Use the substitution y = f
−1
(x) and integration by parts to prove the identity
Z
f
−1
(x) dx = xf −1
(x) −
Z
f (y) dy.

Z is the integral sign.

We are tasked with proving the identity using the substitution \( y = f^{-1}(x) \) and integration by parts:

\[
\int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(y) \, dy
\]

### Step-by-step solution:

#### 1. Substitution \( y = f^{-1}(x) \)
We start with the substitution \( y = f^{-1}(x) \), which implies that:
\[
x = f(y)
\]
Since \( x = f(y) \), the derivative \( \frac{dx}{dy} \) is simply:
\[
\frac{dx}{dy} = f'(y)
\]
Thus, \( dx = f'(y) \, dy \).

#### 2. Applying Integration by Parts
We will use the formula for integration by parts:
\[
\int u \, dv = uv - \int v \, du
\]
Let:
- \( u = f^{-1}(x) \), so \( du = \frac{1}{f'(y)} \, dx \),
- \( dv = dx = f'(y) \, dy \), so \( v = x = f(y) \).

Now, applying integration by parts:
\[
\int f^{-1}(x) \, dx = x f^{-1}(x) - \int x \, d(f^{-1}(x))
\]
But \( x = f(y) \), and \( d(f^{-1}(x)) = \frac{1}{f'(y)} dx = dy \), so the equation becomes:
\[
\int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(y) \, dy
\]

Thus, the identity is proven.

### Final Result:
\[
\boxed{\int f^{-1}(x) \, dx = x f^{-1}(x) - \int f(y) \, dy}
\]

Would you like further clarification or details on any part of the proof? 

Here are 5 related questions to explore the concept further:
1. How is the substitution \( y = f^{-1}(x) \) applied in other types of integrals?
2. Can you use the same method to derive other inverse function integrals?
3. How does integration by parts help in simplifying the integral involving inverse functions?
4. What are the conditions for \( f^{-1}(x) \) to exist and be differentiable?
5. What is the geometric interpretation of the integral of \( f^{-1}(x) \)?

**Tip:** When using substitution, always track how both the function and its derivative transform to maintain consistency throughout the problem.

Use the substitution y = f^−1(x) and integration by parts to prove the identity ∫f^−1(x) dx = x f^−1(x) − ∫f(y) dy.

Explore how to prove the identity ∫f^−1(x) dx = x f^−1(x) − ∫f(y) dy using substitution and integration by parts. This detailed step-by-step solution covers advanced calculus concepts such as inverse functions and integration by parts, ideal for undergraduate students.

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