If `int_(0)^(x^(2)(1+x))f(t)dt=x`, then the value of f(2) is.

Options

A. `1//2`
B. `1//3`
C. `1//4`
D. `1//5`

Explanation

To solve the problem, we need to find the value of \( f(2) \) given the equation:

\[ \int_{0}^{x^2(1+x)} f(t) \, dt = x \]

### Step 1: Differentiate both sides with respect to \( x \)

Using the Fundamental Theorem of Calculus and the Chain Rule, we differentiate the left-hand side:

\[ \frac{d}{dx} \left( \int_{0}^{x^2(1+x)} f(t) \, dt \right) = f(x^2(1+x)) \cdot \frac{d}{dx}(x^2(1+x)) \]

Now, we differentiate \( x^2(1+x) \):

\[ \frac{d}{dx}(x^2(1+x)) = \frac{d}{dx}(x^2 + x^3) = 2x + 3x^2 \]

So, we have:

\[ f(x^2(1+x)) \cdot (2x + 3x^2) \]

Now, differentiate the right-hand side:

\[ \frac{d}{dx}(x) = 1 \]

Setting both sides equal gives us:

\[ f(x^2(1+x)) \cdot (2x + 3x^2) = 1 \]

### Step 2: Solve for \( f(x^2(1+x)) \)

From the equation above, we can isolate \( f(x^2(1+x)) \):

\[ f(x^2(1+x)) = \frac{1}{2x + 3x^2} \]

### Step 3: Substitute \( x = 1 \) to find \( f(2) \)

We need to find \( f(2) \). First, we need to find the value of \( x \) such that \( x^2(1+x) = 2 \):

Let \( x = 1 \):

\[ 1^2(1+1) = 1 \cdot 2 = 2 \]

Now, substitute \( x = 1 \) into the expression for \( f(x^2(1+x)) \):

\[ f(2) = f(1^2(1+1)) = f(2) = \frac{1}{2(1) + 3(1^2)} = \frac{1}{2 + 3} = \frac{1}{5} \]

### Final Answer

Thus, the value of \( f(2) \) is:

\[ \boxed{\frac{1}{5}} \]