Approximate $$$\int\limits_{0}^{2} e^{- 5 x^{2}}\, dx$$$ with $$$n = 4$$$ using the right endpoint approximation

Approximate the integral $$$\int\limits_{0}^{2} e^{- 5 x^{2}}\, dx$$$ with $$$n = 4$$$ using the right endpoint approximation.

The right Riemann sum (also known as the right endpoint approximation) uses the right endpoint of a subinterval for computing the height of the approximating rectangle:

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

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

We have that $$$f{\left(x \right)} = e^{- 5 x^{2}}$$$, $$$a = 0$$$, $$$b = 2$$$, 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 right endpoints of the subintervals.

$$$f{\left(x_{1} \right)} = f{\left(\frac{1}{2} \right)} = e^{- \frac{5}{4}}\approx 0.28650479686019$$$

$$$f{\left(x_{2} \right)} = f{\left(1 \right)} = e^{-5}\approx 0.006737946999085$$$

$$$f{\left(x_{3} \right)} = f{\left(\frac{3}{2} \right)} = e^{- \frac{45}{4}}\approx 0.000013007297654$$$

$$$f{\left(x_{4} \right)} = f{\left(2 \right)} = e^{-20}\approx 2.061154 \cdot 10^{-9}$$$

Finally, just sum up the above values and multiply by $$$\Delta x = \frac{1}{2}$$$: $$$\frac{1}{2} \left(0.28650479686019 + 0.006737946999085 + 0.000013007297654 + 2.061154 \cdot 10^{-9}\right) = 0.146627876609041.$$$

$$$\int\limits_{0}^{2} e^{- 5 x^{2}}\, dx\approx 0.146627876609041$$$A