Question of the Day

20. The closed interval \([a, b]\) is partitioned into \(n\) equal subintervals, each of width \(\Delta x\), by the numbers \(x_0, x_1, \dots, x_n\) where \(0 < a = x_0 < x_1 < \cdots < x_{n-1} < x_n = b\). What is \[ \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{\sqrt{x_i}} \, \Delta x \, ? \]

\( \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{a}} \)

\( \frac{\sqrt{b} - \sqrt{a}}{2} \)

\( 2(\sqrt{b} - \sqrt{a}) \)

\( \sqrt{b} - \sqrt{a} \)

Answer is: option3

\( 2(\sqrt{b} - \sqrt{a}) \)

Solution:

We are given a Riemann sum:

\[ \lim_{n \to \infty} \sum_{i=1}^n \frac{1}{\sqrt{x_i}} \Delta x \]

This sum approximates the definite integral of a function over an interval. Specifically, the sum becomes the definite integral:

\[ \int_a^b \frac{1}{\sqrt{x}} \, dx \]

The function is \( f(x) = \frac{1}{\sqrt{x}} \)

The interval is from \( a \) to \( b \)

We compute:

\[ \int_a^b \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_a^b = 2\sqrt{b} - 2\sqrt{a} \]

So, the correct answer is: \[ \text{(C) } 2(\sqrt{b} - \sqrt{a}) \]

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AP Calculus BC Tags