Simpson's Rule Calculator for a Function

Approximate the integral $$$\int\limits_{0}^{1} \frac{1}{\sqrt[3]{x^{5} + 7}}\, dx$$$ with $$$n = 4$$$ using the Simpson's rule.

The Simpson's 1/3 rule (also known as the parabolic rule) uses parabolas to approximate the area:

$$$\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \frac{\Delta x}{3} \left(f{\left(x_{0} \right)} + 4 f{\left(x_{1} \right)} + 2 f{\left(x_{2} \right)} + 4 f{\left(x_{3} \right)} + 2 f{\left(x_{4} \right)}+\dots+4 f{\left(x_{n-3} \right)} + 2 f{\left(x_{n-2} \right)} + 4 f{\left(x_{n-1} \right)} + f{\left(x_{n} \right)}\right)$$$

where $$$\Delta x = \frac{b - a}{n}$$$.

We have that $$$f{\left(x \right)} = \frac{1}{\sqrt[3]{x^{5} + 7}}$$$, $$$a = 0$$$, $$$b = 1$$$, and $$$n = 4$$$.

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

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

Now, just evaluate the function at these endpoints.

$$$f{\left(x_{0} \right)} = f{\left(0 \right)} = \frac{7^{\frac{2}{3}}}{7}\approx 0.52275795857471$$$

$$$4 f{\left(x_{1} \right)} = 4 f{\left(\frac{1}{4} \right)} = \frac{32 \sqrt[3]{2} \cdot 7169^{\frac{2}{3}}}{7169}\approx 2.09093460413808$$$

$$$2 f{\left(x_{2} \right)} = 2 f{\left(\frac{1}{2} \right)} = \frac{4 \sqrt[3]{15} \cdot 2^{\frac{2}{3}}}{15}\approx 1.043964704311697$$$

$$$4 f{\left(x_{3} \right)} = 4 f{\left(\frac{3}{4} \right)} = \frac{32 \sqrt[3]{2} \cdot 7411^{\frac{2}{3}}}{7411}\approx 2.067923042238355$$$

$$$f{\left(x_{4} \right)} = f{\left(1 \right)} = \frac{1}{2} = 0.5$$$

Finally, just sum up the above values and multiply by $$$\frac{\Delta x}{3} = \frac{1}{12}$$$: $$$\frac{1}{12} \left(0.52275795857471 + 2.09093460413808 + 1.043964704311697 + 2.067923042238355 + 0.5\right) = 0.518798359105237.$$$

$$$\int\limits_{0}^{1} \frac{1}{\sqrt[3]{x^{5} + 7}}\, dx\approx 0.518798359105237$$$A