Answer is: option2
\( \int_0^2 \sqrt{x} \, dx \)Solution:
Given:
\[ \lim_{n \to \infty} 2n \sum_{k=1}^{n} \sqrt{\frac{2k}{n}} \]
Riemann Sum Form:
\[ \lim_{n \to \infty} \sum_{k=1}^{n} \left( \frac{b - a}{n} \right) f \left( a + \frac{(b - a)k}{n} \right) \]
The coefficient \( \frac{2}{n} \) in the given limit corresponds to \( \frac{b - a}{n} \) in the Riemann sum.
\[ \frac{b - a}{n} = \frac{2}{n} \Rightarrow b - a = 2 \]
The argument of the function \( \sqrt{\frac{2k}{n}} \) corresponds to: \[ f\left(a + \frac{(b - a)k}{n}\right) \]
Set \( a = 0 \):
Given \( b - a = 2 \) and choosing \( a = 0 \), we get \( b = 2 \).
Substitute \( a = 0 \) into the function argument:
\[ f\left(0 + \frac{2k}{n}\right) = \sqrt{\frac{2k}{n}} \Rightarrow f\left(\frac{2k}{n}\right) = \sqrt{\frac{2k}{n}} \]
\[ f(x) = \sqrt{x} \]
Substituting \( f(x) = \sqrt{x}, a = 0 \), and \( b = 2 \) into the Riemann sum: \[ \lim_{n \to \infty} 2n \sum_{k=1}^{n} \sqrt{\frac{2k}{n}} = \int_0^2 \sqrt{x} \, dx \]
This corresponds to option (B)