# Approximate $\int\limits_{-1}^{3} \left(15 x^{2} + 24 x + 1\right)\, dx$ with $n = 3$ using the left endpoint approximation

The calculator will approximate the integral of $15 x^{2} + 24 x + 1$ from $-1$ to $3$ with $n = 3$ subintervals using the left endpoint approximation, with steps shown.

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Approximate the integral $\int\limits_{-1}^{3} \left(15 x^{2} + 24 x + 1\right)\, dx$ with $n = 3$ 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 $f{\left(x \right)} = 15 x^{2} + 24 x + 1$, $a = -1$, $b = 3$, and $n = 3$.

Therefore, $\Delta x = \frac{3 - \left(-1\right)}{3} = \frac{4}{3}$.

Divide the interval $\left[-1, 3\right]$ into $n = 3$ subintervals of the length $\Delta x = \frac{4}{3}$ with the following endpoints: $a = -1$, $\frac{1}{3}$, $\frac{5}{3}$, $3 = b$.

Now, just evaluate the function at the left endpoints of the subintervals.

$f{\left(x_{0} \right)} = f{\left(-1 \right)} = -8$

$f{\left(x_{1} \right)} = f{\left(\frac{1}{3} \right)} = \frac{32}{3}\approx 10.666666666666667$

$f{\left(x_{2} \right)} = f{\left(\frac{5}{3} \right)} = \frac{248}{3}\approx 82.666666666666667$

Finally, just sum up the above values and multiply by $\Delta x = \frac{4}{3}$: $\frac{4}{3} \left(-8 + 10.666666666666667 + 82.666666666666667\right) = 113.777777777777779.$

$\int\limits_{-1}^{3} \left(15 x^{2} + 24 x + 1\right)\, dx\approx 113.777777777777779$A