Answer is: option1
\( \int_0^1 \sqrt{x} \, dx \)Solution:
The given expression can be written as:
\[ \sum_{k=1}^{30} \left( \frac{1}{30} \right) \sqrt{\frac{k}{30}} \]
This matches the general form of a Riemann sum:
\[ \sum_{k=1}^{n} \left( \frac{b-a}{n} \right) f\left(a + \frac{(b-a)k}{n} \right) \]
where:
- \(n = 30\) (number of subintervals),
- \(\frac{b-a}{n} = \frac{1}{30}\) (width of each subinterval),
- \(f \left(a + \frac{(b-a)k}{n} \right) = \sqrt{\frac{k}{30}}\) (function evaluated at sample points).
From \(\frac{b-a}{30} = \frac{1}{30}\), we get \(b - a = 1\).
The sample points are \(\frac{k}{30}\), which suggests \(a = 0\) (since when \(k = 1\), the first point is \(\frac{1}{30}\)).
Thus, \(b = 1\) (since \(b - a = 1\) and \(a = 0\)).
The function evaluated at the sample points is:
\[ f\left(\frac{k}{30}\right) = \sqrt{\frac{k}{30}} \Longrightarrow f(x) = \sqrt{x} \]
The Riemann sum approximates:
\[ \int_a^b f(x)\,dx = \int_0^1 \sqrt{x}\,dx \]
This corresponds to option (A).