Problem 58 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(F_{0}(x)=x \sin x\) and \(F_{n+1}(x)=\int F_{n}(x) d x\). (a) Determine \(F_{1}(x), F_{2}(x), F_{3}(x)\), and \(F_{4}(x)\). (b) On the basis of part (a), conjecture the form of \(F_{16}(x)\).

Step-by-Step Solution

Answer

F_1(x) = -x cos(x) + sin(x), F_2(x) = -x sin(x), F_3(x) = -x cos(x) + sin(x), F_4(x) = -x sin(x), F_16(x) = -x sin(x).

Step 1: Define F_0(x)

Given that \(F_{0}(x) = x \sin(x)\). This is the base function from which we will derive the succeeding functions by integration.

Step 2: Find F_1(x)

We need to compute \(F_1(x)\), which is the integral of \(F_0(x)\):\[F_1(x) = \int x \sin(x) \, dx\]Using integration by parts, where \(u = x\) and \(dv = \sin(x)\,dx\):\- \(du = dx\) and \(v = -\cos(x)\).Apply the integration by parts formula \(\int u\,dv = uv - \int v\,du\):\[F_1(x) = -x\cos(x) + \int \cos(x) \, dx = -x\cos(x) + \sin(x) + C_1\]

Step 3: Find F_2(x)

Compute \(F_2(x)\) by integrating \(F_1(x)\):\[F_2(x) = \int (-x\cos(x) + \sin(x)) \, dx\]Integrate term by term, using integration by parts for \(-x\cos(x)\):- Let \(u = x\) and \(dv = -\cos(x)\,dx\): \(du = dx\) and \(v = -\sin(x)\).Apply again:\[\int -x\cos(x) \, dx = -x\sin(x) + \int \sin(x) \, dx = -x\sin(x) + \cos(x) + C_2\]\[\int \sin(x) \, dx = -\cos(x)\]So, \(F_2(x) = -x\sin(x) + \cos(x) - \cos(x) + C_2 = -x\sin(x) + C_2\).

Step 4: Find F_3(x)

Integrate \(F_2(x)\) to find \(F_3(x)\):\[F_3(x) = \int -x\sin(x) \, dx\]Using integration by parts, where \(u = x\) and \(dv = -\sin(x)\,dx\): \(du = dx\) and \(v = \cos(x)\).Apply:\[\int -x\sin(x) \, dx = -x\cos(x) + \int \cos(x) \, dx = -x\cos(x) + \sin(x) + C_3\]

Step 5: Find F_4(x)

Compute \(F_4(x)\) by integrating \(F_3(x)\):\[F_4(x) = \int (-x\cos(x) + \sin(x)) \, dx\]Integrate term by term:From Step 2: \(-x\cos(x) + \sin(x)\) integrates to:\[-x\sin(x) + \cos(x) + C_4\]So, \(F_4(x) = -x\sin(x) + \cos(x) + C_4\).

Step 6: Analyze the pattern

Observe the pattern:- \(F_0(x) = x\sin(x)\)- \(F_1(x) = -x\cos(x) + \sin(x)\)- \(F_2(x) = -x\sin(x)\)- \(F_3(x) = -x\cos(x) + \sin(x)\)- \(F_4(x) = -x\sin(x)\)The pattern alternates between forms similar to \(F_1(x)\) and \(F_2(x)\).

Step 7: Conjecture F_16(x)

From the pattern observed:- (Odd n, like \(F_1\) and \(F_3\)) : \(-x\cos(x) + \sin(x)\)- (Even n, like \(F_2\) and \(F_4\)) : \(-x\sin(x)\)Since 16 is even, \(F_16(x)\) would follow the pattern for even indices:\[F_{16}(x) = -x\sin(x)\]

Key Concepts

Definite Integrals

Definite integrals offer a way to calculate the accumulated area under a curve over a specific interval. Unlike indefinite integrals, which provide a general solution with a constant of integration, definite integrals deliver an exact numerical value. This numerical value represents the net area under the curve, taking into account areas above and below the x-axis by considering them with positive and negative signs, respectively.

Here's a quick look at how a definite integral is structured:

• It involves an integral symbol, the function to be integrated, and the limits of integration, which denote the interval over which the area is calculated.
• The result is sensitive to the limits of integration: changing these limits changes the area computed.

The methods used to solve definite integrals are often the same as those used for indefinite integrals, such as substitution or integration by parts, though definite integrals will include evaluating the antiderivative at the given bounds.

In exercises like the one given, definite integrals weren't directly calculated, but understanding them provides a deeper grasp of indefinite integrals since the processes are intertwined.

Indefinite Integrals

Indefinite integrals are essentially the antiderivative of a function. When solving an indefinite integral, the result is a family of functions with an added constant of integration, denoted as \( C \). This constant arises because differentiating a constant yields zero, making it impossible to detect through differentiation alone which specific constant was in the original function set.

There are various techniques to solve indefinite integrals:

• Basic Antiderivatives: Recognize the antiderivatives of basic functions such as sin(x) and cos(x).
• Integration by Parts: A technique useful when handling products of functions, often using the formula \( \int u \, dv = uv - \int v \, du \).
• Substitution: Simplifies the integral by substituting part of the integral with a single variable, useful when dealing with composite functions.

In the problem provided, we repeatedly found indefinite integrals using integration by parts. Each step included finding an antiderivative, demonstrating how each integration contributes to forming the series of functions \( F_n(x) \). Understanding indefinite integrals is crucial, as it lays the groundwork for mastering definite integrals.

Pattern Recognition in Calculus

Pattern recognition is a powerful tool in calculus, especially when dealing with sequences or series of functions. When evaluating problems like the exercise, recognizing patterns can simplify predicting the behavior of future functions in a sequence.

For example, in the given exercise, a pattern emerged in the functions \( F_n(x) \):

• The functions alternated between two main forms, \(-x\cos(x) + \sin(x)\) for odd indices and \(-x\sin(x)\) for even indices.
• Recognizing this pattern enabled the conjecture of \( F_{16}(x) \) without calculating every intermediary function.

This type of recognition helps not only in predicting future values but also in verifying the consistency and correctness of solutions. Patterns can exist in many forms:

• Alternating behaviors as seen in this exercise.
• Repeating cycles or motifs across intervals.
• Symmetries or other predictable behaviors.

By identifying patterns, students can reduce complex problems into simpler ones, enhancing problem-solving efficiency and comprehension.