Approximate $$$\int\limits_{0}^{1} e^{- x^{2}}\, dx$$$ with $$$n = 4$$$ using the right endpoint approximation

Approximate the integral $$$\int\limits_{0}^{1} e^{- x^{2}}\, dx$$$ 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 $$$f{\left(x \right)} = e^{- x^{2}}$$$, $$$a = 0$$$, $$$b = 1$$$, 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 the right endpoints of the subintervals.

$$$f{\left(x_{1} \right)} = f{\left(\frac{1}{4} \right)} = e^{- \frac{1}{16}}\approx 0.939413062813476$$$

$$$f{\left(x_{2} \right)} = f{\left(\frac{1}{2} \right)} = e^{- \frac{1}{4}}\approx 0.778800783071405$$$

$$$f{\left(x_{3} \right)} = f{\left(\frac{3}{4} \right)} = e^{- \frac{9}{16}}\approx 0.569782824730923$$$

$$$f{\left(x_{4} \right)} = f{\left(1 \right)} = e^{-1}\approx 0.367879441171442$$$

Finally, just sum up the above values and multiply by $$$\Delta x = \frac{1}{4}$$$: $$$\frac{1}{4} \left(0.939413062813476 + 0.778800783071405 + 0.569782824730923 + 0.367879441171442\right) = 0.663969027946811.$$$

$$$\int\limits_{0}^{1} e^{- x^{2}}\, dx\approx 0.663969027946811$$$A