Code: In the code below in trapz we have taken x value at space of 1 from -3 to 1
clc
clear
syms x
f = 2*x^2 + 2*x - 1;
I_int = int(f, -3, 1)
x = -3:1;
y = @(x)2.*x.^2 + 2.*x - 1;
I_trapz = trapz(x, y(x))
I_integral = integral(y, -3, 1)
Output:
I_int =
20/3
I_trapz =
8
I_integral =
6.6667
Code: But if we take smaller step size in x for trapz it gives better accuracy
clc
clear
syms x
f = 2*x^2 + 2*x - 1;
I_int = int(f, -3, 1)
x = -3:0.01:1; %just taken a smaller step size
y = @(x)2.*x.^2 + 2.*x - 1;
I_trapz = trapz(x, y(x))
I_integral = integral(y, -3, 1)
Output:
I_int =
20/3
I_trapz =
6.6668
I_integral =
6.6667
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Produce following function in MATLAB
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eeeceved. 3) calculatelmpulse: consumes a series (represented by a column-vector of floats)-...
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