Answer is: option2
\( \int_0^2 x \, dx \)Solution:
The given expression can be written as:
\[ \sum_{k=1}^{20} \left( \frac{1}{10} \right) \left( \frac{k}{10} \right) \]
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),
- \( \frac{b - a}{20} = \frac{1}{10} \) (width of each subinterval),
- \( f\left( a + \frac{(b-a)k}{n} \right) = \frac{k}{10} \) (function evaluated at sample points).
From \( \frac{b - a}{20} = \frac{1}{10} \), we get \( b - a = 2 \).
The sample points are \( \frac{k}{10} \), which suggests \( a = 0 \) (since when \( k = 1 \), the first point is \( \frac{1}{10} \)).
Thus, \( b = 2 \) (since \( b - a = 2 \) and \( a = 0 \)).
The function evaluated at the sample points is:
\[ f\left( \frac{k}{10} \right) = \frac{k}{10} \Rightarrow f(x) = x \]
The Riemann sum approximates:
\[ \int_a^b f(x) \, dx = \int_0^2 x \, dx \]
This corresponds to option (B).