Answer is: option3
\( \int_0^1 x^2 \, dx \)Solution:
The given expression can be written as:
\[ \sum_{k=1}^{20} \left( \frac{1}{20} \right) \left( \frac{k}{20} \right)^2 \]
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:
- \(n = 20\) (number of subintervals),
- \(\frac{b-a}{n} = \frac{1}{20}\) (width of each subinterval),
- \(f \left( a + \frac{(b-a)k}{n} \right) = \left( \frac{k}{20} \right)^2\) (function evaluated at sample points).
From \(\frac{b-a}{20} = \frac{1}{20}\), we get \(b - a = 1\).
The sample points are \(\frac{k}{20}\), which suggests \(a = 0\) (since when \(k = 1\), the first point is \(\frac{1}{20}\)).
Thus, \(b = 1\) (since \(b - a = 1\) and \(a = 0\)).
The function evaluated at the sample points is:
\[ f\left(\frac{k}{20}\right) = \left(\frac{k}{20}\right)^2 \Rightarrow f(x) = x^2 \]
The Riemann sum approximates:
\[ \int_a^b f(x)\,dx = \int_0^1 x^2\,dx \]
This corresponds to option (C).