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

The calculator will approximate the integral of $e^{- x^{2}}$ from $0$ to $1$ with $n = 4$ subintervals using the right endpoint approximation, with steps shown.

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

### Solution

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