This function gets increasingly oscillatory near the right endpoint.

f = @(x) (x+1).^2.*cos((2*x+1)./(x-4.3));

fplot(f,[0 4],2000)

xlabel('x'), ylabel('f(x)') % ignore this line

title('A wiggly integrand')

Accordingly, the trapezoid rule is more accurate on the left half of the interval than on the right half.

n_ = 50*2.^(0:3)';

T_ = [];

for i = 1:length(n_)

n = n_(i);

T_(i,1) = trapezoid(f,0,2,n);

T_(i,2) = trapezoid(f,2,4,n);

end

left_val = integral(f,0,2,'abstol',1e-14,'reltol',1e-14);

right_val = integral(f,2,4,'abstol',1e-14,'reltol',1e-14);

table(n_,T_(:,1)-left_val,T_(:,2)-right_val,...

'variablenames',{'n','left_error','right_error'})

ans = 4×3 table

	n	left_error	right_error
1	50	-2.4911e-03	5.0423e-01
2	100	-6.2271e-04	9.6004e-02
3	200	-1.5568e-04	2.2547e-02
4	400	-3.8919e-05	5.5542e-03

Both the picture and the numbers suggest that more nodes should be used on the right half of the interval than on the left half.