Determine the function representing the height of the region.

The height $h(x)$ of the region between the two curves is given by the upper curve minus the lower curve.

$$h(x) = (2x^2 + 1) - (x^2 + 1) = 2x^2 + 1 - x^2 - 1 = x^2$$

So, we need to estimate the area under the curve $y = x^2$ from $x = 0$ to $x = 5$.

The interval is from $x = 0$ to $x = 5$, and we are using 5 strips.

$$h = \frac{\text{upper limit} - \text{lower limit}}{\text{number of strips}} = \frac{5 - 0}{5} = 1$$

The x-values are $x_0 = 0, x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4, x_5 = 5$.
The corresponding y-values are:

$$y_0 = 0^2 = 0$$ $$y_1 = 1^2 = 1$$ $$y_2 = 2^2 = 4$$ $$y_3 = 3^2 = 9$$ $$y_4 = 4^2 = 16$$ $$y_5 = 5^2 = 25$$

The trapezium rule formula is:

$$\text{Area} \approx \frac{h}{2} [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})]$$

Substitute the values:

$$\text{Area} \approx \frac{1}{2} [y_0 + y_5 + 2(y_1 + y_2 + y_3 + y_4)]$$ $$\text{Area} \approx \frac{1}{2} [0 + 25 + 2(1 + 4 + 9 + 16)]$$ $$\text{Area} \approx \frac{1}{2} [25 + 2(30)]$$ $$\text{Area} \approx \frac{1}{2} [25 + 60]$$ $$\text{Area} \approx \frac{1}{2} [85]$$ $$\text{Area} \approx 42.5$$

The estimated area is $\text{42.5}$.