# Approximate $\int\limits_{1}^{6} \sqrt{x^{3} + 1}\, dx$ with $n = 10$ using the right endpoint approximation

The calculator will approximate the integral of $\sqrt{x^{3} + 1}$ from $1$ to $6$ with $n = 10$ subintervals using the right endpoint approximation, with steps shown.

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Approximate the integral $\int\limits_{1}^{6} \sqrt{x^{3} + 1}\, dx$ with $n = 10$ using the right endpoint approximation.

### Solution

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

$\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \Delta x \left(f{\left(x_{1} \right)} + f{\left(x_{2} \right)} + f{\left(x_{3} \right)}+\dots+f{\left(x_{n-1} \right)} + f{\left(x_{n} \right)}\right)$

where $\Delta x = \frac{b - a}{n}$.

We have that $f{\left(x \right)} = \sqrt{x^{3} + 1}$, $a = 1$, $b = 6$, and $n = 10$.

Therefore, $\Delta x = \frac{6 - 1}{10} = \frac{1}{2}$.

Divide the interval $\left[1, 6\right]$ into $n = 10$ subintervals of the length $\Delta x = \frac{1}{2}$ with the following endpoints: $a = 1$, $\frac{3}{2}$, $2$, $\frac{5}{2}$, $3$, $\frac{7}{2}$, $4$, $\frac{9}{2}$, $5$, $\frac{11}{2}$, $6 = b$.

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

$f{\left(x_{1} \right)} = f{\left(\frac{3}{2} \right)} = \frac{\sqrt{70}}{4}\approx 2.091650066335189$

$f{\left(x_{2} \right)} = f{\left(2 \right)} = 3$

$f{\left(x_{3} \right)} = f{\left(\frac{5}{2} \right)} = \frac{\sqrt{266}}{4}\approx 4.077376607575023$

$f{\left(x_{4} \right)} = f{\left(3 \right)} = 2 \sqrt{7}\approx 5.291502622129181$

$f{\left(x_{5} \right)} = f{\left(\frac{7}{2} \right)} = \frac{3 \sqrt{78}}{4}\approx 6.623820649745885$

$f{\left(x_{6} \right)} = f{\left(4 \right)} = \sqrt{65}\approx 8.06225774829855$

$f{\left(x_{7} \right)} = f{\left(\frac{9}{2} \right)} = \frac{\sqrt{1474}}{4}\approx 9.598176910226233$

$f{\left(x_{8} \right)} = f{\left(5 \right)} = 3 \sqrt{14}\approx 11.224972160321824$

$f{\left(x_{9} \right)} = f{\left(\frac{11}{2} \right)} = \frac{\sqrt{2678}}{4}\approx 12.937349032935611$

$f{\left(x_{10} \right)} = f{\left(6 \right)} = \sqrt{217}\approx 14.730919862656235$

Finally, just sum up the above values and multiply by $\Delta x = \frac{1}{2}$: $\frac{1}{2} \left(2.091650066335189 + 3 + 4.077376607575023 + 5.291502622129181 + 6.623820649745885 + 8.06225774829855 + 9.598176910226233 + 11.224972160321824 + 12.937349032935611 + 14.730919862656235\right) = 38.819012830111865.$

$\int\limits_{1}^{6} \sqrt{x^{3} + 1}\, dx\approx 38.819012830111865$A