Problem 76
Question
Use integration by parts to show that if \(f\) has an inverse with continuous first derivative, then. $$\int f^{-1}(x) d x=x f^{-1}(x)-\int x\left(f^{-1}\right)^{\prime}(x) d x$$
Step-by-Step Solution
Verified Answer
Using integration by parts with \(u = f^{-1}(x)\) and \(dv = dx\), we find \(du = \left(f^{-1}\right)^\prime(x) dx\) and \(v = x\). Applying the formula, we get:
\[\int f^{-1}(x) dx = x f^{-1}(x) - \int x\left(f^{-1}\right)^\prime(x) dx.\]
This proves the given expression.
1Step 1: Recall the necessary formulas
First, we need to recall the formula for the integration by parts:
\[\int u dv = uv - \int v du,\]
where \(u\) and \(v\) are differentiable functions.
Secondly, we need to remember the formula for the derivative of an inverse function \(f^{-1}(x)\), which is:
\[\left(f^{-1}\right)^\prime(x) = \frac{1}{f^\prime\left(f^{-1}(x)\right)}.\]
2Step 2: Choose u and dv for integration by parts
To choose the appropriate functions for integration by parts, let's rewrite the given integral using \(y = f^{-1}(x)\) as:
\[\int y\left(\frac{dx}{dy}\right) dy.\]
Here, we choose \(u = y\) and \(\frac{dv}{dy} = \frac{dx}{dy}\).
3Step 3: Find du and v
First, let's differentiate \(u\) with respect to \(y\) to find du:
\(du = \frac{d(y)}{dy} dy = dy\).
Now, let's integrate \(dv\) with respect to \(y\) to find \(v\):
\(v = \int \frac{dx}{dy} dy = x\).
4Step 4: Apply the integration by parts formula
According to our previous choices, applying the integration by parts formula yields:
\[\int y\left(\frac{dx}{dy}\right) dy = yx - \int x dy.\]
5Step 5: Convert back to the original variable x
Now, let's substitute \(y = f^{-1}(x)\) and use the formula for the derivative of an inverse function to find \(dy\):
\(dy = \frac{dx}{\left(f^{-1}\right)^\prime(x)}\).
Plug these expressions back into the equation:
\[\int f^{-1}(x) dx = x f^{-1}(x) - \int x\left(f^{-1}\right)^\prime(x) dx.\]
This completes the proof that
\[\int f^{-1}(x) dx = x f^{-1}(x) - \int x\left(f^{-1}\right)^\prime(x) dx.\]
Key Concepts
Inverse FunctionsCalculusIntegration Techniques
Inverse Functions
An inverse function is essentially a function that reverses the effect of the original function. If we have a function \(f(x)\), and its inverse \(f^{-1}(x)\), applying \(f\) then \(f^{-1}\) will return the initial value: \(f(f^{-1}(x)) = x\). Not all functions have inverses, but those that do must pass the horizontal line test, meaning they must be one-to-one.
The derivative of an inverse function, as used in our problem, is given by:\[\left(f^{-1}\right)^' (x) = \frac{1}{f^' (f^{-1}(x))}\]
This is derived from the basic property of inverse functions: if \(y = f(x)\) then \(x = f^{-1}(y)\). Differentiating both sides with respect to \(x\) and rearranging gives the formula. Understanding this concept is crucial when applying calculus to problems involving inverse functions.
The derivative of an inverse function, as used in our problem, is given by:\[\left(f^{-1}\right)^' (x) = \frac{1}{f^' (f^{-1}(x))}\]
This is derived from the basic property of inverse functions: if \(y = f(x)\) then \(x = f^{-1}(y)\). Differentiating both sides with respect to \(x\) and rearranging gives the formula. Understanding this concept is crucial when applying calculus to problems involving inverse functions.
Calculus
Calculus is the branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It is a foundational tool in understanding the behavior of functions and mathematical modeling. Calculus consists of two main parts: differentiation and integration.
Differentiation assesses how functions change, while integration involves the accumulation of quantities, such as areas under curves. In this exercise, we're specifically dealing with integration. Integration by parts is an advanced technique that is based on the product rule for differentiation. It breaks down the integral of a product of functions into simpler parts. This simplifies the process of dealing with complex integrals, such as those involving inverse functions.
Differentiation assesses how functions change, while integration involves the accumulation of quantities, such as areas under curves. In this exercise, we're specifically dealing with integration. Integration by parts is an advanced technique that is based on the product rule for differentiation. It breaks down the integral of a product of functions into simpler parts. This simplifies the process of dealing with complex integrals, such as those involving inverse functions.
Integration Techniques
Integration techniques are methods used to find integrals and solve problems involving accumulation. Among many tools, integration by parts is essential for dealing with products of functions. It's guided by the formula:\[ \int u \, dv = uv - \int v \, du \].
This method works by identifying parts of the integral to differentiate and integrate separately. For instance, in our exercise:
This method works by identifying parts of the integral to differentiate and integrate separately. For instance, in our exercise:
- We identify \(u\) as the inverse function \(y\), making it easier to differnetiate
- \(dv\) is represented as \(\frac{dx}{dy}\) to be integrated, yielding \(v = x\)
- Substituting \(u\), \(du\), \(v\), and \(dv\) allows us to simplify the integral
Other exercises in this chapter
Problem 74
If \(P\) is a polynomial of degree \(k\). then $$\int P(x) e^{x} d x=\left[P(x)-P^{\prime}(x)+\cdots \pm P^{(i)}(x)\right] e^{x}+C$$ Verify this statement. For
View solution Problem 75
Use the statement in Exercise 74 to calculate: (a) \(\int\left(x^{2}-3 x+1\right) e^{x} d x\) (b) \(\int\left(x^{3}-2 x\right) e^{x} d x\)
View solution Problem 77
Show that if \(f\) and \(g\) have continuous sceond derivatives and \(f(a)=g(a)=f(b)=g(b)=0,\) then $$\int_{a}^{b} f(x) y^{\prime \prime}(x) d x=\int_{a}^{b} g(
View solution Problem 78
You are familiar with the identity $$f(b)-f(a)=\int_{a}^{b} f^{\prime}(x) d x$$ (a) Assume that \(f\) has a continuous second derivative. Use integration by par
View solution