Let `f(x): R^+ to R^+` is an invertible function such that `f^(prime)(x)>0a n df"(x)>0AAx in [1,5]dot` If `f(1)=1` and `f(5)=5` and area under the curve `y=f(x)` on x-axis from `x=1tox=5i s8` sq. units, then area bounded by `y=f^(-1)(x)` on x-axis from `x=1tox=5` is

Options

A. 12
B. 16
C. 18
D. 20

Answer

To solve the problem step by step, we will analyze the given information and use the properties of the function and its inverse.

### Step 1: Understand the function and its properties
We are given a function \( f: \mathbb{R}^+ \to \mathbb{R}^+ \) that is invertible, with the following properties:
- \( f'(x) > 0 \) for all \( x \in [1, 5] \) (indicating that \( f(x) \) is strictly increasing)
- \( f(1) = 1 \)
- \( f(5) = 5 \)
- The area under the curve \( y = f(x) \) from \( x = 1 \) to \( x = 5 \) is 8 square units.

### Step 2: Visualize the area under \( f(x) \)
The area under the curve \( y = f(x) \) from \( x = 1 \) to \( x = 5 \) can be represented as:
\[ \text{Area} = \int_{1}^{5} f(x) \, dx = 8 \]

### Step 3: Relationship between \( f(x) \) and \( f^{-1}(x) \)
The area under the curve \( y = f^{-1}(x) \) from \( x = 1 \) to \( x = 5 \) can be found using the property of inverse functions. The area under \( f^{-1}(x) \) from \( x = a \) to \( x = b \) can be calculated as:
\[ \text{Area} = \int_{a}^{b} f^{-1}(x) \, dx \]
For the functions \( f(x) \) and \( f^{-1}(x) \), we have the relationship:
\[ \int_{1}^{5} f(x) \, dx + \int_{1}^{5} f^{-1}(x) \, dx = 5 \cdot 5 - 1 \cdot 1 \]
This is because the area of the square formed by the points (1,1) and (5,5) is \( 25 \) square units.

### Step 4: Calculate the area under \( f^{-1}(x) \)
Using the relationship derived:
\[ \int_{1}^{5} f(x) \, dx + \int_{1}^{5} f^{-1}(x) \, dx = 25 - 1 \]
Substituting the known area under \( f(x) \):
\[ 8 + \int_{1}^{5} f^{-1}(x) \, dx = 24 \]
Now, solving for \( \int_{1}^{5} f^{-1}(x) \, dx \):
\[ \int_{1}^{5} f^{-1}(x) \, dx = 24 - 8 = 16 \]

### Conclusion
Thus, the area bounded by \( y = f^{-1}(x) \) on the x-axis from \( x = 1 \) to \( x = 5 \) is \( 16 \) square units.

### Final Answer
The area bounded by \( y = f^{-1}(x) \) on the x-axis from \( x = 1 \) to \( x = 5 \) is \( 16 \) square units.

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