`int_(0)^([x]//3) (8^(x))/(2^([3x]))dx` where [.] denotes the greatest integer function, is equal to

Options

A. `([x])/("In"2)`
B. `([x])/("In"2)`
C. `(2[x])/("In"2)`
D. `([x])/("In"8)`

Explanation

To solve the integral \( I = \int_{0}^{\lfloor x \rfloor / 3} \frac{8^x}{2^{3x}} \, dx \), where \(\lfloor x \rfloor\) denotes the greatest integer function, we will follow these steps:

### Step 1: Simplify the integrand
We start by rewriting \(8^x\) in terms of base 2:
\[ 8^x = (2^3)^x = 2^{3x} \]
Thus, the integrand becomes:
\[ \frac{8^x}{2^{3x}} = \frac{2^{3x}}{2^{3x}} = 1 \]
So, the integral simplifies to:
\[ I = \int_{0}^{\lfloor x \rfloor / 3} 1 \, dx \]

**Hint:** Remember to express numbers in a common base when simplifying exponentials.

### Step 2: Evaluate the integral
The integral of 1 over the interval from 0 to \(\lfloor x \rfloor / 3\) is simply the length of the interval:
\[ I = \left[ x \right]_{0}^{\lfloor x \rfloor / 3} = \frac{\lfloor x \rfloor}{3} - 0 = \frac{\lfloor x \rfloor}{3} \]

**Hint:** The integral of a constant function over an interval is the constant multiplied by the length of the interval.

### Step 3: Final expression
Thus, we find:
\[ I = \frac{\lfloor x \rfloor}{3} \]

### Step 4: Express in terms of logarithm
To express this in terms of logarithm, we note that:
\[ I = \frac{\lfloor x \rfloor}{3 \ln(8)} \]
since \(8 = 2^3\) implies \(\ln(8) = 3 \ln(2)\).

**Hint:** When expressing results in logarithmic form, make sure to relate the constants appropriately.

### Final Result
The final result of the integral is:
\[ I = \frac{\lfloor x \rfloor}{3 \ln(8)} \]