Question

find a limit equal to the area shown. ( y = sqrt{x} ) (graph with x - axis from 0 to 6, y - axis from 0 to 2, shaded region from ( x = 2 ) to ( x = 6 ) under the curve ( y = sqrt{x} ))

Explanation:

Step1: Determine the interval and width of subintervals

The area is between \( x = 2 \) and \( x = 6 \), so the interval is \([a, b]=[2, 6]\). The number of subintervals is \( n \), so the width of each subinterval \( \Delta x=\frac{b - a}{n}=\frac{6 - 2}{n}=\frac{4}{n} \).

Step2: Determine the sample points (using right endpoints here)

For a right - endpoint Riemann sum, the \( i \) - th sample point \( x_i=a + i\Delta x=2 + i\cdot\frac{4}{n}=2+\frac{4i}{n} \).

Step3: Determine the function

The function is \( f(x)=\sqrt{x} \). The area under the curve \( y = f(x) \) from \( x = a \) to \( x = b \) can be approximated by the Riemann sum \( \sum_{i = 1}^{n}f(x_i)\Delta x \), and as \( n
ightarrow\infty \), the limit of the Riemann sum is the definite integral (which represents the area). So the limit equal to the area is \( \lim_{n
ightarrow\infty}\sum_{i = 1}^{n}\sqrt{2+\frac{4i}{n}}\cdot\frac{4}{n} \)

Answer:

\(\lim_{n
ightarrow\infty}\sum_{i = 1}^{n}\sqrt{2+\frac{4i}{n}}\cdot\frac{4}{n}\)