How do you use the trapezoidal rule with n=3 to approximate the area between the curve y=x^2 and the x-axis for 1 ≤ t ≤ 4?

1 Answer
Sep 3, 2015

Answer:

#"Area"=22.125" square units"#

Explanation:

NB : Assuming that the interval is #color(blue)(1<=x<=4)#and not #color(red)(1<=t<=4)#

The trapezoidal rule says that the area can be found using the formula:
#"Area"=h/2(y_1+y_n+2(y_2+y_3+...+y_(n-1)))#

where #h# is the step length(the length of a single strip), #y_1,y_2,...,y_n# are the #y# values corresponding to each #x# value taken from the interval #[1,4]#

In this problem, since #n=3#,
#"Area"=h/2(y_1+y_3+2(y_2))#

First,
#h=(x_n-x_1)/(n-1)#

#x_1# and #x_n# are respectively the first and last #x# values in the interval

#h=(4-1)/(3-1)=3/2#

Next,
#y_1=(x_1)^2=(1)^2=1#
#y_2=(1+3/2)^2=(5/2)^2=25/4#
#y_3=(5/2+3/2)^2=(4)^2=16#

NB : In order to get the next #x# we add #3/2# to the previous.
Example : #x_2=1+3/2=5/2#

Finally,

#"Area"=(3/2)/2(1+16+2(25/4))~=color(blue)(22.125)#