$\displaystyle \int_{0}^{1}13\cos(x)^{2} dx$

n = 4

values of intervals

are $\displaystyle y_{0} = 13$

$\displaystyle y_{1} = 12.9997525 $

$\displaystyle y_{2} = 12.99901002 $

$\displaystyle y_{3} = 12.99777261 $

$\displaystyle y_{4} = 12.996040 $

change in $\displaystyle x = \dfrac{b - a}{n} = \dfrac{1 - 0}{4} = 1/4$

By the trapezoidal rule

$\displaystyle \dfrac{change-in-x}{2}(y_{0} + 2(y_{1}) + 2(y_{2}) +2(y_3}) + y_{4})$

I got 12.998638 (rounded to 6 decimal places but this is wrong.