Let the equation of a curve passing through the point (0,1) be given b `y=intx^2e^(x^3)dx`. If the equation of the curve is written in the form `x=f(y)`, then f(y) is

Options

A. `sqrt(log_(e)(3y-2))`
B. `root(3)(log_(e)(3y-2))`
C. `root(3)(log_(e)(2-3y))`
D. none of these

Explanation

To solve the problem, we need to find the function \( f(y) \) such that \( x = f(y) \) for the curve defined by the integral \( y = \int x^2 e^{x^3} \, dx \) and passing through the point (0, 1).

### Step-by-Step Solution:

1. **Set up the integral**:
\[ y = \int x^2 e^{x^3} \, dx \]

2. **Use substitution**:
Let \( t = x^3 \). Then, differentiate to find \( dt \):
\[ dt = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{dt}{3x^2} \]
Therefore, we can express \( x^2 \, dx \) as:
\[ x^2 \, dx = \frac{dt}{3} \]

3. **Substitute into the integral**:
Substitute \( x^2 \, dx \) and \( e^{x^3} \):
\[ y = \int x^2 e^{x^3} \, dx = \int e^t \frac{dt}{3} = \frac{1}{3} \int e^t \, dt \]

4. **Integrate**:
The integral of \( e^t \) is \( e^t \):
\[ y = \frac{1}{3} e^t + C = \frac{1}{3} e^{x^3} + C \]

5. **Determine the constant \( C \)**:
We know the curve passes through the point (0, 1):
\[ 1 = \frac{1}{3} e^{0} + C \quad \Rightarrow \quad 1 = \frac{1}{3} + C \quad \Rightarrow \quad C = 1 - \frac{1}{3} = \frac{2}{3} \]

6. **Substitute \( C \) back into the equation**:
\[ y = \frac{1}{3} e^{x^3} + \frac{2}{3} \]

7. **Rearranging the equation**:
Multiply through by 3 to eliminate the fraction:
\[ 3y = e^{x^3} + 2 \]
Rearranging gives:
\[ e^{x^3} = 3y - 2 \]

8. **Take the natural logarithm**:
\[ x^3 = \ln(3y - 2) \]

9. **Solve for \( x \)**:
Taking the cube root gives:
\[ x = \sqrt[3]{\ln(3y - 2)} \]

Thus, the function \( f(y) \) is:
\[ f(y) = \sqrt[3]{\ln(3y - 2)} \]

### Final Answer:
\[ f(y) = \sqrt[3]{\ln(3y - 2)} \]