# Approximate $\int\limits_{-4}^{0} \left(x^{3} + 1\right)\, dx$ with $n = 4$ using the left endpoint approximation

The calculator will approximate the integral of $x^{3} + 1$ from $-4$ to $0$ with $n = 4$ subintervals using the left endpoint approximation, with steps shown.

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Approximate the integral $\int\limits_{-4}^{0} \left(x^{3} + 1\right)\, dx$ with $n = 4$ 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 $f{\left(x \right)} = x^{3} + 1$, $a = -4$, $b = 0$, and $n = 4$.

Therefore, $\Delta x = \frac{0 - \left(-4\right)}{4} = 1$.

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

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

$f{\left(x_{0} \right)} = f{\left(-4 \right)} = -63$

$f{\left(x_{1} \right)} = f{\left(-3 \right)} = -26$

$f{\left(x_{2} \right)} = f{\left(-2 \right)} = -7$

$f{\left(x_{3} \right)} = f{\left(-1 \right)} = 0$

Finally, just sum up the above values and multiply by $\Delta x = 1$: $1 \left(-63 - 26 - 7 + 0\right) = -96$.

$\int\limits_{-4}^{0} \left(x^{3} + 1\right)\, dx\approx -96$A