Approximate $$$\int\limits_{0}^{4} x^{2}\, dx$$$ with $$$n = 4$$$ using the left endpoint approximation

Approximate the integral $$$\int\limits_{0}^{4} x^{2}\, dx$$$ with $$$n = 4$$$ 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 $$$f{\left(x \right)} = x^{2}$$$, $$$a = 0$$$, $$$b = 4$$$, and $$$n = 4$$$.

Therefore, $$$\Delta x = \frac{4 - 0}{4} = 1$$$.

Divide the interval $$$\left[0, 4\right]$$$ into $$$n = 4$$$ subintervals of the length $$$\Delta x = 1$$$ with the following endpoints: $$$a = 0$$$, $$$1$$$, $$$2$$$, $$$3$$$, $$$4 = b$$$.

Now, just evaluate the function at the left endpoints of the subintervals.

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

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

$$$f{\left(x_{2} \right)} = f{\left(2 \right)} = 4$$$

$$$f{\left(x_{3} \right)} = f{\left(3 \right)} = 9$$$

Finally, just sum up the above values and multiply by $$$\Delta x = 1$$$: $$$1 \left(0 + 1 + 4 + 9\right) = 14$$$.

$$$\int\limits_{0}^{4} x^{2}\, dx\approx 14$$$A