Let y = g(x) be the inverse of a bijective mapping f : `R rarr R f(x) = 3x^(3) + 2x`. The area bounded by graph of g(x), the x-axis and the ordinate at x = 5 is

Options

A. `5/4`
B. `7/4`
C. `9/4`
D. `13/4`

Answer

To find the area bounded by the graph of the inverse function \( g(x) \), the x-axis, and the ordinate at \( x = 5 \), we can follow these steps:

### Step 1: Understand the Functions
Given the function \( f(x) = 3x^3 + 2x \), we need to find its inverse \( g(x) \). Since \( g(x) \) is the inverse of \( f(x) \), we will use the property of inverse functions: the area under \( g(x) \) from \( x = 0 \) to \( x = 5 \) is equal to the area under \( f(x) \) from \( y = 0 \) to \( y = 5 \).

### Step 2: Find the Value of \( f(1) \)
To find the area, we need to determine where \( f(x) = 5 \):
\[ f(x) = 3x^3 + 2x = 5 \]
This simplifies to:
\[ 3x^3 + 2x - 5 = 0 \]
By trial, we can check \( x = 1 \):
\[ f(1) = 3(1)^3 + 2(1) = 3 + 2 = 5 \]
Thus, \( f(1) = 5 \), which means \( g(5) = 1 \).

### Step 3: Set Up the Area Calculation
The area \( A \) under the curve \( g(x) \) from \( x = 0 \) to \( x = 5 \) can be computed using the formula:
\[ A = \int_0^1 (5 - f(x)) \, dx \]
Where \( f(x) \) is the original function.

### Step 4: Compute the Area
Now we can compute the area:
\[ A = \int_0^1 (5 - (3x^3 + 2x)) \, dx \]
This simplifies to:
\[ A = \int_0^1 (5 - 3x^3 - 2x) \, dx \]

### Step 5: Integrate
Now we perform the integration:
\[ A = \int_0^1 5 \, dx - \int_0^1 3x^3 \, dx - \int_0^1 2x \, dx \]
Calculating each integral:
1. \(\int_0^1 5 \, dx = 5x \big|_0^1 = 5(1) - 5(0) = 5\)
2. \(\int_0^1 3x^3 \, dx = \frac{3}{4}x^4 \big|_0^1 = \frac{3}{4}(1^4) - \frac{3}{4}(0^4) = \frac{3}{4}\)
3. \(\int_0^1 2x \, dx = x^2 \big|_0^1 = (1^2) - (0^2) = 1\)

### Step 6: Combine the Results
Now, substituting back into the area formula:
\[ A = 5 - \frac{3}{4} - 1 = 5 - 1 - \frac{3}{4} = 4 - \frac{3}{4} = \frac{16}{4} - \frac{3}{4} = \frac{13}{4} \]

### Final Answer
Thus, the area bounded by the graph of \( g(x) \), the x-axis, and the ordinate at \( x = 5 \) is:
\[ \boxed{\frac{13}{4}} \text{ square units} \]