Approximate $$$\int\limits_{-4}^{0} \left(x^{3} + 1\right)\, dx$$$ with $$$n = 4$$$ using the left endpoint approximation

Approximate the integral $$$\int\limits_{-4}^{0} \left(x^{3} + 1\right)\, dx$$$ 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 $$$f{\left(x \right)} = x^{3} + 1$$$, $$$a = -4$$$, $$$b = 0$$$, and $$$n = 4$$$.

Therefore, $$$\Delta x = \frac{0 - \left(-4\right)}{4} = 1$$$.

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

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

$$$f{\left(x_{0} \right)} = f{\left(-4 \right)} = -63$$$

$$$f{\left(x_{1} \right)} = f{\left(-3 \right)} = -26$$$

$$$f{\left(x_{2} \right)} = f{\left(-2 \right)} = -7$$$

$$$f{\left(x_{3} \right)} = f{\left(-1 \right)} = 0$$$

Finally, just sum up the above values and multiply by $$$\Delta x = 1$$$: $$$1 \left(-63 - 26 - 7 + 0\right) = -96$$$.

$$$\int\limits_{-4}^{0} \left(x^{3} + 1\right)\, dx\approx -96$$$A