Integral Problem: Calculating f_1^e

[bugatti79]

Folks,

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 $\displaystyle f_1^e=\frac{6.25h_e}{2}(1 +\frac{x_{e+1}+2x_e}{3})$

 

[mathman]

It doesn't look right. Check your function definitions.