How do you use the trapezoidal rule with n=4 to approximate the area between the curve  (5t + 6) from 3 to 6?

Jan 3, 2017

${\int}_{3}^{6} 5 t + 6 \setminus \mathrm{dt} = 85.5$

Explanation:

The required area is a trapezium with sides $21$ and $36$ and width $3$, so we can calculate the exact area as $A = \frac{1}{2} \left(21 + 36\right) 3 = 85.5$. As the trapezium rule estimates the area under a curve using trapeziums we would expect to get an exact answer.

The values of $f \left(x\right) = 5 x + 6$ are tabulated as follows (using Excel) working to 5dp

Using the trapezoidal rule:

${\int}_{a}^{b} y \mathrm{dx} \approx \frac{h}{2} \left\{\left({y}_{0} + {y}_{n}\right) + 2 \left({y}_{1} + {y}_{2} + \ldots + {y}_{n - 1}\right)\right\}$

We have:

${\int}_{3}^{6} 5 t + 6 \setminus \mathrm{dt} \approx \frac{0.75}{2} \left\{21 + 36 + 2 \left(24.75 + 28.5 + 32.25\right)\right\}$
$\text{ } = 0.375 \left\{57 + 2 \left(85.5\right)\right\}$
$\text{ } = 0.375 \left\{57 + 171\right\}$
$\text{ } = 0.375 \left\{228\right\}$
$\text{ " = 85.5 " }$, as predicted