Let `f` be a one to one continuous function such that `f(2)=3` and `f(5)=6`. Given `int_(2)^(5)f(x)dx=17`, then find the value of `int_(3)^(7)f^(-1)(x)dx`.

Explanation

To solve the problem, we need to find the value of the integral \( \int_{3}^{7} f^{-1}(x) \, dx \) given the properties of the function \( f \) and the integral \( \int_{2}^{5} f(x) \, dx = 17 \).

### Step-by-Step Solution:

1. **Understand the relationship between \( f \) and \( f^{-1} \)**:
We know that if \( y = f(x) \), then \( x = f^{-1}(y) \). This means we can change the variable of integration when dealing with the inverse function.

2. **Change of Variables**:
We can express the integral of \( f^{-1}(x) \) in terms of \( f(x) \):
\[ \int_{3}^{7} f^{-1}(x) \, dx = \int_{f^{-1}(3)}^{f^{-1}(7)} y \, f'(f^{-1}(y)) \, dy \]
Here, \( f^{-1}(3) = 2 \) and \( f^{-1}(7) = 5 \) based on the given values \( f(2) = 3 \) and \( f(5) = 6 \).

3. **Set up the Integral**:
Thus, we can rewrite the integral:
\[ \int_{3}^{7} f^{-1}(x) \, dx = \int_{2}^{5} f^{-1}(f(x)) f'(x) \, dx \]
This simplifies to:
\[ \int_{2}^{5} x \, f'(x) \, dx \]

4. **Integration by Parts**:
We apply integration by parts where:
- Let \( u = x \) and \( dv = f'(x) \, dx \)
- Then \( du = dx \) and \( v = f(x) \)

The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Applying this, we get:
\[ \int_{2}^{5} x \, f'(x) \, dx = \left[ x f(x) \right]_{2}^{5} - \int_{2}^{5} f(x) \, dx \]

5. **Evaluate the Boundary Terms**:
Now we evaluate the boundary terms:
\[ \left[ x f(x) \right]_{2}^{5} = 5 f(5) - 2 f(2) = 5 \cdot 6 - 2 \cdot 3 = 30 - 6 = 24 \]

6. **Substitute the Known Integral**:
We know from the problem statement that \( \int_{2}^{5} f(x) \, dx = 17 \). Therefore:
\[ \int_{2}^{5} x f'(x) \, dx = 24 - 17 = 7 \]

7. **Final Result**:
Thus, the value of the integral \( \int_{3}^{7} f^{-1}(x) \, dx \) is:
\[ \boxed{7} \]