# Math Help - Integrating a linear polynomial approx

1. ## Integrating a linear polynomial approx

hi,

thanks for reading.can anyone help me out?
see the attached file.i am working out for many hours how to integrate P1(t). does anyone know?
how do you get -1/2hf(..... ) + 3/2hf(....) ?
thats all i need to know!
i tried integrating MANY times. it looks easy.so what am i not understanding?

thanks all

2. Hi

The first integral is

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}$

Substituting $t_{i+1} = t_i + h$

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}$

3. Originally Posted by running-gag
Hi

The first integral is

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i+1}^2}{2}- t_i t_{i+1}+\frac{t_i^2}{2}$

Substituting $t_{i+1} = t_i + h$

$\left[\frac{t^2}{2}-t_it\right]_{t_i}^{t_{i+1}} = \frac{t_{i}^2}{2} + h t_i + \frac{h^2}{2} - t_i^2 - h t_i + \frac{t_i^2}{2} = \frac{h^2}{2}$

aha...! thats where confusion happened... h can be anything =S
a difference between one point and another. thanks! =D