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I don't know how to find the following integral by numeric $$ \int_{g(\alpha)}^{h(\alpha)}f(\alpha,x)dx$$ where,say, $ f(\alpha ,x)=1-\alpha x+\sqrt{x}\alpha^2$ and $g(\alpha)=2\alpha^2-1$, $h(\alpha)=\sqrt{\alpha}+\alpha$ . I want to calculate the integral above in interval $1/40<\alpha<1/8$ in steps of 0.01 and create a table for computations.

Thanks.

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    $\begingroup$ Have you tried NIntegrate[]? $\endgroup$
    – Feyre
    Feb 13, 2017 at 15:04
  • $\begingroup$ You might consider accepting one of the answers, if one of them answers your question. $\endgroup$
    – Michael E2
    May 15, 2017 at 10:16

3 Answers 3

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Try this:

f[alpha_, x_] := 1 - x alpha + Sqrt[x] alpha^2;
g[alpha_] := 2 alpha^2 - 1;
h[alpha_] := Sqrt[alpha] + alpha;
Table[NIntegrate[f[alpha, x], {x, g[alpha], h[alpha]}], {alpha, 1/40.,1/8.,0.01}]
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f[a_, x_] = 1 - a x + Sqrt[x] a^2;

g[a_] = 2 a^2 - 1;

h[a_] = Sqrt[a] + a;

int[a_] = 
 Assuming[{1/40 < a < 1/8}, Integrate[f[a, x], {x, g[a], h[a]}]]

(*  1 + Sqrt[a] + (3*a)/2 - (5*a^2)/2 - 
   a^(5/2) - (5*a^3)/2 + 2*a^5 + 
   (2/3)*a^2*(a*Sqrt[Sqrt[a] + a] + 
        I*(1 - 2*a^2)^(3/2) + 
        Sqrt[a^(3/2) + a^2])  *)

Plot[Evaluate@ReIm[int[a]], {a, 1/40, 1/8},
 PlotLegends -> {Re, Im},
 AxesLabel -> {a, None},
 Epilog -> Inset[
   LogPlot[
    Evaluate@ReIm[int[a]], {a, 1/40, 1/8},
    PlotLabel -> "LogPlot"],
   {0.09, 2/3}]]

enter image description here

Table[{a, int[a]}, {a, 1/40., 1/8., 0.01}] // Grid

enter image description here

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  • $\begingroup$ i've said that the $\alpha$ values are discrete and obtain a numerical integration then each computations of $\alpha$ is inserted into a table. $\endgroup$
    – Ali184
    Feb 14, 2017 at 5:43
  • $\begingroup$ very thanks for your complete answer. $\endgroup$
    – Ali184
    Feb 14, 2017 at 7:07
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If it is possible to get a closed form for the integral that is preferable as it simplifies the downstream processing. Note: this is not always possible.

Integrate[f[alpha, x], {x, g[alpha], h[alpha]}, 
 Assumptions -> 1/40 <= alpha <= 1/8]

(* 1 + Sqrt[alpha] + (3 alpha)/2 - (5 alpha^2)/2 - alpha^(
 5/2) - (5 alpha^3)/2 + 2 alpha^5 + 
 2/3 alpha^2 (alpha Sqrt[Sqrt[alpha] + alpha] + 
    I (1 - 2 alpha^2)^(3/2) + Sqrt[alpha^(3/2) + alpha^2]) *)

Now define a function, f2

f2[alpha_] := 
 1 + Sqrt[alpha] + (3 alpha)/2 - (5 alpha^2)/2 - alpha^(5/2) - (
  5 alpha^3)/2 + 2 alpha^5 + 
  2/3 alpha^2 (alpha Sqrt[Sqrt[alpha] + alpha] + 
     I (1 - 2 alpha^2)^(3/2) + Sqrt[alpha^(3/2) + alpha^2])

To generate the table

Table[{alpha, N@f2[alpha]}, {alpha, 1/40, 1/8, 1/40}]

(* {{1/40, 1.19395 + 0.000415886 I}, {1/20, 
  1.29172 + 0.00165418 I}, {3/40, 1.37048 + 0.0036869 I}, {1/10, 
  1.43738 + 0.00646767 I}, {1/8, 1.49509 + 0.00993222 I}} *)
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