# How do you Use the trapezoidal rule with n=6 to approximate the integral int_0^3dx/(1+x^2+x^4)dx?

Sep 16, 2014

The answer is $\frac{4643}{5187}$.

The trapezoidal rule is just a formula. From what we are given, we have:

$a = 0$
$b = 3$
$n = 6$
$f \left(x\right) = \frac{1}{1 + {x}^{2} + {x}^{4}}$
$h = \frac{b - a}{n} = \frac{1}{2}$

The formula is:

$T = \frac{h}{2} \left[f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + 2 f \left({x}_{3}\right) + 2 f \left({x}_{4}\right) + 2 f \left({x}_{5}\right) + f \left({x}_{6}\right)\right]$
$f \left({x}_{0}\right) = f \left(0\right) = \frac{1}{1}$
$f \left({x}_{1}\right) = f \left(\frac{1}{2}\right) = \frac{1}{1 + \frac{1}{4} + \frac{1}{16}} = \frac{16}{21}$
$f \left({x}_{2}\right) = f \left(1\right) = \frac{1}{1 + 1 + 1} = \frac{1}{3}$
$f \left({x}_{3}\right) = f \left(\frac{3}{2}\right) = \frac{1}{1 + \frac{9}{4} + \frac{81}{16}} = \frac{16}{133}$
$f \left({x}_{4}\right) = f \left(2\right) = \frac{1}{1 + 4 + 16} = \frac{1}{21}$
$f \left({x}_{5}\right) = f \left(\frac{5}{2}\right) = \frac{1}{1 + \frac{25}{4} + \frac{625}{16}} = \frac{16}{741}$
$f \left({x}_{6}\right) = f \left(3\right) = \frac{1}{1 + 9 + 81} = \frac{1}{91}$
$T = \frac{1}{4} \left[1 + \frac{32}{21} + \frac{2}{3} + \frac{32}{133} + \frac{2}{21} + \frac{32}{741} + \frac{1}{91}\right]$
$T = \frac{4643}{5187} \approx .89512$

Using numeric integration on a graphing calculator, we get $T \approx .89537$. So our answer is good for 3 sig figs.