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

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

Related calculator: Midpoint Rule Calculator for a Table

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Approximate the integral $\int\limits_{-5}^{5} x^{2}\, dx$ with $n = 2$ 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 $f{\left(x \right)} = x^{2}$, $a = -5$, $b = 5$, and $n = 2$.

Therefore, $\Delta x = \frac{5 - \left(-5\right)}{2} = 5$.

Divide the interval $\left[-5, 5\right]$ into $n = 2$ subintervals of the length $\Delta x = 5$ with the following endpoints: $a = -5$, $0$, $5 = b$.

Now, just evaluate the function at the midpoints of the subintervals.

$f{\left(\frac{x_{0} + x_{1}}{2} \right)} = f{\left(\frac{-5 + 0}{2} \right)} = f{\left(- \frac{5}{2} \right)} = \frac{25}{4} = 6.25$

$f{\left(\frac{x_{1} + x_{2}}{2} \right)} = f{\left(\frac{0 + 5}{2} \right)} = f{\left(\frac{5}{2} \right)} = \frac{25}{4} = 6.25$

Finally, just sum up the above values and multiply by $\Delta x = 5$: $5 \left(6.25 + 6.25\right) = 62.5$.

$\int\limits_{-5}^{5} x^{2}\, dx\approx 62.5$A