rectangle, the center point rule and the tangent trapezoid rule are only different geometric interpretations of the same quadrature formula. ${\displaystyle (c,f(c))}$ ${\displaystyle (c,f(c))}$

The midpoint rule is exact for polynomial functions of degree at most 1 (i.e. for affine-linear functions) and therefore of order 2.

With the composite center point rule or the composite tangent trapezoidal formula , the interval is now divided into equidistant sub-intervals of the width . Then the center rule is carried out for each of the sub-intervals and the areas are summed up. This leads to the equation: ${\displaystyle [a,b]}$ ${\displaystyle n}$ ${\displaystyle h=(b-a)/n}$

${\displaystyle \int \limits _{a}^{b}f(x)\,dx\approx h\cdot \sum _{k=1}^{n}f\!\left(a-{\frac {h}{2}}+k\cdot h\right)}$.

example

Let a function (the natural logarithm) be integrated in the interval . This would require the calculation of the integral . The general solution is: ${\displaystyle f\colon \mathbb {R} _{>0}\to \mathbb {R} ,\ x\mapsto \ln(x)}$ ${\displaystyle [2,6]}$ ${\displaystyle \int \limits _{2}^{6}f(x)\,dx=\int \limits _{2}^{6}\ln(x)\,dx}$

${\displaystyle \int \limits \ln(x)\,dx=x\ln(x)-x+C}$.

So is ${\displaystyle \int \limits _{2}^{6}f(x)\,dx=5{,}3642\ldots }$

Using the compound center rule with four sub-intervals results in the following:

1. Breakdown of the interval into four sub-intervals: and with the interval centers 2.5, 3.5, 4.5 and 5.5. ${\displaystyle [2,6]}$ ${\displaystyle [2,3],[3,4],[4,5]}$ ${\displaystyle [5,6]}$
2. Calculation of: ${\displaystyle {\begin{aligned}{\frac {6-2}{4}}\cdot (f(2{,}5)+f(3{,}5)+f(4{,}5)+f(5{,}5))&={\frac {4}{4}}(\ln(2{,}5)+\ln(3{,}5)+\ln(4{,}5)+\ln(5{,}5))\\&\approx 0{,}9163+1{,}2528+1{,}5041+1{,}7047=5{,}3779.\end{aligned}}}$
3. So it applies . ${\displaystyle \int \limits _{2}^{6}f(x)\,dx\approx 5{,}3779}$

Web links

• Numerical integration ( Java applet for displaying various integration methods)

Individual evidence

1. ^ Hans Petter Langtangen: A Primer on Scientific Programming with Python . 3. Edition. Springer, Berlin / Heidelberg 2012, ISBN 978-3-642-30293-0 .