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

May 6, 2015

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

This gives us 4 trapezoidal areas
${A}_{T 1} = \left(\frac{\sqrt{2} + \sqrt{2.5}}{2}\right) \times 0.5$
${A}_{T 2} = \left(\frac{\sqrt{2.5} + \sqrt{3.0}}{2}\right) \times 0.5$
${A}_{T 3} = \left(\frac{\sqrt{3.0} + \sqrt{3.5}}{2}\right) \times 0.5$
${A}_{T 4} = \left(\frac{\sqrt{3.5} + \sqrt{4.0}}{2}\right) \times 0.5$

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