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Chapter 11: Problem 1

Zeigen Sie, dass sich zwei verschiedene Stammfunktionen \(F_{1}\) und \(F_{2}\) einer gegebenen Funktion \(f\) h枚chstens um eine additive Konstante unterscheiden.

Short Answer

Expert verified
Answer: From the solution, we can conclude that the two different antiderivatives of a given function can differ only by an additive constant because their difference, represented by the function G(x), has a derivative of zero, indicating that G(x) is a constant function.

Step by step solution

01

Define the Antiderivatives of the Given Function

Given a function \(f\), we have two antiderivatives, \(F_1\) and \(F_2\), such that: (1) \(\frac{dF_1}{dx} = f(x)\) (2) \(\frac{dF_2}{dx} = f(x)\)
02

Establish the Relationship between \(F_1\) and \(F_2\)

Now, let's consider a new function \(G(x) = F_1(x) - F_2(x)\). We aim to show that this new function \(G(x)\) only differs by a constant. To prove this, we will take the derivative of \(G(x)\) with respect to \(x\).
03

Derivative of \(G(x)\)

Applying the derivative operator on both sides of the equation, we get: \(\frac{dG(x)}{dx} = \frac{d(F_1(x) - F_2(x))}{dx}\) Using the properties of derivatives, we know that the derivative of a subtraction is the subtraction of the derivatives: \(\frac{dG(x)}{dx} = \frac{dF_1(x)}{dx} - \frac{dF_2(x)}{dx}\)
04

Substitute the Given Antiderivatives and Simplify

Recall the given antiderivatives \(F_1\) and \(F_2\), for which we have: \(\frac{dF_1(x)}{dx} = f(x)\) \(\frac{dF_2(x)}{dx} = f(x)\) Substitute these values back into the equation: \(\frac{dG(x)}{dx} = f(x) - f(x)\) Simplifying the expression gives: \(\frac{dG(x)}{dx} = 0\)
05

Conclude that \(G(x)\) is a Constant Function

Since the derivative of \(G(x)\) is zero, we can conclude that \(G(x)\) is a constant function. This constant function represents the difference between the two antiderivatives \(F_1(x)\) and \(F_2(x)\). So, we have finally proven that the two different antiderivatives of a given function \(f\) can differ only by an additive constant.

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