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$\displaystyle L^i(x)= f(x_{i-1}) + \frac{f(x_i) - f(x_{i-1})}{x_i - x_{i-1}} (x-x_{i-1})$

Show that this linear approximation gives the trapezoidal rule.

I know the formula of the trapezoid rule, but not how and where to use this L at all.

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This result comes directly from the Taylor Series expansion (http://mathworld.wolfram.com/TaylorSeries.html) of f(x) , $T_f(x)$ about the point $x_{i-1}$:

$$T_f(x)=f(x_{i-1})+\frac{df}{dx}(x-x_{i-1})+...$$

The linear approximation is the result of ignoring all higher than first order derivatives, giving:

$$L^i(x)=f(x_{i-1})+\frac{df}{dx}(x-x_{i-1})$$ Now the first order derivative is approximated by a forward finite difference approximation of a derivative (http://en.wikipedia.org/wiki/Finite_difference):

$$\frac{df}{dx}\approx\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}$$

Putting it all together gives you the desired result:

$$L^i(x)\approx f(x_{i-1})+\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}(x-x_{i-1})$$

EDIT : All higher order Newton-Cotes approximations are derived by the same method, just using higher order approximations of the Taylor series.

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