IDNLearn.com: Your trusted platform for finding reliable answers. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.
Sagot :
Splitting up [0, 3] into [tex]n[/tex] equally-spaced subintervals of length [tex]\Delta x=\frac{3-0}n = \frac3n[/tex] gives the partition
[tex]\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right][/tex]
where the right endpoint of the [tex]i[/tex]-th subinterval is given by the sequence
[tex]r_i = \dfrac{3i}n[/tex]
for [tex]i\in\{1,2,3,\ldots,n\}[/tex].
Then the definite integral is given by the infinite Riemann sum
[tex]\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.