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Integrating a list of values

List operation on specific elementsEfficient way to obtain values of a function defined by an IntegralNumerical integration of modified bessel functionIntegrating an interpolating functionNumerical Integral with Boolean as part of argumentHow to set a line coordinate for a symbolic line integral over a curve?Integrating curve peaksStrange results by integrating Abs[Sin[a - t]]Integrating over colorsIntegrating only over positive values of an oscillating function

2

$begingroup$

The data given here

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1]

generates the following curve

ListPlot[data]

I want to know, how to compute the integral of this curve using only the data given above.

list-manipulation calculus-and-analysis numerical-integration

share|improve this question

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If you only have a list of function values, you need to give the step size as well.
  $endgroup$
  – J. M. is away♦
  Apr 1 at 12:56
  

  
  
  
  

add a comment | 

2

$begingroup$

The data given here

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1]

generates the following curve

ListPlot[data]

I want to know, how to compute the integral of this curve using only the data given above.

list-manipulation calculus-and-analysis numerical-integration

share|improve this question

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If you only have a list of function values, you need to give the step size as well.
  $endgroup$
  – J. M. is away♦
  Apr 1 at 12:56
  

  
  
  
  

add a comment | 

$begingroup$

The data given here

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1]

generates the following curve

ListPlot[data]

I want to know, how to compute the integral of this curve using only the data given above.

list-manipulation calculus-and-analysis numerical-integration

share|improve this question

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

$endgroup$

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1]

ListPlot[data]

share|improve this question

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

share|improve this question

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

edited Apr 1 at 12:55

J. M. is away♦

98.9k10311467

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

asked Apr 1 at 12:48

Tobias FritznTobias Fritzn

1945

4 Answers
4

active

oldest

votes

2

$begingroup$

Using Tai's method:

ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data

Alternatively

a = Table[x, Clip[Sin[x], 0., 1.], x, 0, 2 π, 0.1];
Integrate[Interpolation[a][x], x, a[[1, 1]], a[[-1, 1]]]

2.00038

share|improve this answer

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For Interpolation you can also play with the InterpolationOrder option to increase the accuracy (sometimes). In this case it doesn't do much though.
  $endgroup$
  – Roman
  Apr 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
  $endgroup$
  – Henrik Schumacher
  Apr 1 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want the integral as a plot, a curve. Any way of doing that?
  $endgroup$
  – Tobias Fritzn
  Apr 1 at 14:16
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:

0.1*Total[data]

to get the numerical integral. To visualize the integral and plot it you can ListPlot:

0.1*Accumulate[data]

Hence:

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1];
ListPlot[data, 0.1*Accumulate[data]]

share|improve this answer

edited Apr 1 at 16:13

answered Apr 1 at 12:54

bill sbill s

54.9k377158

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is the function Accumulate related to Integration?
  $endgroup$
  – Tobias Fritzn
  Apr 1 at 15:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
  $endgroup$
  – bill s
  Apr 1 at 16:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, @bill, it works nice. But looking at the definition of Accumulate[], it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
  $endgroup$
  – Tobias Fritzn
  Apr 2 at 9:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is called the Rieman approximation to the integral.
  $endgroup$
  – bill s
  Apr 2 at 16:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate[]. en.wikipedia.org/wiki/Riemann_sum
  $endgroup$
  – Tobias Fritzn
  Apr 2 at 17:10
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.

tuples = Transpose@Range[0, 2 Pi, 0.1], data;

Show[
 Plot[
 NIntegrate[Interpolation[tuples][x], x, 0, xmax, Method -> "Trapezoidal"],
 xmax, 0, 2 Pi, PlotLegends -> "integral"
 ],
 ListPlot[
 Style[tuples, Thick, ColorData[97][2]],
 Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
 PlotLegends -> "data", Joined -> True
 ]
]

share|improve this answer

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

$endgroup$

add a comment | 

0

$begingroup$

It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:

simpsoncoefficients[n_] := SparseArray[1 -> 1, -1 -> 1, i_?EvenQ -> 4, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]

Then integral[data, 0.1] gives 2.00024.

share|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029

$endgroup$

add a comment | 

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2

$begingroup$

Using Tai's method:

ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data

Alternatively

a = Table[x, Clip[Sin[x], 0., 1.], x, 0, 2 π, 0.1];
Integrate[Interpolation[a][x], x, a[[1, 1]], a[[-1, 1]]]

2.00038

share|improve this answer

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For Interpolation you can also play with the InterpolationOrder option to increase the accuracy (sometimes). In this case it doesn't do much though.
  $endgroup$
  – Roman
  Apr 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
  $endgroup$
  – Henrik Schumacher
  Apr 1 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want the integral as a plot, a curve. Any way of doing that?
  $endgroup$
  – Tobias Fritzn
  Apr 1 at 14:16
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Using Tai's method:

ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data

Alternatively

a = Table[x, Clip[Sin[x], 0., 1.], x, 0, 2 π, 0.1];
Integrate[Interpolation[a][x], x, a[[1, 1]], a[[-1, 1]]]

2.00038

share|improve this answer

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For Interpolation you can also play with the InterpolationOrder option to increase the accuracy (sometimes). In this case it doesn't do much though.
  $endgroup$
  – Roman
  Apr 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Jepp. The reseason is the kink in the middle of the integral. This way, one cannot profit from higher order quadrature rules. Trapezoidal rule is almost optimal.
  $endgroup$
  – Henrik Schumacher
  Apr 1 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I want the integral as a plot, a curve. Any way of doing that?
  $endgroup$
  – Tobias Fritzn
  Apr 1 at 14:16
  
  

  
  
  
  

add a comment | 

$begingroup$

Using Tai's method:

ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data

Alternatively

a = Table[x, Clip[Sin[x], 0., 1.], x, 0, 2 π, 0.1];
Integrate[Interpolation[a][x], x, a[[1, 1]], a[[-1, 1]]]

2.00038

share|improve this answer

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

$endgroup$

ω = ConstantArray[0.1, Length[data]];
ω[[1]] *= 0.5;
ω[[-1]] *= 0.5;
ω.data

a = Table[x, Clip[Sin[x], 0., 1.], x, 0, 2 π, 0.1];
Integrate[Interpolation[a][x], x, a[[1, 1]], a[[-1, 1]]]

share|improve this answer

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

answered Apr 1 at 12:52

Henrik SchumacherHenrik Schumacher

59.5k582166

2

$begingroup$

Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:

0.1*Total[data]

to get the numerical integral. To visualize the integral and plot it you can ListPlot:

0.1*Accumulate[data]

Hence:

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1];
ListPlot[data, 0.1*Accumulate[data]]

share|improve this answer

edited Apr 1 at 16:13

answered Apr 1 at 12:54

bill sbill s

54.9k377158

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is the function Accumulate related to Integration?
  $endgroup$
  – Tobias Fritzn
  Apr 1 at 15:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
  $endgroup$
  – bill s
  Apr 1 at 16:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, @bill, it works nice. But looking at the definition of Accumulate[], it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
  $endgroup$
  – Tobias Fritzn
  Apr 2 at 9:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is called the Rieman approximation to the integral.
  $endgroup$
  – bill s
  Apr 2 at 16:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate[]. en.wikipedia.org/wiki/Riemann_sum
  $endgroup$
  – Tobias Fritzn
  Apr 2 at 17:10
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:

0.1*Total[data]

to get the numerical integral. To visualize the integral and plot it you can ListPlot:

0.1*Accumulate[data]

Hence:

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1];
ListPlot[data, 0.1*Accumulate[data]]

share|improve this answer

edited Apr 1 at 16:13

answered Apr 1 at 12:54

bill sbill s

54.9k377158

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is the function Accumulate related to Integration?
  $endgroup$
  – Tobias Fritzn
  Apr 1 at 15:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  As you are accumulating the data, you are integrating the function up to that point. So this is the answer to your statement that you "want the integral as a plot."
  $endgroup$
  – bill s
  Apr 1 at 16:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, @bill, it works nice. But looking at the definition of Accumulate[], it is not immediately clear how it should give an integral when multiplied by the stepsize. I mean given the definition Accumulate[i,j,k]=i,i+j,i+j+k, how does this lead to integration?
  $endgroup$
  – Tobias Fritzn
  Apr 2 at 9:12
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is called the Rieman approximation to the integral.
  $endgroup$
  – bill s
  Apr 2 at 16:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Correct me if I am wrong. In a Rieman sum, nth term is not the sum of all (n-1) terms, which seems to be the case with Accumulate[]. en.wikipedia.org/wiki/Riemann_sum
  $endgroup$
  – Tobias Fritzn
  Apr 2 at 17:10
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Assuming the stepsize is 0.1 as suggested by the construction of the Table, you can calculate:

0.1*Total[data]

to get the numerical integral. To visualize the integral and plot it you can ListPlot:

0.1*Accumulate[data]

Hence:

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1];
ListPlot[data, 0.1*Accumulate[data]]

share|improve this answer

edited Apr 1 at 16:13

answered Apr 1 at 12:54

bill sbill s

54.9k377158

$endgroup$

0.1*Total[data]

0.1*Accumulate[data]

data = Table[Clip[Sin[x], 0, 1], x, 0, 2 [Pi], 0.1];
ListPlot[data, 0.1*Accumulate[data]]

share|improve this answer

edited Apr 1 at 16:13

answered Apr 1 at 12:54

bill sbill s

54.9k377158

answered Apr 1 at 12:54

bill sbill s

54.9k377158

answered Apr 1 at 12:54

bill sbill s

54.9k377158

1

$begingroup$

You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.

tuples = Transpose@Range[0, 2 Pi, 0.1], data;

Show[
 Plot[
 NIntegrate[Interpolation[tuples][x], x, 0, xmax, Method -> "Trapezoidal"],
 xmax, 0, 2 Pi, PlotLegends -> "integral"
 ],
 ListPlot[
 Style[tuples, Thick, ColorData[97][2]],
 Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
 PlotLegends -> "data", Joined -> True
 ]
]

share|improve this answer

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

$endgroup$

add a comment | 

1

$begingroup$

You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.

tuples = Transpose@Range[0, 2 Pi, 0.1], data;

Show[
 Plot[
 NIntegrate[Interpolation[tuples][x], x, 0, xmax, Method -> "Trapezoidal"],
 xmax, 0, 2 Pi, PlotLegends -> "integral"
 ],
 ListPlot[
 Style[tuples, Thick, ColorData[97][2]],
 Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
 PlotLegends -> "data", Joined -> True
 ]
]

share|improve this answer

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

$endgroup$

add a comment | 

$begingroup$

You mention that you want the integral as a plot in a comment; I wonder if the following is what you had in mind. Here I am using your definition of data, and assuming a $0.1$ step size, as hinted at by your Table expression.

tuples = Transpose@Range[0, 2 Pi, 0.1], data;

Show[
 Plot[
 NIntegrate[Interpolation[tuples][x], x, 0, xmax, Method -> "Trapezoidal"],
 xmax, 0, 2 Pi, PlotLegends -> "integral"
 ],
 ListPlot[
 Style[tuples, Thick, ColorData[97][2]],
 Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
 PlotLegends -> "data", Joined -> True
 ]
]

share|improve this answer

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

$endgroup$

tuples = Transpose@Range[0, 2 Pi, 0.1], data;

Show[
 Plot[
 NIntegrate[Interpolation[tuples][x], x, 0, xmax, Method -> "Trapezoidal"],
 xmax, 0, 2 Pi, PlotLegends -> "integral"
 ],
 ListPlot[
 Style[tuples, Thick, ColorData[97][2]],
 Mesh -> All, MeshStyle -> Directive[Black, PointSize[0.01]],
 PlotLegends -> "data", Joined -> True
 ]
]

share|improve this answer

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

answered Apr 1 at 16:06

MarcoBMarcoB

38.6k557115

0

$begingroup$

It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:

simpsoncoefficients[n_] := SparseArray[1 -> 1, -1 -> 1, i_?EvenQ -> 4, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]

Then integral[data, 0.1] gives 2.00024.

share|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029

$endgroup$

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It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:

simpsoncoefficients[n_] := SparseArray[1 -> 1, -1 -> 1, i_?EvenQ -> 4, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]

Then integral[data, 0.1] gives 2.00024.

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edited Apr 2 at 5:50

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029

$endgroup$

add a comment | 

$begingroup$

It seems that Simpson's rule has not been mentioned yet, which is the result from a 2nd-order interpolation and will have a smaller error than that from a 1st-order one. So according to the formula, the inputs are the List of samples of the function data and the step size h:

simpsoncoefficients[n_] := SparseArray[1 -> 1, -1 -> 1, i_?EvenQ -> 4, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]

Then integral[data, 0.1] gives 2.00024.

share|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029

$endgroup$

simpsoncoefficients[n_] := SparseArray[1 -> 1, -1 -> 1, i_?EvenQ -> 4, n, 2]
integral[data_, h_] := (h/3) simpsoncoefficients[Length[#]].# &[data]

share|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029

answered Apr 2 at 4:12

Αλέξανδρος ΖεγγΑλέξανδρος Ζεγγ

4,49011029