How to compute the difference of the integrals of two functions (f(x) and j(x)) only over the portions where f(x)>j(x)

I would like to compute the difference of integrals of two functions (f(x), shown in thick black, and j(x), shown in thick red (portion shaded in light red), only over the portion where 0<x<86400 and where f(x)>j(x), as indicated in the graph.

Any quick solution? It's been a bit a struggle.

Thank you in advance.

Best regards,

Tim.

(Code given below).

%Part of no interest for the question (given for the code to run)
Tmean_out=30;
Tmax_out=40;
Tadaptive=32;
meg=((2*pi)/86400);
DeltaT_out=Tmax_out-Tmean_out;
Tmin_out=Tmean_out-DeltaT_out;
Tswing_out=DeltaT_out*2;
disp(DeltaT_out);

DeltaT_in_wo=0.15*DeltaT_out;
Tmean_in_wo=1.5+Tmean_out;
Tmax_in_wo=Tmean_in_wo+DeltaT_in_wo;
Tmin_in_wo=Tmean_in_wo-DeltaT_in_wo;

Tmin_in_w=Tmin_in_wo-(0.5*(abs(Tmin_in_wo-Tmin_out)));
Tmax_in_w=Tmax_in_wo-(0.25*(abs(Tmax_in_wo-Tmax_out)));
Tmean_in_w=(Tmin_in_w+Tmax_in_w)/2;
DeltaT_in_w=Tmax_in_w-Tmean_in_w;

%part of interest for the question:
f=@(x) Tmean_out+(DeltaT_out*cos(meg*(x)));
j=@(x) Tmean_in_wo+(DeltaT_in_wo*cos(meg*x));

Answer 1

[Torsten]

Integrate

g(x) = max(f(x)-j(x),0)

Answer 2

[Star Strider]

Perhaps this —

%Part of no interest for the question (given for the code to run)
Tmean_out=30;
Tmax_out=40;
Tadaptive=32;
meg=((2*pi)/86400);
DeltaT_out=Tmax_out-Tmean_out;
Tmin_out=Tmean_out-DeltaT_out;
Tswing_out=DeltaT_out*2;
disp(DeltaT_out);

    10

DeltaT_in_wo=0.15*DeltaT_out;
Tmean_in_wo=1.5+Tmean_out;
Tmax_in_wo=Tmean_in_wo+DeltaT_in_wo;
Tmin_in_wo=Tmean_in_wo-DeltaT_in_wo;

Tmin_in_w=Tmin_in_wo-(0.5*(abs(Tmin_in_wo-Tmin_out)));
Tmax_in_w=Tmax_in_wo-(0.25*(abs(Tmax_in_wo-Tmax_out)));
Tmean_in_w=(Tmin_in_w+Tmax_in_w)/2;
DeltaT_in_w=Tmax_in_w-Tmean_in_w;

%part of interest for the question:
f=@(x) Tmean_out+(DeltaT_out*cos(meg*(x)));
j=@(x) Tmean_in_wo+(DeltaT_in_wo*cos(meg*x));

figure
fplot(f, [0 8.64E4], 'DisplayName','f(x)')
hold on
fplot(j, [0 8.64E4], 'DisplayName','j(x)')
hold off
grid
legend('Location','best')

intf = integral(f, 0, 8.64E4)                               % 'f' Integral

intf = 2592000

intj = integral(j, 0, 8.64E4)                               % 'j' Integral

intj = 2.7216e+06

fminusj = integral(@(x)f(x)-j(x), 0, 8.64E4)                % 'f-j' Integral

fminusj = -129600

fgtj = integral(@(x)(f(x)-j(x)).*(f(x)>j(x)), 0, 8.64E4)    % 'f>j' Integral

fgtj = 1.7262e+05

EDIT — (23 Dec 2022 at 13:56)

Changed limits to [0 8.64E4].

.

[TIMOTHEE]

Thank you Star!

This definitely works as well,

I would "accept" all answers if I could ..

Best,

Tim.

[Star Strider]

My pleasure!

Answer 3

[John D'Errico]

I don't see why it should be a struggle. Start at the beginning. Break the problem down into small pieces. Then solve each of them. Here is an example problem.

You have two curves.

1. Find the point of intersection.
2. Integrate. That might be an analytical solution, it might be numerical.

F1 = @cos;
F2 = @(x) expm1(x);

% plot them
fplot(F1,[-10,2])
hold on
fplot(F2,[-10,2])

expm1 is the function exp(x)-1, a useful special function that seems to get overlooked. So I thought I'd use it here as an example. There are four regions that might be of interest. First, find the points of intersection.

F21 = @(x) F1(x) - F2(x);

We can use numerical tools to find the intersection point, so fzero, or even vpasolve. There will be 4 points of interest.

xstart = [-10 -9.42;-9.42 -8;-4 -3;-3 -2;0 1];
xcross = nan(size(xstart,1),1);
fval = xcross;
for i = 1:size(xstart,1)
  [xcross(i),fval(i)] = fzero(F21,xstart(i));
end
format long g
[xcross,fval]

ans = 5×2

-9.43740239690328 0 -9.41199210388775 0 -3.40059290398226 1.11022302462516e-16 -2.78912964643395 0 0.60134676772582 -1.11022302462516e-16

And just perform the integration. Symbolic tools or numerical tools both can apply.

areaBetween = zeros(4,1);
subplot(4,1,1)
for i = 1:4
  areaBetween(i) = integral(F21,xcross(i),xcross(i+1));
  subplot(4,1,i)
  fplot(F21,[xcross(i),xcross(i+1)])
end

areaBetween

areaBetween = 4×1

-1.367117078358e-06 6.2470272332719 -0.0179826714572249 2.5383406282387

Of course, I've produced a signed area, depending on which curve is greater in the corresponding region.

[TIMOTHEE]

Thank you John,

The step by step process will surely come in handy with other similar issues.

Always super helpful to have in one's toolbox.

Best regards,

Tim.