If f (x) is monotonic differentiable function on [a,b] then `int_(a)^(b) f(x) dx + int_(f (a))^(f(b)) f^(-1) (x) dx`=

Options

A. `b f(a)- af(b)`
B. `bf(b)- af (a)`
C. `f(a ) + f(b)`
D. cannot be found

Answer

To solve the problem, we need to show that:

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

### Step-by-Step Solution:

1. **Start with the given integrals:**
\[ I_1 = \int_{a}^{b} f(x) \, dx \]
\[ I_2 = \int_{f(a)}^{f(b)} f^{-1}(x) \, dx \]

2. **Use integration by parts on \( I_1 \):**
We can express the integral using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Let \( u = f(x) \) and \( dv = dx \). Then, \( du = f'(x) \, dx \) and \( v = x \).

Thus, we have:
\[ I_1 = \left[ x f(x) \right]_{a}^{b} - \int_{a}^{b} x f'(x) \, dx \]
This simplifies to:
\[ I_1 = b f(b) - a f(a) - \int_{a}^{b} x f'(x) \, dx \]

3. **Change of variables in \( I_2 \):**
For \( I_2 \), we perform a change of variables. Let \( x = f(t) \), then \( dx = f'(t) \, dt \). The limits change as follows:
- When \( x = f(a) \), \( t = a \)
- When \( x = f(b) \), \( t = b \)

Thus, we can rewrite \( I_2 \):
\[ I_2 = \int_{a}^{b} t f'(t) \, dt \]

4. **Combine \( I_1 \) and \( I_2 \):**
Now we can combine the two integrals:
\[ I_1 + I_2 = \left( b f(b) - a f(a) - \int_{a}^{b} x f'(x) \, dx \right) + \int_{a}^{b} t f'(t) \, dt \]
Since \( x \) and \( t \) are dummy variables, we can replace \( t \) with \( x \):
\[ I_1 + I_2 = b f(b) - a f(a) - \int_{a}^{b} x f'(x) \, dx + \int_{a}^{b} x f'(x) \, dx \]
The integrals cancel out:
\[ I_1 + I_2 = b f(b) - a f(a) \]

5. **Final Result:**
Therefore, we conclude that:
\[ \int_{a}^{b} f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(x) \, dx = b f(b) - a f(a) \]