If `I_(1)=int_(1-x)^(x) x sin{x(1-x)}dx` and `I_(2)=int_(1-x)^(x) sin{x(1-x)}dx`, then

Options

A. `I_(1)2I_(2)`
B. `2I_(1)=I_(2)`
C. `I_(1)=I_(2)`
D. none of these

Answer

To solve the problem, we need to find the relationship between the integrals \( I_1 \) and \( I_2 \) defined as follows:

\[ I_1 = \int_{1-x}^{x} x \sin(x(1-x)) \, dx \]
\[ I_2 = \int_{1-x}^{x} \sin(x(1-x)) \, dx \]

### Step 1: Apply the property of definite integrals

We can use the property of definite integrals which states that:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

In our case, we have \( a = 1 - x \) and \( b = x \). Therefore, we can rewrite \( I_1 \) as:

\[ I_1 = \int_{1-x}^{x} x \sin(x(1-x)) \, dx = \int_{1-x}^{x} (1-x + x - x) \sin(x(1-x)) \, dx \]

### Step 2: Simplify the integral

Using the property mentioned, we can express \( I_1 \) as:

\[ I_1 = \int_{1-x}^{x} (1-x) \sin((1-x)x) \, dx \]

### Step 3: Split the integral

Now we can separate the integral into two parts:

\[ I_1 = \int_{1-x}^{x} (1-x) \sin((1-x)x) \, dx + \int_{1-x}^{x} x \sin((1-x)x) \, dx \]

### Step 4: Relate \( I_1 \) and \( I_2 \)

Notice that the second integral is actually \( I_2 \):

\[ I_2 = \int_{1-x}^{x} \sin(x(1-x)) \, dx \]

Thus, we can express \( I_1 \) in terms of \( I_2 \):

\[ I_1 = I_2 - I_1 \]

### Step 5: Solve for \( I_1 \)

Rearranging gives us:

\[ 2I_1 = I_2 \]

### Conclusion

Thus, the relationship between \( I_1 \) and \( I_2 \) is:

\[ I_2 = 2I_1 \]