# How can i find the areas under single parts of a plot?

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Pietro Fiondella el 2 de Jul. de 2022
Respondida: Voss el 2 de Jul. de 2022
I have this vector P wich represent the istant power used by a certain load in 24 h.
I would like to obtain a vector a vector E [1x24] whit the area under the segment given by 2 consecutive element of P.
I tryed this
%Power
P=[ 0.6495 0.7393 0.7551 0.7947 0.7974 0.8713 1.0482 1.2621 1.6027 1.7189 1.5710 1.3070 1.4416 1.5684 1.608 1.6053 1.6581 2.2865 2.4925 2.6087 2.3288 2.1941 1.9671 1.6265];
%Hours
H=[0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23];
%I can plot these 2 ad obtain
plot(H,P);
Now i can find the whole area of the plot whit "trapz" or the cumulative area whit "cumtrapz" like this
trapz(H,P)
ans = 35.3648
cumtrapz(H,P)
ans = 1×24
0 0.6944 1.4416 2.2165 3.0126 3.8469 4.8066 5.9618 7.3942 9.0550 10.6999 12.1389 13.5132 15.0182 16.6064 18.2131 19.8448 21.8171 24.2066 26.7572 29.2259 31.4874 33.5680 35.3648
As i said I would like to obtain instead a vector E [1x23] whit the area under the segment given by 2 consecutive element of P.
For example for the first 2 element
E1=trapz(P(1,1:2))
E1 = 0.6944
E2=trapz(P(1,2:3))
E2 = 0.7472
I would like to avoid to write manually each element of E, is it possible?
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Voss el 2 de Jul. de 2022
%Power
P=[ 0.6495 0.7393 0.7551 0.7947 0.7974 0.8713 1.0482 1.2621 1.6027 1.7189 1.5710 1.3070 1.4416 1.5684 1.608 1.6053 1.6581 2.2865 2.4925 2.6087 2.3288 2.1941 1.9671 1.6265];
%Hours
H=[0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23];
diff(cumtrapz(H,P))
ans = 1×23
0.6944 0.7472 0.7749 0.7961 0.8343 0.9597 1.1551 1.4324 1.6608 1.6449 1.4390 1.3743 1.5050 1.5882 1.6066 1.6317 1.9723 2.3895 2.5506 2.4688 2.2614 2.0806 1.7968
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