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

1 Answer
Sep 16, 2014

The answer is #4643/5187#.

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

#a=0#
#b=3#
#n=6#
#f(x)=1/(1+x^2+x^4)#
#h=(b-a)/n=1/2#

The formula is:

#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)=1/1#
#f(x_1)=f(1/2)=1/(1+1/4+1/16)=16/21#
#f(x_2)=f(1)=1/(1+1+1)=1/3#
#f(x_3)=f(3/2)=1/(1+9/4+81/16)=16/133#
#f(x_4)=f(2)=1/(1+4+16)=1/21#
#f(x_5)=f(5/2)=1/(1+25/4+625/16)=16/741#
#f(x_6)=f(3)=1/(1+9+81)=1/91#
#T=1/4[1+32/21+2/3+32/133+2/21+32/741+1/91]#
#T=4643/5187~~.89512#

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