Approximate $$$\int\limits_{-1}^{3} \left(15 x^{2} + 24 x + 1\right)\, dx$$$ with $$$n = 3$$$ using the left endpoint approximation

Approximate the integral $$$\int\limits_{-1}^{3} \left(15 x^{2} + 24 x + 1\right)\, dx$$$ with $$$n = 3$$$ 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 $$$f{\left(x \right)} = 15 x^{2} + 24 x + 1$$$, $$$a = -1$$$, $$$b = 3$$$, and $$$n = 3$$$.

Therefore, $$$\Delta x = \frac{3 - \left(-1\right)}{3} = \frac{4}{3}$$$.

Divide the interval $$$\left[-1, 3\right]$$$ into $$$n = 3$$$ subintervals of the length $$$\Delta x = \frac{4}{3}$$$ with the following endpoints: $$$a = -1$$$, $$$\frac{1}{3}$$$, $$$\frac{5}{3}$$$, $$$3 = b$$$.

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

$$$f{\left(x_{0} \right)} = f{\left(-1 \right)} = -8$$$

$$$f{\left(x_{1} \right)} = f{\left(\frac{1}{3} \right)} = \frac{32}{3}\approx 10.666666666666667$$$

$$$f{\left(x_{2} \right)} = f{\left(\frac{5}{3} \right)} = \frac{248}{3}\approx 82.666666666666667$$$

Finally, just sum up the above values and multiply by $$$\Delta x = \frac{4}{3}$$$: $$$\frac{4}{3} \left(-8 + 10.666666666666667 + 82.666666666666667\right) = 113.777777777777779.$$$

$$$\int\limits_{-1}^{3} \left(15 x^{2} + 24 x + 1\right)\, dx\approx 113.777777777777779$$$A