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I am trying to repeat what is in the book to calculate ##f_1^e##

Given

##\displaystyle f_i^e=\int_{x_e}^{x_{e+1}} f \psi_i^e (x) dx##

where ##\psi_1^e(x)=1- \frac{x}{h_e}##

##x_{e+1}-x_e=h_e##

##f(x)=6.25(1+x)##

I calculate ##\displaystyle f_1^e=\int_{x_e}^{x_{e+1}} 6.25(1+x) (1-x/h_e) dx=6.25\int_{x_e}^{x_{e+1}} (1-x/h_e +x - x^2/h_e) dx##

##\displaystyle=6.25(x-\frac{x^2}{2h_e}+\frac{x^2}{2}-\frac{x^3}{3h_e})|_{x_e}^{x_{e+1}}##

I am not sure how my work can arrive at the book answer below...?

The book calculates ##\displaystyle f_1^e=\frac{6.25h_e}{2}(1 +\frac{x_{e+1}+2x_e}{3})##

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# Integral Problem

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