Problem 12 - University Calculus: Early Transcendentals

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. 2\. a. \(\frac{1}{3 x}\) b. \(\frac{2}{5 x}\) c. \(1+\frac{4}{3 x}-\frac{1}{x^{2}}\)

Step-by-Step Solution

Answer

a. \(\frac{1}{3}\ln|x| + C\); b. \(\frac{2}{5}\ln|x| + C\); c. \(x + \frac{4}{3}\ln|x| + \frac{1}{x} + C\).

Step 1: Understanding Antiderivatives

An antiderivative of a function is another function whose derivative is the original function. To find an antiderivative, you need to think about a function that can be differentiated to produce the given function.

Step 2: Applying Basic Antidifferentiation

For a function of the form \(\frac{1}{x}\), an antiderivative is \(\ln|x|\). Therefore, the antiderivative of \(\frac{1}{3x}\) can be written as \(\frac{1}{3}\ln|x| + C\), where \(C\) is the constant of integration.

Step 3: Solving Part (a)

For \(\frac{1}{3x}\), the antiderivative is \(\frac{1}{3}\ln|x| + C\).

Step 4: Solving Part (b)

Similarly, for \(\frac{2}{5x}\), the antiderivative is \(\frac{2}{5}\ln|x| + C\). This is because the constant \(\frac{2}{5}\) is factored out from the integrand.

Step 5: Separating Terms in Part (c)

The given function in part (c) is a sum of terms: \(1 + \frac{4}{3x} - \frac{1}{x^{2}}\). You can find the antiderivative of each term separately and then add them together.

Step 6: Antidifferentiating Each Term

- For \(1\), the antiderivative is \(x\).- For \(\frac{4}{3x}\), the antiderivative is \(\frac{4}{3}\ln|x|\).- For \(-\frac{1}{x^{2}}\), rewrite it as \(-x^{-2}\). The antiderivative is \(\frac{1}{x}\), because integrating \(-x^{-2}\) gives \(x^{-1}\).

Step 7: Solving Part (c)

Combine the antiderivatives of each term: The antiderivative is \(x + \frac{4}{3}\ln|x| + \frac{1}{x} + C\), where \(C\) is the constant of integration.

Step 8: Checking Antiderivatives by Differentiation

Differentiate each of the antiderivatives to verify that they match the original function.- Differentiate \(\frac{1}{3}\ln|x| + C\) to get \(\frac{1}{3x}\).- Differentiate \(\frac{2}{5}\ln|x| + C\) to get \(\frac{2}{5x}\).- Differentiate \(x + \frac{4}{3}\ln|x| + \frac{1}{x} + C\) to get \(1 + \frac{4}{3x} - \frac{1}{x^{2}}\).The antiderivatives are correct.

Key Concepts

Understanding Integration

Integration is a fundamental concept in calculus. It essentially involves finding an antiderivative or a family of functions whose derivative is the given function. Integration is the reverse process of differentiation. When we talk about "finding an antiderivative," we mean looking for a function that will yield the given function when differentiated. This process often involves recognizing patterns and applying certain known formulas or rules.

A common integration formula to remember is the integral of \( \frac{1}{x} \) which is \( \ln|x| \). This tells us that the natural logarithm function has a special relationship with integrals. Moreover, integration always involves a constant of integration, represented as \( C \). This constant exists because differentiating a constant yields zero, and hence we never see it in the original function.

In practice, we sometimes need to break down complex functions into simpler parts, integrating each component separately and combining results. For example, with a function like \( 1 + \frac{4}{3x} - \frac{1}{x^{2}} \, \) each term is integrated individually to find the overall antiderivative. This step-by-step approach is both systematic and straightforward.

Exploring Differentiation

Differentiation is the process of finding the derivative of a function. It is a critical concept in calculus that measures how a function changes. In simple terms, the derivative tells us the rate of change or the slope of the function at a specific point.

Differentiation and integration are inverse operations. This means that if you differentiate a function and then integrate the result, you'll end up back where you started. This property is essential for checking antiderivatives. Once you've found an antiderivative, differentiating it should give you back the original function. For instance, differentiating \( \frac{1}{3}\ln|x| + C \) gives \( \frac{1}{3x} \), thus verifying the original function.

When applying differentiation, you need to be comfortable with several rules, such as the power rule and the chain rule, among others. These rules help break down more complicated differentiation tasks into manageable steps, much like the breakdown seen in integration.

The Role of Natural Logarithms

Natural logarithms are logarithms with base \( e \), where \( e \approx 2.71828 \). They are particularly significant in calculus due to their intrinsic relationship with the exponential function. The derivative of \( \ln|x| \) is \( \frac{1}{x} \, \) a fact that makes the natural logarithm function crucial when dealing with integrals involving \( \frac{1}{x} \).

This relationship is leveraged in integration. When integrating functions of the form \( \frac{1}{x} \), the result naturally involves the logarithmic function \( \ln|x| \. \) For a function like \( \frac{1}{3x} \, \) the antiderivative becomes \( \frac{1}{3}\ln|x| + C \).

Understanding this connection is crucial not just for solving integrals but also for appreciating the profound symmetry in calculus. The natural logarithm provides the necessary bridge between integration and differentiation for certain types of functions. It is both an algebraic and graphically smooth transition between these two operations, underscoring the elegance and interconnectedness of calculus principles.