# Approximate $\int\limits_{1}^{3} \frac{\sin{\left(x \right)}}{x}\, dx$ with $n = 2$ using the midpoint rule

The calculator will approximate the integral of $\frac{\sin{\left(x \right)}}{x}$ from $1$ to $3$ with $n = 2$ subintervals using the midpoint rule, with steps shown.

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Approximate the integral $\int\limits_{1}^{3} \frac{\sin{\left(x \right)}}{x}\, dx$ 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 $f{\left(x \right)} = \frac{\sin{\left(x \right)}}{x}$, $a = 1$, $b = 3$, and $n = 2$.

Therefore, $\Delta x = \frac{3 - 1}{2} = 1$.

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

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

$f{\left(\frac{x_{0} + x_{1}}{2} \right)} = f{\left(\frac{1 + 2}{2} \right)} = f{\left(\frac{3}{2} \right)} = \frac{2 \sin{\left(\frac{3}{2} \right)}}{3}\approx 0.664996657736036$

$f{\left(\frac{x_{1} + x_{2}}{2} \right)} = f{\left(\frac{2 + 3}{2} \right)} = f{\left(\frac{5}{2} \right)} = \frac{2 \sin{\left(\frac{5}{2} \right)}}{5}\approx 0.239388857641583$

Finally, just sum up the above values and multiply by $\Delta x = 1$: $1 \left(0.664996657736036 + 0.239388857641583\right) = 0.904385515377619.$

$\int\limits_{1}^{3} \frac{\sin{\left(x \right)}}{x}\, dx\approx 0.904385515377619$A