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

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

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Approximate the integral $\int\limits_{-5}^{5} x^{2}\, dx$ with $n = 2$ using the left endpoint approximation.

### Solution

The left Riemann sum (also known as the left endpoint approximation) uses the left endpoint of a subinterval for computing the height of the approximating rectangle:

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

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 left endpoints of the subintervals.

$f{\left(x_{0} \right)} = f{\left(-5 \right)} = 25$

$f{\left(x_{1} \right)} = f{\left(0 \right)} = 0$

Finally, just sum up the above values and multiply by $\Delta x = 5$: $5 \left(25 + 0\right) = 125$.

$\int\limits_{-5}^{5} x^{2}\, dx\approx 125$A