Approximate $$$\int\limits_{1}^{6} \sqrt{x^{3} + 1}\, dx$$$ with $$$n = 10$$$ using the right endpoint approximation
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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 $$$\Delta x = \frac{1}{2}$$$: $$$\frac{1}{2} \left(2.091650066335189 + 3 + 4.077376607575023 + 5.291502622129181 + 6.623820649745885 + 8.06225774829855 + 9.598176910226233 + 11.224972160321824 + 12.937349032935611 + 14.730919862656235\right) = 38.819012830111865.$$$
Answer
$$$\int\limits_{1}^{6} \sqrt{x^{3} + 1}\, dx\approx 38.819012830111865$$$A