Let `A_(1) = int_(0)^(x)(int_(0)^(u)f(t)dt) dt` and `A_(2) = int_(0)^(x)f(u).(x-u)` then `(A_(1))/(A_(2))` is equal to :

Options

A. `1/2`
B. 1
C. 2
D. `-1`

Answer

To solve the problem, we need to evaluate the expressions for \( A_1 \) and \( A_2 \) and then find the ratio \( \frac{A_1}{A_2} \).

### Step 1: Define \( A_1 \)
We have:
\[ A_1 = \int_0^x \left( \int_0^u f(t) \, dt \right) du \]
This represents a double integral where we first integrate \( f(t) \) with respect to \( t \) from \( 0 \) to \( u \), and then integrate the result with respect to \( u \) from \( 0 \) to \( x \).

### Step 2: Change the order of integration for \( A_1 \)
To simplify \( A_1 \), we can change the order of integration. The limits will change accordingly:
\[ A_1 = \int_0^x \left( \int_t^x du \right) f(t) \, dt \]
Here, for a fixed \( t \), \( u \) varies from \( t \) to \( x \).

### Step 3: Evaluate the inner integral
The inner integral \( \int_t^x du \) can be computed as:
\[ \int_t^x du = x - t \]
Thus, we can rewrite \( A_1 \) as:
\[ A_1 = \int_0^x f(t) (x - t) \, dt \]

### Step 4: Define \( A_2 \)
Now, we have:
\[ A_2 = \int_0^x f(u) (x - u) \, du \]
This expression is similar to the one we derived for \( A_1 \).

### Step 5: Compare \( A_1 \) and \( A_2 \)
Notice that:
\[ A_1 = \int_0^x f(t) (x - t) \, dt \]
and
\[ A_2 = \int_0^x f(u) (x - u) \, du \]
Both integrals have the same form, just with different dummy variables of integration.

### Step 6: Calculate the ratio \( \frac{A_1}{A_2} \)
Since \( A_1 \) and \( A_2 \) are equal:
\[ \frac{A_1}{A_2} = \frac{\int_0^x f(t) (x - t) \, dt}{\int_0^x f(u) (x - u) \, du} = 1 \]

### Final Answer
Thus, we conclude:
\[ \frac{A_1}{A_2} = 1 \]