# How do you find the area using the trapezoidal approximation method, given sqrtx, on the interval [1,4] with n=3?

Jan 4, 2017

${\int}_{1}^{4} \sqrt{x} \setminus \mathrm{dx} = 4.646$ (4dp)

(For comparison the exact value is $4.6666 \dot{6}$)

#### Explanation:

The values of $f \left(x\right) = \sqrt{x}$ are tabulated as follows (using Excel) working to 6dp.

The Trapezium Rule:

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

uses a series of two consecutive ordinates and a best fit straight line to form trapeziums to approximate the area under a curve, It will have 100% accuracy if $y = f \left(x\right)$ is a straight line and typically provides a good estimate provided $n$ is chosen appropriately.

So,

${\int}_{1}^{4} \sqrt{x} \setminus \mathrm{dx} = \frac{1}{2} \left\{\left(1 + 2\right) + 2 \left(1.414213 + 1.73205\right)\right\}$
$\text{ } = 0.5 \left\{+ 3 + 2 \left(3.146264\right)\right\}$
$\text{ } = 0.5 \left\{+ 3 + 6.292528\right\}$
$\text{ } = 0.5 \left\{+ 9.292528\right\}$
$\text{ } = 4.646264$

Exact Value:

${\int}_{1}^{4} \sqrt{x} \setminus \mathrm{dx} = {\left[\frac{2}{3} {x}^{\frac{3}{2}}\right]}_{1}^{4}$
$\text{ } = \frac{2}{3} \left\{{4}^{\frac{3}{2}} - {1}^{\frac{3}{2}}\right\}$
$\text{ } = \frac{2}{3} \left(8 - 1\right)$
$\text{ } = \frac{14}{3}$
$\text{ } = 4.6666 \dot{6}$