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

The calculator will approximate the integral of $x^{2}$ from $0$ to $4$ with $n = 4$ 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 $\int\limits_{0}^{4} x^{2}\, dx$ 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 $f{\left(x \right)} = x^{2}$, $a = 0$, $b = 4$, and $n = 4$.

Therefore, $\Delta x = \frac{4 - 0}{4} = 1$.

Divide the interval $\left[0, 4\right]$ into $n = 4$ subintervals of the length $\Delta x = 1$ with the following endpoints: $a = 0$, $1$, $2$, $3$, $4 = b$.

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

$f{\left(\frac{x_{0} + x_{1}}{2} \right)} = f{\left(\frac{0 + 1}{2} \right)} = f{\left(\frac{1}{2} \right)} = \frac{1}{4} = 0.25$

$f{\left(\frac{x_{1} + x_{2}}{2} \right)} = f{\left(\frac{1 + 2}{2} \right)} = f{\left(\frac{3}{2} \right)} = \frac{9}{4} = 2.25$

$f{\left(\frac{x_{2} + x_{3}}{2} \right)} = f{\left(\frac{2 + 3}{2} \right)} = f{\left(\frac{5}{2} \right)} = \frac{25}{4} = 6.25$

$f{\left(\frac{x_{3} + x_{4}}{2} \right)} = f{\left(\frac{3 + 4}{2} \right)} = f{\left(\frac{7}{2} \right)} = \frac{49}{4} = 12.25$

Finally, just sum up the above values and multiply by $\Delta x = 1$: $1 \left(0.25 + 2.25 + 6.25 + 12.25\right) = 21$.

$\int\limits_{0}^{4} x^{2}\, dx\approx 21$A