Problem 73 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(p(x)\) is a polynomial of degree \(n\) and \(G_{1}, G_{2}, \ldots, G_{n+1}\), are successive antiderivatives of a function \(g\), then, by repeated integration by parts, \(\int p(x) g(x) d x=p(x) G_{1}(x)-p^{\prime}(x) G_{2}(x)+p^{\prime \prime}(x) G_{3}(x)-\cdots\) \(+(-1)^{n} p^{(n)}(x) G_{n+1}(x)+C\) Use this result to find each of the following: (a) \(\int\left(x^{3}-2 x\right) e^{x} d x\) (b) \(\int\left(x^{2}-3 x+1\right) \sin x d x\)

Step-by-Step Solution

Answer

(a) \((x^3 - 3x^2 + 6x - 6)e^x + C\), (b) \((-x^2 + 3x - 1)\cos x + (2x - 3)\sin x + 2\cos x + C\)."

Step 1: Identifying the Polynomial and Antiderivative Functions

For problem (a), the polynomial is \(p(x) = x^3 - 2x\) and the function \(g(x) = e^x\). For problem (b), the polynomial is \(p(x) = x^2 - 3x + 1\) and the function \(g(x) = \sin x\). The degrees of the polynomials are 3 and 2, respectively.

Step 2: Finding Successive Antiderivatives for (a)

For \(g(x) = e^x\), each antiderivative remains \( G_n(x) = e^x\). Thus, \(G_1(x) = e^x\), \(G_2(x) = e^x\), ..., \(G_4(x) = e^x\).

Step 3: Applying By Parts Formula to (a)

Based on the polynomial \(x^3 - 2x\), using the formula: \[\int(x^3 - 2x)e^x dx = (x^3 - 2x)e^x - (3x^2 - 2)e^x + (6x)e^x - 6e^x + C\] Simplifying gives the result.

Step 4: Finding Successive Antiderivatives for (b)

For \(g(x) = \sin x\), the successive antiderivatives cycle as follows: \(G_1(x) = -\cos x\), \(G_2(x) = -\sin x\), \(G_3(x) = \cos x\).

Step 5: Applying By Parts Formula to (b)

Based on \(x^2 - 3x + 1\), applying the formula, we have: \[\int (x^2 - 3x + 1) \sin x \, dx = (x^2 - 3x + 1)(-\cos x) - (2x - 3)(-\sin x) + 2\cos x + C\] Simplify the expression for the result.

Key Concepts

Polynomial Integration

Polynomial integration involves finding the integral of polynomial expressions, which are mathematical expressions made up of variables raised to different powers, summed with coefficients. Integrating polynomials can seem straightforward, because there are specific rules that govern how each term in the polynomial can be integrated.
For instance, to integrate a polynomial term like \(x^n\), where \(n\) is a non-negative integer, we can use the basic rule:

• Increase the power by one: \(x^{n+1}\)
• Divide by the new exponent: \(\frac{x^{n+1}}{n+1}\)

This rule allows us to systematically find the integral or "antiderivative," which is a function whose derivative yields the original polynomial. However, when combined with other functions, such as exponential or trigonometric functions, integration often requires more advanced techniques, like integration by parts, to solve the problem efficiently.
Understanding this concept provides a foundational backdrop for mastering more complex integration scenarios.

Successive Antiderivatives

Successive antiderivatives refer to repeatedly finding antiderivatives, which are essentially the opposite of taking derivatives. Taking the antiderivative of a function gradually brings us back to its original form. For many problems, especially those involving integration by parts, it is necessary to find several antiderivatives of a function.
When the function is simple, like \(g(x) = e^x\), the antiderivative process is straightforward:

• The derivative of \(e^x\) is \(e^x\), making each successive antiderivative the same: \(G_1(x) = e^x, G_2(x) = e^x, \ldots\)

However, for functions like \(\sin x\), the antiderivatives form a cycle:

• The first antiderivative is \(-\cos x\)
• The second, \(-\sin x\)
• The third, \(\cos x\)

Recognizing such patterns is crucial in integration by parts, allowing us to apply the formula multiple times effectively.

Integration Techniques

Integration techniques are strategies employed to solve integral problems, especially when the standard rules for simple functions do not apply. Common techniques include substitution, partial fraction decomposition, and integration by parts.
Integration by parts is particularly useful for problems involving the product of a polynomial and another function, such as exponential or trigonometric functions. The method is based on the formula:\[\int u \, dv = uv - \int v \, du\]In this formula:

• \(u\) is a polynomial, which is differentiated, reducing its degree.
• \(dv\) is the other function, for which we find successive antiderivatives.

By applying this technique repeatedly, as seen in the step-by-step solution, we simplify complex integrals into manageable terms. This strategy not only helps solve intricate calculus problems but also reinforces the concept of choosing appropriate functions for \(u\) and \(dv\) for efficient problem-solving.

Calculus Problem-Solving

Calculus problem-solving often involves a structured approach, dissecting the problem into smaller, manageable pieces. This method is crucial for solving integrals involving complex functions. The problems given—\( \int (x^3 - 2x)e^x \, dx \) and \( \int (x^2 - 3x + 1) \sin x \, dx \)—show how calculus techniques combine to achieve a solution.
Each step in solving such problems involves identifying:

• The polynomial and its degree, which steers the choice of technique.
• The other function (exponential or trigonometric), establishing what requires antiderivatives.

Applying successive integration by parts helps to systematically deal with the higher-order polynomials, breaking them down successively until reaching a solvable form.
This structured approach, backing each mathematical decision with reasoning, aids in improving overall problem-solving skills in calculus. Mastery of this enables tackling even more sophisticated mathematical challenges with confidence and clarity.