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- Feb 22nd 2006, 12:49 PMfrozenflamesCalculus Help
- Feb 22nd 2006, 09:03 PMCaptainBlackQuote:

Originally Posted by**frozenflames**

$\displaystyle

\int_a^b f(x)\ dx=\frac{h}{2} (y_0 + 2y_1+2y_2+...+2y_{n-1}+y_n)

$

but your data don't have a fixed tabulation interval so we need an alternate

definition of the trapezoidal approximation, which is:

$\displaystyle

\int_a^b f(x)\ dx=\sum_1^{n-1}\ (x_{i+1}-x_i)(f(x_{i+1})+f(x_i))/2

$,

which in this case is:

$\displaystyle

\int_a^b f(x)\ dx=3\times 20+2 \times 35 + 1 \times 30 = 160

$

RonL