Riemann Sum Calculator for a Function

Approximate the integral $$$\int\limits_{0}^{2} \sqrt[3]{x^{4} + 1}\, dx$$$ with $$$n = 4$$$ using the left Riemann sum.

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 $$$f{\left(x \right)} = \sqrt[3]{x^{4} + 1}$$$, $$$a = 0$$$, $$$b = 2$$$, and $$$n = 4$$$.

Therefore, $$$\Delta x = \frac{2 - 0}{4} = \frac{1}{2}$$$.

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

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

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

$$$f{\left(x_{1} \right)} = f{\left(\frac{1}{2} \right)} = \frac{\sqrt[3]{17} \cdot 2^{\frac{2}{3}}}{4}\approx 1.020413775479337$$$

$$$f{\left(x_{2} \right)} = f{\left(1 \right)} = \sqrt[3]{2}\approx 1.259921049894873$$$

$$$f{\left(x_{3} \right)} = f{\left(\frac{3}{2} \right)} = \frac{2^{\frac{2}{3}} \sqrt[3]{97}}{4}\approx 1.82340825744217$$$

Finally, just sum up the above values and multiply by $$$\Delta x = \frac{1}{2}$$$: $$$\frac{1}{2} \left(1 + 1.020413775479337 + 1.259921049894873 + 1.82340825744217\right) = 2.55187154140819.$$$

$$$\int\limits_{0}^{2} \sqrt[3]{x^{4} + 1}\, dx\approx 2.55187154140819$$$A