Solving an equation with integral with one variable

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Sergio Manzetti
Sergio Manzetti am 10 Aug. 2017
Kommentiert: Star Strider am 11 Aug. 2017
Hi, how can I calculate the following equation involving an integral in matlab?
C + Integral_{-4}_{4} e^(x^2)*x^2 dx = 1
where -4 and 4 are the lower and upper limit, and C is the unknown constant.
Thanks!

Akzeptierte Antwort

John BG
John BG am 10 Aug. 2017
Bearbeitet: John BG am 11 Aug. 2017
Hi Sergio
Since the primitive of
exp(x^2)*x^2
is
Ly=x*exp(x^2)/2-.5*1j*(pi)^.4/2*erf(1j*x)
then the integral in the interval [y1 y2] is
L=Ly2-Ly1
this is
y2=4;
Ly2=y2*exp(y2^2)/2-.5*1j*(pi)^.5*erf(sym(1j*y2))
Ly2 =
149084195602433/8388608 - (erf(4i)*3991211251234741i)/4503599627370496
y1=-4;
Ly1=y1*exp(y1^2)/2-.5*1j*(pi)^.5*erf(sym(1j*y1))
Ly1 =
(erf(4i)*3991211251234741i)/4503599627370496 - 149084195602433/8388608
the value of the integral is
L= Ly2-Ly1
L =
149084195602433/4194304 - (erf(4i)*3991211251234741i)/2251799813685248
way around erf() only working for real inputs
L= double(Ly2-Ly1)
L =
3.7843e+07
therefore
C=1-L
C =
-3.7843e+07
Repeating with
format double
the result is
L =
3.784324335121135e+07
Comment:
When attempting direct integration with command sum
x=[-4:.001:4];L=sum(exp(x.^2) .* x.^2)
L =
3.4537e+10
x=[-4:.00001:4];L=sum(exp(x.^2) .* x.^2)
L =
3.4396e+12
x=[-4:.0000001:4];L=sum(exp(x.^2) .* x.^2)
L =
3.4395e+14
The slopes when approaching -4 and 4 are too sharp for command sum.
Grazie tante per la sua attenzione.
if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?
To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link
thanks in advance
John BG

Weitere Antworten (1)

Star Strider
Star Strider am 10 Aug. 2017
Bearbeitet: Star Strider am 10 Aug. 2017
If I understand correctly what you want to do, this works:
f = @(x) exp(x.^2) .* x.^2;
Cs = fzero(@(C) C + integral(f, -4, 4) - 1, 1)
Cs =
-3.4395e+07
Or more simply, since ‘C’ is a constant:
C = 1 - integral(f, -4, 4)
C =
-3.4395e+07
If you are using R2011b or earlier, use quad instead of integral:
C = 1 - quad(f, -4, 4)
The result is the same:
C =
-34395040.4474417
here using format long g.
  2 Kommentare
Sergio Manzetti
Sergio Manzetti am 11 Aug. 2017
Hi Star Strider, I was thinking of something real simple, like on Wolfram Alpha, where you type in "C*integral of f(x) + integral g(x) = 1" . Then you get the output, for the numerical integration of the two integrands, and you get the value of C automatically. Is this feasible in Matlab?
Star Strider
Star Strider am 11 Aug. 2017
Not to my knowledge.

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