The value of `int_1^a[x]f^(prime)(x)dxf^(prime)(x)dx ,w h e r ea >1,a n d[x]` denotes the greatest integer not exceeding `x ,` is
`af(a)-{f(1)f(2)+...+f([a])}` `[a]f(a)-{f(1)+f(2)+...+f([a])}` `[a]f(a)-{f(1)+f(2)+...+fA}`
`af([a])-{f(1)+f(2)+...+fA}`

Options

A. `af(a)-(f(1)+f(2)+………..+f([a]))`
B. `[a]f(a)-(f(1)+f(2))+………..+f([a]))`
C. `[a]f([a])-(f(1)+f(2)+………..+f(a))`
D. `af([a])-(f(1)+f(2)+…………+f(a))`

Explanation

To solve the given integral problem, we need to evaluate the integral

\[ I = \int_1^a [x] f'(x) \, dx \]

where \( [x] \) denotes the greatest integer not exceeding \( x \), and \( a > 1 \).

### Step 1: Break the Integral into Intervals
Since \( [x] \) is a piecewise constant function, we can break the integral into intervals where \( [x] \) is constant. The greatest integer function \( [x] \) changes at each integer. Therefore, we can write:

\[ I = \int_1^2 [x] f'(x) \, dx + \int_2^3 [x] f'(x) \, dx + \ldots + \int_{k}^{a} [x] f'(x) \, dx \]

where \( k = [a] \).

### Step 2: Evaluate Each Integral
On the interval \( [n, n+1) \) for \( n = 1, 2, \ldots, k-1 \), we have \( [x] = n \). Thus, we can rewrite the integral as:

\[ I = \sum_{n=1}^{k-1} \int_n^{n+1} n f'(x) \, dx + \int_k^{a} k f'(x) \, dx \]

### Step 3: Simplify Each Integral
For each integral \( \int_n^{n+1} n f'(x) \, dx \):

\[ \int_n^{n+1} n f'(x) \, dx = n \left[ f(x) \right]_n^{n+1} = n (f(n+1) - f(n)) \]

Thus, we can express \( I \) as:

\[ I = \sum_{n=1}^{k-1} n (f(n+1) - f(n)) + k \int_k^{a} f'(x) \, dx \]

### Step 4: Evaluate the Last Integral
The last integral can be evaluated as:

\[ \int_k^{a} f'(x) \, dx = f(a) - f(k) \]

### Step 5: Combine Everything
Now substituting back, we have:

\[ I = \sum_{n=1}^{k-1} n (f(n+1) - f(n)) + k (f(a) - f(k)) \]

### Step 6: Recognize the Telescoping Nature
The sum \( \sum_{n=1}^{k-1} n (f(n+1) - f(n)) \) can be simplified. This sum telescopes, leading to:

\[ I = k f(a) - \sum_{n=1}^{k} f(n) \]

where \( \sum_{n=1}^{k} f(n) = f(1) + f(2) + \ldots + f(k) \).

### Final Result
Thus, we arrive at the final expression:

\[ I = k f(a) - \sum_{n=1}^{k} f(n) \]

Since \( k = [a] \), we can write:

\[ I = [a] f(a) - \sum_{n=1}^{[a]} f(n) \]

### Conclusion
The value of the integral is:

\[ I = [a] f(a) - \{ f(1) + f(2) + \ldots + f([a]) \} \]