Using n=4 trapezoids, how do you approximate the value of #int sqrt(x+1) dx# from [1,3]?

1 Answer
May 6, 2015

To approximate #int_1^3sqrt(x+1) dx# by a series of 4 trapezoids
we need to evaluate #sqrt(x+1)# at 5 points #x={1.0, 1.5, 2.0, 2.5, 3.0}#
(using a constant #Delta x# of #0.5#)

This gives us 4 trapezoidal areas
#A_(T1) = ((sqrt(2)+sqrt(2.5))/2)xx0.5#
#A_(T2) = ((sqrt(2.5)+sqrt(3.0))/2)xx0.5#
#A_(T3) = ((sqrt(3.0)+sqrt(3.5))/2)xx0.5#
#A_(T4) = ((sqrt(3.5)+sqrt(4.0))/2)xx0.5#

Adding together the 4 trapezoidal areas (with the aid of a calculator) gives an approximation of the integral #=3.445563#