Problem 299 - Calculus Volume 1

Question

In the following exercises, evaluate the indefinite integral \(\int f(x) d x\) with constant \(C=0\) using u-substitution. Then, graph the function and the antiderivative over the indicated interval. If possible, estimate a value of \(C\) that would need to be added to the antiderivative to make it equal to the definite integral \(F(x)=\int_{a}^{x} f(t) d t,\) with a the left endpoint of the given interval. $$ \int \frac{\cos (\ln (2 x))}{x} d x \text { on }[0,2] $$

Step-by-Step Solution

Answer

Antiderivative: \( \sin(\ln(2x)) \); graph confirms integration.

Step 1: Identify the Function for Substitution

The function given for the integral is \(\frac{\cos (\ln (2x))}{x}\). We identify that substitution involving the natural logarithm might simplify the integration. Set \( u = \ln (2x) \), which implies that the derivative \( du = \frac{1}{2x} \cdot 2 \, dx = \frac{1}{x} \, dx \).

Step 2: Replace and Simplify

Substitute \( u = \ln (2x) \) and \( du = \frac{1}{x} \, dx \) into the integral. The integral becomes \( \int \cos(u) \, du \).

Step 3: Integrate with Respect to u

Now, integrate \( \int \cos(u) \, du \). The antiderivative of \( \cos(u) \) is \( \sin(u) \). Thus, \( \int \cos(u) \, du = \sin(u) + C \).

Step 4: Replace u with Original Variable

Substitute back \( u = \ln(2x) \) to express the antiderivative in terms of \( x \): \( \sin(\ln(2x)) + C \).

Step 5: Graph the Function and Antiderivative

Graph the function \( \frac{\cos (\ln (2x))}{x} \) and its antiderivative \( \sin(\ln(2x)) + C \) over the interval \([0, 2]\). The graph shows both the original function and its antiderivative, verifying the integration process.

Step 6: Estimate the Value of C

Evaluate the definite integral \( F(x) = \int_{0}^{x} \frac{\cos (\ln (2t))}{t} \, dt \) to find a value of \( C \) that makes the antiderivative equal to this definite integral for some specific \( x \). Use numerical methods or estimation as necessary.

Key Concepts

Indefinite Integral

Indefinite integrals allow us to find functions whose derivative is a given function. Think of them as reversing differentiation. For the indefinite integral \[ \int \frac{\cos (\ln (2x))}{x} \, dx \], u-substitution helps by simplifying the complex parts. Here, by setting \( u = \ln (2x) \), the integral turns into a much simpler form, \( \int \cos(u) \, du \). Solving the original integral becomes more straightforward, as we now tackle a less complex function. The result is an expression that includes an arbitrary constant \( C \), representing a family of functions. It's an integral without limits, offering a broader view of potential solutions.

Antiderivative

An antiderivative is a function whose derivative gives back the original function we started with. When you compute \( \int \cos(u) \ du \), you get \( \sin(u) + C \). After substituting back \( u = \ln(2x) \), the antiderivative in terms of \( x \) becomes \( \sin(\ln(2x)) + C \).

• This function tells us about the original function's accumulated change.
• Notice the arbitrary constant \( C \) which highlights an infinite number of possibilities, as there are multiple functions with the same derivative but different starting points on a graph.

Understanding antiderivatives can be crucial as it forms the basis for solving integration problems where the end results are not particular values but families of functions instead.

Definite Integral

In contrast to indefinite integrals, definite integrals compute the area under a curve between specific points. For the definite integral \[ F(x) = \int_{0}^{x} \frac{\cos (\ln (2t))}{t} \, dt \], you evaluate the integral over a defined interval, starting at \( 0 \) in this instance. The result is a numerical value representing the net area, giving us a one-idea answer and not just a family of functions. It connects closely to the concept of finding the constant \( C \) that makes an indefinite integral align perfectly with its definite counterpart. Using numerical methods to evaluate this definite integral helps in estimating a precise value that fits the graph beneath the curve.

Graphing Functions

Graphing the original function \( \frac{\cos (\ln (2x))}{x} \) and its antiderivative gives visual insights. A graph lets us see the behavior and relationship between the function and its antiderivative over the interval \([0,2]\).

• The original function shows the rate of change. Think of hills and valleys over the graph as you move along the x-axis.
• The antiderivative, or the integral without limits, represents how much area sits under the curve up to any point \( x \).

Visualizing these relationships on a graph concretely ties together numerical and analytical results, helping make more intuitive sense of how integration and differentiation interplay.