Find area bounded by `y = f^(-1)(x), x = 4, x = 10` and the `x`-axis, given that the area bounded by `y = f(x) , x = 2, x = 6` and the `x`-axis is `30` sq. units, where `f(2) = 4` and `f(6) = 10`, and that `f(x)` is an invertible function.

Answer

To find the area bounded by \( y = f^{-1}(x) \), \( x = 4 \), \( x = 10 \), and the x-axis, we can follow these steps:

### Step 1: Understand the Given Information
We know that the area under the curve \( y = f(x) \) from \( x = 2 \) to \( x = 6 \) is 30 square units. This can be expressed as:
\[ \int_{2}^{6} f(x) \, dx = 30 \]
Additionally, we have the values \( f(2) = 4 \) and \( f(6) = 10 \).

### Step 2: Set Up the Integral for the Inverse Function
We need to find the area under \( y = f^{-1}(x) \) from \( x = 4 \) to \( x = 10 \). The area can be expressed as:
\[ \text{Area} = \int_{4}^{10} f^{-1}(x) \, dx \]

### Step 3: Use the Relationship Between \( f \) and \( f^{-1} \)
Using the property of inverse functions, we know that:
\[ \int_{a}^{b} f^{-1}(x) \, dx = b \cdot f^{-1}(b) - a \cdot f^{-1}(a) - \int_{f^{-1}(a)}^{f^{-1}(b)} f(t) \, dt \]
In our case, \( a = 4 \) and \( b = 10 \). From the given information:
- \( f^{-1}(4) = 2 \) (since \( f(2) = 4 \))
- \( f^{-1}(10) = 6 \) (since \( f(6) = 10 \))

### Step 4: Substitute into the Integral
Substituting into our equation gives:
\[ \text{Area} = 10 \cdot 6 - 4 \cdot 2 - \int_{2}^{6} f(t) \, dt \]

### Step 5: Calculate Each Term
Now we can calculate each term:
- \( 10 \cdot 6 = 60 \)
- \( 4 \cdot 2 = 8 \)
- We already know \( \int_{2}^{6} f(t) \, dt = 30 \)

### Step 6: Combine the Results
Putting it all together:
\[ \text{Area} = 60 - 8 - 30 \]
\[ \text{Area} = 60 - 38 = 22 \]

### Conclusion
Thus, the area bounded by \( y = f^{-1}(x) \), \( x = 4 \), \( x = 10 \), and the x-axis is:
\[ \boxed{22} \text{ square units} \]

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