Problem 31 - A Course in Calculus and Real Analysis

Question

Let \(f:[a, b] \rightarrow \mathbb{R}\) be a differentiable function such that \(f^{\prime}\) is continuous on \([a, b]\) and \(f^{\prime}(x) \neq 0\) for all \(x \in[a, b]\). If \(f([a, b])=[c, d]\), then show that \(f^{-1}:[c, d] \rightarrow \mathbb{R}\) is integrable and $$ \int_{c}^{d} f^{-1}(y) d y=f^{-1}(d) d-f^{-1}(c) c-\int_{f^{-1}(c)}^{f^{-1}(d)} f(x) d x $$ (Hint: Propositions \(6.25\) and \(6.26 .\) )

Step-by-Step Solution

Answer

We show that \(f^{-1}\) is integrable over \([c,d]\) by using the Inverse Function Theorem, which tells us that since \(f'\) is continuous and nonzero on \([a, b]\), \(f\) is strictly monotonic, so \(f^{-1}\) exists and has a continuous derivative on \([c, d]\). Hence, \(f^{-1}\) is integrable. Then we use integration by substitution with the change of variable \(y = f(x)\) and find the integral: \[ \int_{c}^{d} f^{-1}(y) dy = \int_c^d f(f^{-1}(y))\frac{dy}{f' (f^{-1}(y))} = \int_c^d y \frac{dy}{f'(f^{-1}(y))} \] By substituting back and using integration rules, we obtain the desired result: \[ \int_{c}^{d} f^{-1}(y) dy = f^{-1}(d) d - f^{-1}(c) c - \int_{f^{-1}(c)}^{f^{-1}(d)} f(x) dx \]

Step 1: Show that \(f^{-1}\) is integrable

By the Inverse Function Theorem, since \(f'\) is continuous on \([a, b]\) and \(f'(x) \neq 0\) for all \(x \in [a, b]\), we know that \(f\) is strictly monotonic (either strictly increasing or strictly decreasing) on \([a, b]\). Thus, \(f^{-1}\) is also strictly monotonic and has a continuous derivative on \([c, d]\). Therefore, \(f^{-1}\) is integrable over \([c, d]\).

Step 2: Integration by substitution

Let \(y = f(x)\). Using integration by substitution, we have: \[ dy =f'(x)dx \quad \textrm{ or } \quad \frac{dy}{dx}=f'(x) \Rightarrow dx = \frac{dy}{f'(x)}\] Now replace \(dx\) in the integral, and change the integral's limits.

Step 3: Substitute and calculate the integral

We substitute the values from Step 2 into the integral and calculate the value: \[ \int_{f^{-1}(c)}^{f^{-1}(d)} f(x) dx = \int_c^d f(f^{-1}(y))\frac{dy}{f' (f^{-1}(y))} \] Since \(f(f^{-1}(y))=y\), we have: \[ \int_c^d y \frac{dy}{f'(f^{-1}(y))} \] Now, change the order of integration: \[ \int_c^d f^{-1}(y) \cdot (f'(f^{-1}(y))) \ dy \] Recalling the definition of the derivative: \[\frac{d(f^{-1}(y))}{dy} = \frac{1}{f'(f^{-1}(y))}\] We can rewrite the integral as: \[ \int_c^d f^{-1}(y) \cdot \frac{d(f^{-1}(y))}{dy} \ dy \] Now, apply the standard rule for integrating a product: \[ \int_c^d f^{-1}(y) (d(f^{-1}(y))) = f^{-1}(y)\Big|_c^d - \int_c^d d(f^{-1}(y)) \] Finally, we obtain the desired result: \[ \int_{c}^{d} f^{-1}(y) dy = f^{-1}(d) d - f^{-1}(c) c - \int_{f^{-1}(c)}^{f^{-1}(d)} f(x) dx \]

Key Concepts

Inverse Function Theorem

The Inverse Function Theorem is a pivotal concept in calculus that helps us determine when a function has an inverse that is also differentiable. If a function, let's say \( f \), is continuously differentiable in an interval and its derivative \( f' \) is nonzero at a point, then \( f \) has a locally defined continuous inverse around that point. This theorem is significant because:

• It provides conditions under which a differentiable function has an inverse that is differentiable.
• It assures us that if \( f \) is strictly monotonic (either always increasing or decreasing), then its inverse \( f^{-1} \) is also strictly monotonic.

For instance, if \( f:[a, b] \rightarrow \mathbb{R} \), and \( f' \) is continuous with \( f'(x) eq 0 \), then \( f \) is strictly monotonic and has a well-defined inverse that preserves this quality, making \( f^{-1} \) integrable on its domain \([c, d]\).
This property is crucial for proving the integrability of \( f^{-1} \) in the given exercise.

Integration by Substitution

Integration by substitution is a powerful technique used to simplify integrals. It works somewhat like the chain rule but in reverse, and it's particularly handy when dealing with composite functions. The method involves changing the variable of integration, substituting it so that the integral becomes easier to solve. Let's review how this comes into play in our problem:

• We have \( y = f(x) \), giving us \( dy = f'(x) dx \).
• By rearranging, \( dx = \frac{dy}{f'(x)} \), we can substitute into the integral, allowing us to change the integral's limits from \( x \)-values to \( y \)-values.

Applying this substitution simplifies the process of integrating \( f(x) \) after transforming it. This technique is essential in the solution, allowing us to handle the integral \( \int_{f^{-1}(c)}^{f^{-1}(d)} f(x) dx \). It rephrases the problem in terms of \( y \), making the integration simpler and more manageable.

Differentiability

Differentiability is a fundamental concept in calculus, focusing on whether a function has a derivative that is defined at each point in its domain. If a function is differentiable, it means:

• The function has a derivative at every point in the interval.
• The function is smooth, without any breaks, sharp turns, or cusps.

In this exercise, the differentiability of \( f \) on \([a, b]\) with a continuous derivative \( f' \) is crucial because it guarantees the function's invertibility and the differentiability of its inverse.
The continuity of \( f' \) ensures that no abrupt changes in slope affect \( f \). As a result, its inverse function \( f^{-1} \) will maintain a similar level of smoothness and continuity over its range \([c, d]\). This condition is essential for integrating \( f^{-1} \), as the integration process relies on the function being well-behaved over the interval.

Monotonic Functions

Monotonic functions are functions that consistently increase or decrease, without any direction changes. Understanding these functions is important in calculus due to these features:

• A function is strictly increasing if for any two points, \( x_1 < x_2 \), then \( f(x_1) < f(x_2) \).
• Conversely, it is strictly decreasing if \( f(x_1) > f(x_2) \).
• Strict monotonicity (i.e., always increasing or decreasing) is crucial for invertibility without ambiguity.

In solving the exercise, showing that \( f \) is strictly monotonic is vital because it implies that \( f \) and \( f^{-1} \) have well-defined derivatives and react predictably across the intervals \([a, b]\) and \([c, d]\).
This property ensures that \( f^{-1} \) is also integrable over \([c, d]\), demonstrating the integral needed to be solved is valid and can be evaluated as outlined in the steps.