Let `a gt 0 and f(x) ` is monotonic increase such that `f(0)=0 and f(a)=b, "then " int_(0)^(a) f(x) dx +int_(0)^(b) f^(-1) (x) dx ` is equal to

Options

A. (a)`a+b`
B. (b)`ab+b`
C. (c)`ab+a`
D. (d)ab

Answer

To solve the problem, we need to evaluate the expression:

\[ I = \int_{0}^{a} f(x) \, dx + \int_{0}^{b} f^{-1}(x) \, dx \]

where \( f(x) \) is a monotonic increasing function with \( f(0) = 0 \) and \( f(a) = b \).

### Step 1: Define the integrals
Let:
- \( I_1 = \int_{0}^{a} f(x) \, dx \)
- \( I_2 = \int_{0}^{b} f^{-1}(x) \, dx \)

Thus, we have:

\[ I = I_1 + I_2 \]

### Step 2: Understand the relationship between \( f(x) \) and \( f^{-1}(x) \)
Since \( f \) is a monotonic increasing function, it is invertible. The area under the curve of \( f(x) \) from \( 0 \) to \( a \) can be related to the area under the curve of \( f^{-1}(x) \) from \( 0 \) to \( b \).

### Step 3: Visualize the areas
To visualize this, consider the graph of \( f(x) \) and \( f^{-1}(x) \):
- The area under \( f(x) \) from \( 0 \) to \( a \) represents the area under the curve up to the point \( (a, b) \).
- The area under \( f^{-1}(x) \) from \( 0 \) to \( b \) represents the area under the inverse function, which geometrically corresponds to the area of a rectangle formed by the points \( (0,0) \), \( (a,0) \), \( (a,b) \), and \( (0,b) \).

### Step 4: Calculate the total area
The area of the rectangle formed by the dimensions \( a \) and \( b \) is given by:

\[ \text{Area} = \text{Length} \times \text{Breadth} = a \times b \]

### Step 5: Conclude the result
Thus, we can conclude that:

\[ I = \int_{0}^{a} f(x) \, dx + \int_{0}^{b} f^{-1}(x) \, dx = a \cdot b \]

### Final Answer
Therefore, the expression evaluates to:

\[ I = ab \]