I tried to evaluate this integral:
∫exp(7.313 x)/√(x^2+1.7439)dx on the interval 0 to 0.001.Then I had this kind of error : Unexpected MATLAB expression. So how do I write and evaluate it?
Thank you.
MATLAB: How to evaluate the integral
id
Related Solutions
This works for me:
Q = 3.3;D5L = [12.75*0.0254 14*0.0254 16*0.0254 18*0.0254 20*0.0254 22*0.0254 24*0.0254 26*0.0254 28*0.0254 30*0.0254 32*0.0254 34*0.0254 36*0.0254 38*0.0254 40*0.0254 42*0.0254 44*0.0254 46*0.0254 48*0.0254 52*0.0254 56*0.0254 60*0.0254 64*0.0254 68*0.0254 72*0.0254 76*0.0254 80*0.0254];V5L = ((4.*Q)./(pi.*D5L.^2));V = V5L((V5L > 1) & (V5L < 4.5));Do = sqrt((4.*Q)./(pi.*V))producing:
Do = 1.0160 1.0668 1.1176 1.1684 1.2192 1.3208 1.4224 1.5240 1.6256 1.7272 1.8288 1.9304 2.0320EDIT — (27 Jan 2019 at 15:09)
This runs for me without error (in R2018b):
Q=3.3;%gamma=8.43 kN/m3 = 859.62076503 Kg/m3 (*1000/9.81)
rouh=859.62076503;g=9.81;Gamma=(rouh.*g);Zs=0;Ze=0;L=1285000;Ks=(0.000045);Er=(1E-8);viscosity=(3.83.*1E-3);% material grade X70
strength= 482000000;%API 5L
% % D5L=[12.75*0.0254 14*0.0254 16*0.0254 18*0.0254 20*0.0254 22*0.0254 24*0.0254 26*0.0254 28*0.0254 30*0.0254 32*0.0254 34*0.0254 36*0.0254 38*0.0254 40*0.0254 42*0.0254 44*0.0254 46*0.0254 48*0.0254 52*0.0254 56*0.0254 60*0.0254 64*0.0254 68*0.0254 72*0.0254 76*0.0254 80*0.0254]
% % V5L=((4.*Q)./(pi.*D5L.^2));
% % if V=V5L(V5L > 1 & abs(V5L - 4.5<1e-5));
% % end
% % Do=((4.*Q)./(pi.*V)).^(0.5)
D5L = [12.75*0.0254 14*0.0254 16*0.0254 18*0.0254 20*0.0254 22*0.0254 24*0.0254 26*0.0254 28*0.0254 30*0.0254 32*0.0254 34*0.0254 36*0.0254 38*0.0254 40*0.0254 42*0.0254 44*0.0254 46*0.0254 48*0.0254 52*0.0254 56*0.0254 60*0.0254 64*0.0254 68*0.0254 72*0.0254 76*0.0254 80*0.0254];V5L = ((4.*Q)./(pi.*D5L.^2));V = V5L((V5L > 1) & (V5L < 4.5));Do = sqrt((4.*Q)./(pi.*V));Re=(V.*Do./viscosity);Fo=0.01;for I=0:1:1E6; Fn=(1./(-4.*log10((Ks./(3.71.*Do))+((1.26)./(Re.*sqrt(Fo)))))).^2; E=abs((Fn-Fo)/Fn); if E<=Er; Fn=Fn(end); display(E) break end if E>Er; Fo=Fn; endendHloss=((Fn.*L.*Q.^2)./(12.*Do.^5));Hpump=Hloss+Ze-Zs;%calculating Hloss every 128500M
LC=1:128500:1285000;for i=1:1:length(LC); Hpump=Hloss+Ze-Zs; PRESSURE=rouh.*g.*Hpump.*1E-5; numberofpumps=ceil(PRESSURE./90); pressureperpump=(PRESSURE./numberofpumps); HlossN=((Fn.*LC(i).*Q.^2)./(12.*sqrt(Do))); TEL(i,:)=(Hpump-HlossN+Zs); HGL(i,:)=(TEL(i,:)-((V.^2)./(2.*g))); PE=Gamma.*(HGL-Ze); T=((PE(i).*Do)./(2.*strength));endfor ii=1:1:length(Do) subplot(4,4,ii) plot(LC,TEL(:,ii)) hold on plot(LC,HGL(:,ii)) xlabel('LC') ylabel('HGL & TEL')endYour ymax function returns a sym result. If you specify the limits as function handles in your call to integral2 those function handles must return a numeric result, not a symbolic result.
If you want to integrate with a symbolic limit, work solely with symbolic variables and call the int function twice instead of calling integral2.
If you want to be able to specify b when you call integral2 rather than when you define the ymax function:
ymax = @(x, b) b.*sqrt(x);Now when you call integral2 (I'm assuming you've defined fun, xmin, xmax, and ymin before this call) specify b:
% Use b = 1
Q = integral2(fun, xmin, xmax, ymin, @(x) ymax(x, 1));% Use b = x.^2
Q = integral2(fun, xmin, xmax, ymin, @(x) ymax(x, x.^2));
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