# How do you find the area using the trapezoidal approximation method, given (5t + 6) dt , on the interval [3, 6] with n=4?

Jan 31, 2017

$85.5 \setminus {\text{unit}}^{2}$

#### Explanation:

The values of $f \left(t\right) = 5 t + 6$ 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) \ 5t+6 \ dt ~~ 0.75/2 { (21 + 36) + "
 " " 2(24.75 + 28.5 + 32.25) }
$\text{ } = 0.375 \left\{+ 57 + 2 \left(85.5\right)\right\}$
$\text{ } = 0.375 \left\{+ 57 + 171\right\}$
$\text{ } = 0.375 \left\{+ 228\right\}$
$\text{ } = 85.5$

It is always worth thinking about the interpretation of a mathematical question and/or result. In this case we are being asked to find the area under a straight line, which will be a trapezium

graph{5x+6 [-3, 8, -5, 55]}

So we can calculate the exact area:

$A = \left(\frac{1}{2}\right) \left(21 + 36\right) \left(3\right) = 85.5$

It is no coincidence that the above trapezium rule calculation gave us the exact result, because obviously the rule estimates the area of integration with small width trapeziums. In the case of a straight line those trapeziums are exact.