Let `f (x) = x^(3)-3x ^(2)+3x+1 ` and g be the inverse of it , then area bounded by the curve `y = g (x)` wirth x-axis between `x =1` to `x=2` is (in square units):

Options

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

Answer

To find the area bounded by the curve \( y = g(x) \) with the x-axis between \( x = 1 \) and \( x = 2 \), where \( g \) is the inverse of the function \( f(x) = x^3 - 3x^2 + 3x + 1 \), we can follow these steps:

### Step 1: Rewrite \( f(x) \)
We start with the function:
\[ f(x) = x^3 - 3x^2 + 3x + 1 \]
We can rewrite this by adding and subtracting 1:
\[ f(x) = (x^3 - 3x^2 + 3x - 1) + 2 \]
Recognizing the first part as a perfect cube, we can express it as:
\[ f(x) = (x - 1)^3 + 2 \]

### Step 2: Find the Inverse Function \( g(x) \)
Let \( y = f(x) \):
\[ y = (x - 1)^3 + 2 \]
To find the inverse, we solve for \( x \):
\[ y - 2 = (x - 1)^3 \]
Taking the cube root:
\[ x - 1 = (y - 2)^{1/3} \]
Thus,
\[ x = (y - 2)^{1/3} + 1 \]
Now, replacing \( y \) with \( x \) and \( x \) with \( y \) gives us the inverse function:
\[ g(x) = (x - 2)^{1/3} + 1 \]

### Step 3: Set Up the Area Integral
The area \( A \) bounded by the curve \( y = g(x) \) and the x-axis from \( x = 1 \) to \( x = 2 \) is given by:
\[ A = \int_{1}^{2} g(x) \, dx \]
Substituting \( g(x) \):
\[ A = \int_{1}^{2} \left((x - 2)^{1/3} + 1\right) \, dx \]

### Step 4: Evaluate the Integral
We can split the integral:
\[ A = \int_{1}^{2} (x - 2)^{1/3} \, dx + \int_{1}^{2} 1 \, dx \]
The second integral is straightforward:
\[ \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1 \]

Now, we evaluate the first integral:
Let \( u = x - 2 \), then \( du = dx \) and when \( x = 1 \), \( u = -1 \) and when \( x = 2 \), \( u = 0 \):
\[ \int_{1}^{2} (x - 2)^{1/3} \, dx = \int_{-1}^{0} u^{1/3} \, du \]
Calculating this integral:
\[ \int u^{1/3} \, du = \frac{u^{4/3}}{4/3} = \frac{3}{4} u^{4/3} \]
Evaluating from \(-1\) to \(0\):
\[ \left[\frac{3}{4} u^{4/3}\right]_{-1}^{0} = \frac{3}{4}(0) - \frac{3}{4}(-1)^{4/3} = 0 + \frac{3}{4} = \frac{3}{4} \]

### Step 5: Combine the Results
Now we combine the results from both integrals:
\[ A = \frac{3}{4} + 1 = \frac{3}{4} + \frac{4}{4} = \frac{7}{4} \]

### Final Answer
Thus, the area bounded by the curve \( y = g(x) \) with the x-axis between \( x = 1 \) and \( x = 2 \) is:
\[ \boxed{\frac{7}{4}} \text{ square units} \]