# How to you approximate the integral of  (t^3 +t) dx from [0,2] by using the trapezoid rule with n=4?

Apr 28, 2015

Dividing the range $\left[0 , 2\right]$ into 4 trapezoids of equal width gives each trapezoid a width of $\frac{1}{2}$

The area of each trapezoid is
$\text{(average height) " xx " (width)}$

For $f \left(x\right) = \left({t}^{3} + t\right)$
the heights of the trapezoids are
$f \left(0\right) = 0$
$f \left(\frac{1}{2}\right) = \frac{5}{8}$
$f \left(1\right) = 2$
$f \left(\frac{3}{2}\right) = \frac{39}{8}$
$f \left(2\right) = 10$

The sum of the areas of the trapezoids is
${A}_{t} = \left(\frac{0 + \frac{5}{8}}{2} \cdot \frac{1}{2}\right) + \left(\frac{\frac{5}{8} + 2}{2} \cdot \frac{1}{2}\right) + \left(\frac{2 + \frac{39}{8}}{2} \cdot \frac{1}{2}\right) + \left(\frac{\frac{39}{8} + 10}{2} \cdot \frac{1}{2}\right)$

$= \frac{1}{4} \cdot \left(\left(0 + 10\right) + 2 \cdot \left(\frac{5}{8} + 2 + \frac{39}{8}\right)\right)$

$= 6 \frac{1}{4}$

So ${\int}_{0}^{2} \left({t}^{3} + t\right) \mathrm{dt}$
is approximately $6 \frac{1}{4}$