# cumsum function question for integration

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haohaoxuexi1 2022 年 9 月 21 日

The code is below.
close all;
clear all;
clc;
format long;
syms t;
inter_val = 0.0001;
tn=0:inter_val:20;
t=tn;
y_cos = sin(t);
int_y_cos=cumsum(y_cos*inter_val);
figure(1);
plot(t, y_cos, 'k');
box on; grid on; figure(2);
plot(t, int_y_cos, 'r');
box on; grid on; I want to know why figure(2) didn't come out with cos plot but with offset?
But switch y_cos = sin(t) to y_cos = cos(t), it can give a perfect sin plot?
Thanks,

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### 採用された回答

Torsten 2022 年 9 月 21 日

cumsum(y_cos*inter_val);
is a Riemann sum for the integral of sin(t) - integrated between 0 and x.
But the integral of sin(t) from 0 to x is 1-cos(x), not cos(x).

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### その他の回答 (1 件)

Cris LaPierre 2022 年 9 月 21 日
It has to do with the fact that sin(t) starts at 0 and is positive until pi. cumsum is just adding up the Y values, so it oscillates between 0 and 2. With cos(t), it is only positive for pi/2, so cumsum only gets to half the value before it starts decreasing, or 1. It then is negative for pi, decreasing the value by 2, reaching -1. It therefore oscillates between -1 and 1.

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