Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove the formula $$ \int \frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)} d x=\frac{f(x)}{g(x)}+C $$

Step-by-Step Solution

Answer

The integral results in \( \frac{f(x)}{g(x)} + C \) by applying the quotient rule in reverse.

Step 1: Identify the Structure of the Integral

We observe that the integral \( \int \frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)} \, dx \) fits the pattern for the quotient rule in calculus, which is used for differentiation. This suggests that the expression is the derivative of a quotient, where the numerator is the result of differentiating \( \frac{f(x)}{g(x)} \).

Step 2: Recall the Quotient Rule for Derivatives

The quotient rule states that the derivative \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \) for two differentiable functions \( u(x) \) and \( v(x) \). By comparing the given function with \( \frac{u'v - uv'}{v^2} \), we suggest that \( u = f(x) \) and \( v = g(x) \).

Step 3: Recognize the Integral as the Antiderivative

The integral \( \int \frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)} \, dx \) represents the inverse operation of finding the derivative using the quotient rule. Thus, when we integrate, we should return to the original quotient \( \frac{f(x)}{g(x)} \).

Step 4: Apply Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if a function is the derivative of another function, then its integral gives back the original function plus a constant, \( C \). Therefore, we have \( \int \frac{g(x) f^{\prime}(x)-f(x) g^{\prime}(x)}{g^{2}(x)} \, dx = \frac{f(x)}{g(x)} + C \).

Key Concepts

Quotient Rule

The Quotient Rule in calculus is a technique used to differentiate functions that are expressed as a quotient, or a division of two functions. Instead of dividing and then differentiating, the quotient rule provides a direct way to find the derivative. It is particularly useful when dealing with more complicated expressions that can't be simplified. The rule states that for two differentiable functions, say \( u(x) \) and \( v(x) \), the derivative of their quotient \( \frac{u}{v} \) is given by the formula:

• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \)

In our exercise, we identified a structure within the integral that fits this exact pattern. By recognizing that the integral represents the derivative of a quotient, we were able to see how applying the quotient rule confirmed our result \( \frac{f(x)}{g(x)} \). It's a clear demonstration of how recognizing and employing the quotient rule helps simplify and solve complex calculus problems.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a cornerstone of calculus, bridging the concepts of differentiation and integration. It essentially tells us how differentiation and integration are reverse processes. This theorem has two parts:

• The first part asserts that if you have a continuous function that is the derivative of another function over an interval, its integral over this interval will give you the change in the original function over that interval.
• The second part allows you to compute the integral of a function using one of its antiderivatives. Specifically, if \( F \) is an antiderivative of \( f \) on an interval \([a, b] \), then:
• \( \int_a^b f(x) \, dx = F(b) - F(a) \)

In the problem at hand, the Fundamental Theorem of Calculus helps us to understand why integrating the derivative returns us to our original function up to a constant \( C \). This is why when we integrate the given derivative, we end up with \( \frac{f(x)}{g(x)} + C \). It shows the power and elegance of this theorem in reversing differentiation.

Antiderivative

An antiderivative is essentially the "reverse" of a derivative. Finding an antiderivative means determining a function whose derivative is the given function. In simpler terms, if you have a function \( f(x) \) and you're looking for another function \( F(x) \) such that \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).Antiderivatives play a crucial role in integration. When you compute an indefinite integral, you are essentially finding an antiderivative of the integrand. The solution of an indefinite integral usually includes a constant \( C \), which represents an entire family of functions, all differing by a constant. This constant is crucial because differentiation of constants yields zero, so it doesn't appear when taking derivatives but must be considered in integration.In our exercise, by recognizing the structure as a derivative of a known function, identifying its antiderivative led to concluding that the integral of the given expression is \( \frac{f(x)}{g(x)} + C \). Antiderivatives help us to connect differential and integral calculus smoothly.