Consider the functions
`f(x) = g(x)` and `g(x) = [1/x]`
Where `[.]` is the greatest integer function.
What is `int_(1/3)^(1/2) g(x)dx` equal to ?

Options

A. `1/6`
B. `1/3`
C. `5/18`
D. `5/36`

Explanation

To solve the integral \( \int_{1/3}^{1/2} g(x) \, dx \) where \( g(x) = \left\lfloor \frac{1}{x} \right\rfloor \), we will follow these steps:

### Step 1: Determine the function \( g(x) \) in the interval \([1/3, 1/2]\)

First, we need to evaluate \( g(x) \) at the endpoints of the interval:

- For \( x = \frac{1}{3} \):
\[ g\left(\frac{1}{3}\right) = \left\lfloor \frac{1}{\frac{1}{3}} \right\rfloor = \left\lfloor 3 \right\rfloor = 3 \]

- For \( x = \frac{1}{2} \):
\[ g\left(\frac{1}{2}\right) = \left\lfloor \frac{1}{\frac{1}{2}} \right\rfloor = \left\lfloor 2 \right\rfloor = 2 \]

### Step 2: Identify the value of \( g(x) \) in the interval \([1/3, 1/2]\)

Next, we need to find out how \( g(x) \) behaves in the interval \( \left[\frac{1}{3}, \frac{1}{2}\right] \).

- For \( x \) in the interval \( \left(\frac{1}{3}, \frac{1}{2}\right) \), we can check the value of \( g(x) \):
- If \( x \) is slightly greater than \( \frac{1}{3} \) (e.g., \( x = 0.34 \)):
\[ g(0.34) = \left\lfloor \frac{1}{0.34} \right\rfloor \approx \left\lfloor 2.94 \right\rfloor = 2 \]
- If \( x \) is slightly less than \( \frac{1}{2} \) (e.g., \( x = 0.49 \)):
\[ g(0.49) = \left\lfloor \frac{1}{0.49} \right\rfloor \approx \left\lfloor 2.04 \right\rfloor = 2 \]

Thus, within the interval \( \left(\frac{1}{3}, \frac{1}{2}\right) \), \( g(x) \) takes the value 2.

### Step 3: Set up the integral

Now we can set up the integral:
\[ \int_{1/3}^{1/2} g(x) \, dx = \int_{1/3}^{1/2} 2 \, dx \]

### Step 4: Evaluate the integral

Since \( 2 \) is a constant, we can factor it out:
\[ \int_{1/3}^{1/2} 2 \, dx = 2 \int_{1/3}^{1/2} 1 \, dx \]

Now, evaluate the integral:
\[ \int_{1/3}^{1/2} 1 \, dx = x \bigg|_{1/3}^{1/2} = \left(\frac{1}{2} - \frac{1}{3}\right) \]

Calculating this:
\[ \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

Thus, we have:
\[ 2 \int_{1/3}^{1/2} 1 \, dx = 2 \cdot \frac{1}{6} = \frac{1}{3} \]

### Final Result

Therefore, the value of the integral \( \int_{1/3}^{1/2} g(x) \, dx \) is:
\[ \boxed{\frac{1}{3}} \]