The solution for l4 is mathematically given as
L_{4}=0.5431
What is the solution for l4?
[tex]&\text { Given } f(x)=a \cos ^{2}(x) \quad\left[\frac{\pi}{8}, \frac{\pi}{2}\right] \text { \& } n=4\\\\&\Delta a=\frac{b-a}{n}=\frac{\pi / 2-\pi / 8}{9}=\frac{3 \pi}{32}\\\\&x_{0}=\pi / 8, x_{1}=\frac{\pi}{8}+\frac{3 \pi}{32}=\frac{7 \pi}{32}\\\\&x_{2}=\frac{5 \pi}{16}, \quad x_{3}=\frac{13 \pi}{32}, x_{4}=\pi / 2\\\\&f\left(x_{0}\right)=f(1 / 8)=0.8535\\[/tex]
[tex]&f\left(x_{1}\right)=f\left(\frac{7 \pi}{32}\right)=0.5975\\\\\&f\left(x_{2}\right)=f\left(\frac{5 \pi}{16}\right)=0.3086\\\\\&f\left(a_{3}\right)=f\left(\frac{13 \pi}{32}\right)=0.0842\\\\\&L_{4}=\sum_{k_{0}^{=0}}^{3} f\left(x_{k}\right) \Delta x\\\\&=\Delta \\\\x\left[f\left(x_{0}\right)+f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)\right]\\\\\&=\frac{3 \pi}{32}[0.8535+0.5975+0.3056+0.0842]\\\\&L_{4}=0.5431[/tex]
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