How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{3} \frac{d x}{1 + x^{2} + x^{4}} d x$ $int_0^3dx/(1+x^2+x^4)dx$ ?

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How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{3} \frac{d x}{1 + x^{2} + x^{4}} d x$ $int_0^3dx/(1+x^2+x^4)dx$ ?

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Answer 1 · 2014-09-16T06:41:23

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

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

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

The formula is:

$T = \frac{h}{2} [f (x_{0}) + 2 f (x_{1}) + 2 f (x_{2}) + 2 f (x_{3}) + 2 f (x_{4}) + 2 f (x_{5}) + f (x_{6})]$ $T=h/2[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+2f(x_4)+2f(x_5)+f(x_6)]$
$f (x_{0}) = f (0) = \frac{1}{1}$ $f(x_0)=f(0)=1/1$
$f (x_{1}) = f (\frac{1}{2}) = \frac{1}{1 + \frac{1}{4} + \frac{1}{16}} = \frac{16}{21}$ $f(x_1)=f(1/2)=1/(1+1/4+1/16)=16/21$
$f (x_{2}) = f (1) = \frac{1}{1 + 1 + 1} = \frac{1}{3}$ $f(x_2)=f(1)=1/(1+1+1)=1/3$
$f (x_{3}) = f (\frac{3}{2}) = \frac{1}{1 + \frac{9}{4} + \frac{81}{16}} = \frac{16}{133}$ $f(x_3)=f(3/2)=1/(1+9/4+81/16)=16/133$
$f (x_{4}) = f (2) = \frac{1}{1 + 4 + 16} = \frac{1}{21}$ $f(x_4)=f(2)=1/(1+4+16)=1/21$
$f (x_{5}) = f (\frac{5}{2}) = \frac{1}{1 + \frac{25}{4} + \frac{625}{16}} = \frac{16}{741}$ $f(x_5)=f(5/2)=1/(1+25/4+625/16)=16/741$
$f (x_{6}) = f (3) = \frac{1}{1 + 9 + 81} = \frac{1}{91}$ $f(x_6)=f(3)=1/(1+9+81)=1/91$
$T = \frac{1}{4} [1 + \frac{32}{21} + \frac{2}{3} + \frac{32}{133} + \frac{2}{21} + \frac{32}{741} + \frac{1}{91}]$ $T=1/4[1+32/21+2/3+32/133+2/21+32/741+1/91]$
$T = \frac{4643}{5187} \approx .89512$ $T=4643/5187~~.89512$

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

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How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{3} \frac{d x}{1 + x^{2} + x^{4}} d x$ $int_0^3dx/(1+x^2+x^4)dx$ ?

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How do you Use the trapezoidal rule with $n = 6$ $n=6$ to approximate the integral $\int_{0}^{3} \frac{d x}{1 + x^{2} + x^{4}} d x$ $int_0^3dx/(1+x^2+x^4)dx$ ?