# How do you find the area using the trapezoidal approximation method, given f(x)=5 sqrt(1+sqrt(x)) , on the interval [0, 4] with n=8?

Jul 7, 2017

${\int}_{0}^{4} \setminus 5 \sqrt{1 + \sqrt{x}} \setminus \mathrm{dx} \approx 30.2149$

#### Explanation:

We have:

$f \left(x\right) = 5 \sqrt{1 + \sqrt{x}}$

We want to calculate over the interval $\left[0 , 4\right]$ with $8$ strips; thus:

$\Delta x = \frac{4 - 0}{8} = \frac{1}{2}$

Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). The values of the function are tabulated as follows;

Trapezium Rule

 int_0^4 \ 5sqrt(1+sqrt(x)) \ dx ~~ 0.5/2 * { 5 + 8.6603 +
 " " 2*(6.5328 + 7.0711 + 7.4578 +
 " " 7.7689 + 8.033 + 8.2645 + 8.4718) }

$\text{ } = 0.25 \cdot \left\{13.6603 + 2 \cdot \left(53.5997\right)\right\}$
$\text{ } = 0.25 \cdot \left\{13.6603 + 107.1994\right\}$
$\text{ } = 0.25 \cdot 120.8597$
$\text{ } = 30.2149$

Actual Value

For comparison of accuracy:

${\int}_{0}^{4} \setminus 5 \sqrt{1 + \sqrt{x}} \setminus \mathrm{dx} = 30.3794795877687 \ldots$