# How do you use the Trapezoidal Rule and the Simpson's Rule when n=4 when approximating the integral  (5t + 6) dt from [3,6]?

Sep 23, 2015

${\int}_{3}^{6} \left(5 t + 6\right) \mathrm{dt} \approx 85.5$

#### Explanation:

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} \approx \frac{\Delta x}{2} \left(f \left({t}_{0}\right) + 2 f \left({t}_{1}\right) + 2 f \left({t}_{2}\right) + \ldots + 2 f \left({t}_{n - 1}\right) + f \left({t}_{n}\right)\right)$
Where $\Delta t = \frac{b - a}{2}$
We have $a = 3 , b = 6 , n = 4$
Therefore, $\Delta t = \frac{6 - 3}{4} = \frac{3}{4}$
Next, we divide [3,6] into four equal intervals with the length $\frac{3}{4}$
This will give us the following endpoints:
$3 , \frac{15}{4} , \frac{9}{2} , \frac{21}{4} , 6$

Now we simply evaluate the function at these endpoints;
$f \left({t}_{0}\right) = f \left(a\right) = f \left(3\right) = 21$
$2 f \left({t}_{1}\right) = 2 f \left(\frac{15}{4}\right) = \frac{99}{2} = 49.5$
$2 f \left({t}_{2}\right) = 2 f \left(\frac{9}{2}\right) = 57$
$2 f \left({t}_{3}\right) = 2 f \left(\frac{21}{4}\right) = \frac{129}{2} = 64.5$
$f \left({t}_{4}\right) = f \left(b\right) = f \left(6\right) = 36$

Now, we just sum up the values and multiply by $\frac{\Delta t}{2} = \frac{3}{8}$
$= \frac{3}{8} \left(21 + 49.5 + 57 + 64.5 + 36\right) = 85.5$

Hope this helps :)