How do you use the trapezoidal rule with n=4 to approximate the area between the curve #1/(1 + x^2) # from 0 to 6?

1 Answer
Jul 28, 2015

Answer:

Use the formula : #Area=h/2(y_1+y_n+2(y_2+y_3+...+y_(n-1)))#
to obtain the result :
#Area=4314/3145~=1.37#

Explanation:

#h# is the step length
We find the step length using the following formula : #h=(b-a)/(n-1)#

#a# is the minimum value of #x# and #b# is the maximum value of #x#. In our case #a=0# and #b=6#

#n# is the number of strips. Hence #n=4#

#=>h=(6-0)/(4-1)=2#

So, the values of #x# are #0,2,4,6#

#"NB :"# Starting from #x=0# we add the step length #h=2# to get the next value of #x# up to #x=6#

In order to find #y_1# up to #y_n#(or #y_4#) we plug-in each value of #x# to get the corresponding #y#

For example : to get #y_1# we plug-in #x=0# in #y=1/(1+x^2)#

#=>y_1=y=1/(1+(0)^2)=1#

For #y_2# we plug-in #x=2# to have : #y_2=1/(1+(2)^2)=1/5#

Similarly,

#y_3=1/(1+(4)^2)=1/17#

#y_4=1/(1+(6)^2)=1/37#

Next, we use the formula,

#Area=h/2(y_1+y_n+2(y_2+y_3+...+y_(n-1)))#

#=>Area=2/2[1+1/5+2(1/17+1/37)]=(3145+629+370+170)/3145=color(blue)(4314/3145)#