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

Options

A. `37/72`
B. `2/3`
C. `17/72`
D. `37/144`

Explanation

To solve the problem, we need to evaluate the integral \( \int_{\frac{1}{3}}^{1} f(x) \, dx \) where \( f(x) = g(x) \) and \( g(x) = \left\lfloor \frac{1}{x} \right\rfloor \), with \( \left\lfloor . \right\rfloor \) denoting the greatest integer function.

### Step 1: Determine the function \( g(x) \) over the interval \( \left[\frac{1}{3}, 1\right] \)

1. For \( x \in \left[\frac{1}{3}, \frac{1}{2}\right) \):
\[ g(x) = \left\lfloor \frac{1}{x} \right\rfloor \]
- When \( x = \frac{1}{3} \), \( \frac{1}{x} = 3 \) so \( g\left(\frac{1}{3}\right) = 3 \).
- When \( x = \frac{1}{2} \), \( \frac{1}{x} = 2 \) so \( g\left(\frac{1}{2}\right) = 2 \).
- Therefore, \( g(x) = 3 \) for \( x \in \left[\frac{1}{3}, \frac{1}{2}\right) \).

2. For \( x \in \left[\frac{1}{2}, 1\right] \):
- When \( x = \frac{1}{2} \), \( \frac{1}{x} = 2 \) so \( g\left(\frac{1}{2}\right) = 2 \).
- When \( x = 1 \), \( \frac{1}{x} = 1 \) so \( g(1) = 1 \).
- Therefore, \( g(x) = 2 \) for \( x \in \left[\frac{1}{2}, 1\right) \).

### Step 2: Break the integral into two parts

We can split the integral as follows:
\[ \int_{\frac{1}{3}}^{1} f(x) \, dx = \int_{\frac{1}{3}}^{\frac{1}{2}} g(x) \, dx + \int_{\frac{1}{2}}^{1} g(x) \, dx \]

### Step 3: Evaluate each integral

1. **First integral** \( \int_{\frac{1}{3}}^{\frac{1}{2}} g(x) \, dx \):
\[ g(x) = 3 \quad \text{for } x \in \left[\frac{1}{3}, \frac{1}{2}\right) \]
\[ \int_{\frac{1}{3}}^{\frac{1}{2}} g(x) \, dx = \int_{\frac{1}{3}}^{\frac{1}{2}} 3 \, dx = 3 \left[ x \right]_{\frac{1}{3}}^{\frac{1}{2}} = 3 \left( \frac{1}{2} - \frac{1}{3} \right) = 3 \left( \frac{3 - 2}{6} \right) = 3 \cdot \frac{1}{6} = \frac{1}{2} \]

2. **Second integral** \( \int_{\frac{1}{2}}^{1} g(x) \, dx \):
\[ g(x) = 2 \quad \text{for } x \in \left[\frac{1}{2}, 1\right) \]
\[ \int_{\frac{1}{2}}^{1} g(x) \, dx = \int_{\frac{1}{2}}^{1} 2 \, dx = 2 \left[ x \right]_{\frac{1}{2}}^{1} = 2 \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1 \]

### Step 4: Combine the results

Now, we combine both integrals:
\[ \int_{\frac{1}{3}}^{1} f(x) \, dx = \frac{1}{2} + 1 = \frac{3}{2} \]

### Final Answer
Thus, the value of \( \int_{\frac{1}{3}}^{1} f(x) \, dx \) is \( \frac{3}{2} \).