Approximate $$$\int\limits_{-5}^{5} x^{2}\, dx$$$ with $$$n = 2$$$ using the left endpoint approximation

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

The left Riemann sum (also known as the left endpoint approximation) uses the left endpoint of a subinterval for computing the height of the approximating rectangle:

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

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 left endpoints of the subintervals.

$$$f{\left(x_{0} \right)} = f{\left(-5 \right)} = 25$$$

$$$f{\left(x_{1} \right)} = f{\left(0 \right)} = 0$$$

Finally, just sum up the above values and multiply by $$$\Delta x = 5$$$: $$$5 \left(25 + 0\right) = 125$$$.

$$$\int\limits_{-5}^{5} x^{2}\, dx\approx 125$$$A