Approximate $$$\int\limits_{-5}^{5} x^{2}\, dx$$$ with $$$n = 2$$$ using the midpoint rule

Approximate the integral $$$\int\limits_{-5}^{5} x^{2}\, dx$$$ with $$$n = 2$$$ using the midpoint rule.

The midpoint rule (also known as the midpoint approximation) uses the midpoint of a subinterval for computing the height of the approximating rectangle:

$$$\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \Delta x \left(f{\left(\frac{x_{0} + x_{1}}{2} \right)} + f{\left(\frac{x_{1} + x_{2}}{2} \right)} + f{\left(\frac{x_{2} + x_{3}}{2} \right)}+\dots+f{\left(\frac{x_{n-2} + x_{n-1}}{2} \right)} + f{\left(\frac{x_{n-1} + x_{n}}{2} \right)}\right)$$$

where $$$\Delta x = \frac{b - a}{n}$$$.

We have that $$$f{\left(x \right)} = x^{2}$$$, $$$a = -5$$$, $$$b = 5$$$, and $$$n = 2$$$.

Therefore, $$$\Delta x = \frac{5 - \left(-5\right)}{2} = 5$$$.

Divide the interval $$$\left[-5, 5\right]$$$ into $$$n = 2$$$ subintervals of the length $$$\Delta x = 5$$$ with the following endpoints: $$$a = -5$$$, $$$0$$$, $$$5 = b$$$.

Now, just evaluate the function at the midpoints of the subintervals.

$$$f{\left(\frac{x_{0} + x_{1}}{2} \right)} = f{\left(\frac{-5 + 0}{2} \right)} = f{\left(- \frac{5}{2} \right)} = \frac{25}{4} = 6.25$$$

$$$f{\left(\frac{x_{1} + x_{2}}{2} \right)} = f{\left(\frac{0 + 5}{2} \right)} = f{\left(\frac{5}{2} \right)} = \frac{25}{4} = 6.25$$$

Finally, just sum up the above values and multiply by $$$\Delta x = 5$$$: $$$5 \left(6.25 + 6.25\right) = 62.5$$$.

$$$\int\limits_{-5}^{5} x^{2}\, dx\approx 62.5$$$A