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NumericalIntegralApproximation (1.0.0) current version: 2.0.0 »

Source Notebook

Determine the value of an integral using a numerical method

Contributed by: Jason Martinez

ResourceFunction["NumericalIntegralApproximation"][f,{x,x_min,x_max},method]

gives a numerical approximation to the integral using the specified method.

Details and Options

ResourceFunction["NumericalIntegralApproximation"] takes the same options as NIntegrate in most cases except that both Method and MaxRecursion are unavailable. In addition it takes an "Intervals" option. "Intervals" specifies the number of subintervals to divide the integral into.

ResourceFunction["NumericalIntegralApproximation"] supports the following methods:

"Midpoint"

midpoint rule

"RightHand"

right Riemann sum

"LeftHand"

left Riemann sum

"Simpson"

Simpson's rule

"Trapezoidal"

trapezoidal rule

"Boole"

Boole's rule

"LeftHand", "Midpoint" and "RightHand" methods disregard all option settings except "Intervals".

By default, "Intervals" takes the value Automatic, corresponding to a single interval.

Examples

Basic Examples (2)

Integrate expressions using classic numerical methods such as Simpson’s rule:

In[1]:=

$ResourceFunction["NumericalIntegralApproximation"][ x^2 + Sqrt[x], {x, 1, 5}, "Simpson"]$

Out[1]=

Compare left hand, right hand, and midpoint integrations:

In[2]:=

Out[2]=

In[3]:=

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In[4]:=

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Scope (1)

Compute integrals with trapezoidal or Boole rules:

In[5]:=

$ResourceFunction["NumericalIntegralApproximation", ResourceVersion->"1.0.0"][(x^2 - 2)/x, {x, 1, 2}, "Trapezoidal"]$

Out[5]=

In[6]:=

$ResourceFunction["NumericalIntegralApproximation", ResourceVersion->"1.0.0"][(x^2 - 2)/x, {x, 1, 2}, "Boole"]$

Out[6]=

Options (1)

Increase accuracy by using multiple intervals:

In[7]:=

$ResourceFunction["NumericalIntegralApproximation"][ x^3 - x^2, {x, 1, 3}, "LeftHand", "Intervals" -> 10]$

Out[7]=

Applications (2)

Examine how increasing the number of intervals affects the result:

In[8]:=

$results = Table[ResourceFunction["NumericalIntegralApproximation"][ x^3 - x^2, {x, 1, 3}, "Midpoint", "Intervals" -> i], {i, 50}];$

In[9]:=

Out[9]=

Compare the results to the exact answer:

In[10]:=

$Integrate[x^3 - x^2, {x, 1, 3}]$

Out[10]=

In[11]:=

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Version History

2.0.0 – 09 August 2022
1.0.0 – 08 November 2019

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License Information

This work is licensed under a Creative Commons Attribution 4.0 International License