Problem 24 - Calculus and its applications

Question

Evaluate. (Be sure to check by differentiating!) $$ \int \frac{d x}{4-x} $$

Step-by-Step Solution

Answer

The integral evaluates to \(-\ln |4-x| + C\).

Step 1: Identify the Integral Form

The integral given is \( \int \frac{dx}{4-x} \). Recognize that this is a standard form of the integral of \( \frac{1}{u} \) where \( u = 4-x \).

Step 2: Find the Derivative of \( u \)

Set \( u = 4-x \). Differentiate \( u \) with respect to \( x \) to find \( du \). This gives \( du = -dx \). This implies \( dx = -du \).

Step 3: Substitute in Terms of \( u \)

Substitute \( u = 4-x \) and \( dx = -du \) into the integral to get \( \int \frac{-du}{u} \).

Step 4: Evaluate the Integral

The integral \( \int \frac{-du}{u} \) evaluates to \(-\ln |u| + C\), where \( C \) is the integration constant. So the integral becomes \(-\ln |4-x| + C\).

Step 5: Verify by Differentiation

Differentiate \(-\ln |4-x| + C\) with respect to \( x \). Using the chain rule, the derivative is \(-\frac{1}{4-x} \times (-1) = \frac{1}{4-x}\), which matches the original integrand.

Key Concepts

Integration Techniques

Integration is a powerful tool in calculus used to find areas under curves, among other applications. Different techniques can be applied depending on the form of the function to integrate. In this exercise, the integral \[\int \frac{dx}{4-x}\]is tackled using substitution. **Substitution Method:**The substitution, or change of variables technique, simplifies the integral by transforming it into a more recognizable form. In this case, we let \[u = 4-x\]and determine the relationship between the differentials: \[du = -dx\].This allows us to rewrite the integral in terms of \( u \):

• Original Integral: \[ \int \frac{dx}{4-x} \]
• In Terms of \( u \): \[ \int \frac{-du}{u} \]

The integral \( \int \frac{du}{u} \)is a standard form, and its solution is well-known, which makes evaluating it straightforward. This technique is immensely helpful with integrands that appear complex at first glance.

Chain Rule

The chain rule is a fundamental concept in calculus for differentiating compositions of functions. This rule states that if a variable \( y \)depends on \( u \), which in turn depends on \( x \), then the derivative of \( y \) with respect to \( x \) can be computed as: \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.\]**Using the Chain Rule for Verification:**Once the integral is solved and simplified as \(-\ln |4-x| + C\),the chain rule helps in verifying our solution by differentiation. Differentiating \(-\ln |4-x| + C\)involves finding:

• The derivative of \( -\ln |u| \) with respect to \( u \), which is \(-\frac{1}{u}\).
• The derivative of \( u = 4-x \) with respect to \( x \), which is \(-1\).

Applying the chain rule, the derivative becomes:\[-\frac{1}{4-x} \times (-1) = \frac{1}{4-x},\]confirming it matches the original function \( \frac{1}{4-x} \). This demonstrates the correctness of the integration process.

Definite and Indefinite Integrals

Integrals are classified primarily as either definite or indefinite. **Indefinite Integrals:**An indefinite integral, like the one in this exercise,begins with a general expression and includes an arbitrary constant \( C \):\[\int \frac{1}{u} \, du = \ln |u| + C.\]This constant arises because integration is essentially the reverse of differentiation, and many functions can have the same derivative.**Definite Integrals:**Definite integrals, on the other hand, have upper and lower limits and result in a numerical value representing an area under a curve. One does not need to include \( C \) in this case. The definite integral adapts the Fundamental Theorem of Calculus, which relates it to antiderivatives. In this specific task, though only an indefinite integral was required, understanding both forms helps you apply these ideas to various contexts and assures versatility in dealing with integration. Thus, knowing their distinctions expands the understanding of integral calculus as a whole.