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Question: ApproximateInt- numerical-simpson

Question:ApproximateInt- numerical-simpson

9009134 110 Maple 2018
integration numeric approximateint
June 19 2019
0

how I can use ApproximateInt for the integral?

approximate_int.mw
 

restart

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"f[1,1](r,theta):=(sin(-4.700000000 10^(-6)+4.700000000 r)-0.1369508410 sinh(-4.700000000 10^(-6)+4.700000000 r)) cos(6 theta):"
> 

"L[1, 1](r, theta):=-2* (((∂)^2)/(∂r^2) f[1,1](r,theta))+7* f[1,1](r,theta)+5 *f[1,1](r,theta)-(2 *6 (((∂)^2)/(∂theta^2) f[1,1](r,theta)))/r+(0.6 (((∂)^4)/(∂r^2∂theta^2) f[1,1](r,theta)))/4+(.5 (((∂)^4)/(∂theta^4) f[1,1](r,theta)))/4"

proc (r, theta) options operator, arrow, function_assign; -2*(diff(f[1, 1](r, theta), r, r))+12*f[1, 1](r, theta)-12*(diff(f[1, 1](r, theta), theta, theta))/r+.6*(diff(f[1, 1](r, theta), r, r, theta, theta))/4+.5*(diff(f[1, 1](r, theta), theta, theta, theta, theta))/4 end proc

 (1)
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for w to 1 do for s to 1 do k[w, s] := (int(int(L[w, s](r, theta)*f[w, 1](r, theta), theta = 0 .. 2*Pi), r = 0 .. 1))/(int(int(f[w, 1](r, theta)^2, theta = 0 .. 2*Pi), r = 0 .. 1)); print([w, s] = %) end do end do

[1, 1] = 0.3929199233e-1*(int(0.1005309649e-16*(2329569981.*r*cos(4.700000000*r)^2-0.9913063750e15*r*cos(4.700000000*r)*sin(4.700000000*r)+0.1054581250e21*r*sin(4.700000000*r)^2-328995293.4*r*cos(4.700000000*r)*cosh(4.700000000*r)+0.6999899860e14*r*cos(4.700000000*r)*sinh(4.700000000*r)+0.6999899860e14*r*sin(4.700000000*r)*cosh(4.700000000*r)-0.1489340396e20*r*sin(4.700000000*r)*sinh(4.700000000*r)+1363855.810*r*cosh(4.700000000*r)^2-0.5803641743e12*r*cosh(4.700000000*r)*sinh(4.700000000*r)+0.6174086961e17*r*sinh(4.700000000*r)^2+2982150000.*cos(4.700000000*r)^2-0.1269000000e16*cos(4.700000000*r)*sin(4.700000000*r)+0.1350000000e21*sin(4.700000000*r)^2-816815901.0*cos(4.700000000*r)*cosh(4.700000000*r)+0.1737906172e15*cos(4.700000000*r)*sinh(4.700000000*r)+0.1737906172e15*sin(4.700000000*r)*cosh(4.700000000*r)-0.3697672707e20*sin(4.700000000*r)*sinh(4.700000000*r)+55931812.29*cosh(4.700000000*r)^2-0.2380077119e14*cosh(4.700000000*r)*sinh(4.700000000*r)+0.2531996935e19*sinh(4.700000000*r)^2)/r, r = 0 .. 1))

 (2)
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