# Approximate $\int\limits_{0}^{2} e^{- 5 x^{2}}\, dx$ with $n = 4$ using the midpoint rule

The calculator will approximate the integral of $e^{- 5 x^{2}}$ from $0$ to $2$ with $n = 4$ subintervals using the midpoint rule, with steps shown.

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

### Solution

The midpoint rule (also known as the midpoint approximation) uses the midpoint of a subinterval for computing the height of the approximating rectangle:

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

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 midpoints of the subintervals.

$f{\left(\frac{x_{0} + x_{1}}{2} \right)} = f{\left(\frac{0 + \frac{1}{2}}{2} \right)} = f{\left(\frac{1}{4} \right)} = e^{- \frac{5}{16}}\approx 0.731615628946642$

$f{\left(\frac{x_{1} + x_{2}}{2} \right)} = f{\left(\frac{\frac{1}{2} + 1}{2} \right)} = f{\left(\frac{3}{4} \right)} = e^{- \frac{45}{16}}\approx 0.060054667895308$

$f{\left(\frac{x_{2} + x_{3}}{2} \right)} = f{\left(\frac{1 + \frac{3}{2}}{2} \right)} = f{\left(\frac{5}{4} \right)} = e^{- \frac{125}{16}}\approx 0.000404645169326$

$f{\left(\frac{x_{3} + x_{4}}{2} \right)} = f{\left(\frac{\frac{3}{2} + 2}{2} \right)} = f{\left(\frac{7}{4} \right)} = e^{- \frac{245}{16}}\approx 2.23802919 \cdot 10^{-7}$

Finally, just sum up the above values and multiply by $\Delta x = \frac{1}{2}$: $\frac{1}{2} \left(0.731615628946642 + 0.060054667895308 + 0.000404645169326 + 2.23802919 \cdot 10^{-7}\right) = 0.396037582907098.$

$\int\limits_{0}^{2} e^{- 5 x^{2}}\, dx\approx 0.396037582907098$A