If the value of definite integral `int_1^a x*a^-[log_ a x]dx,` where `a >1` and [x] denotes the greatest integer, is, `(e-1)/2` then the value of equal to

Options

A. `sqrte`
B. e
C. `sqrt(e+1)`
D. `e-1`

Answer

To solve the definite integral \( \int_1^a x \cdot a^{-\log_a x} \, dx \) where \( a > 1 \) and \( [x] \) denotes the greatest integer, and find the value of \( a \) such that the integral equals \( \frac{e-1}{2} \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_1^a x \cdot a^{-\log_a x} \, dx \]
Using the property of logarithms, we can rewrite \( a^{-\log_a x} \) as:
\[ a^{-\log_a x} = \frac{1}{x} \]
Thus, the integral simplifies to:
\[ I = \int_1^a x \cdot \frac{1}{x} \, dx = \int_1^a 1 \, dx \]

### Step 2: Evaluate the Integral
Now, we evaluate the integral:
\[ I = \int_1^a 1 \, dx = [x]_1^a = a - 1 \]

### Step 3: Set the Integral Equal to the Given Value
According to the problem, we have:
\[ a - 1 = \frac{e - 1}{2} \]

### Step 4: Solve for \( a \)
Now, we solve for \( a \):
\[ a = \frac{e - 1}{2} + 1 = \frac{e + 1}{2} \]

### Step 5: Check the Value of \( a \)
We need to check if this value of \( a \) satisfies the condition \( a > 1 \):
Since \( e \) is approximately 2.718, we have:
\[ \frac{e + 1}{2} \approx \frac{2.718 + 1}{2} = \frac{3.718}{2} \approx 1.859 > 1 \]
Thus, \( a > 1 \) is satisfied.

### Final Answer
The value of \( a \) is:
\[ \boxed{\frac{e + 1}{2}} \]