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) \(\int\left(x^{2}-3 x+1\right) \sin x d x\)

Short answer

(a) \(x^3 e^x - 3x^2 e^x + 4xe^x - 4e^x + C\); (b) \(-x^2 \cos x + 3x \cos x + 2x \sin x + \cos x - 3\sin x + C\).

Step-by-step solution

Identify the Polynomial Terms for Part (a)

We have \(p(x) = x^3 - 2x\). The degree of the polynomial is 3. The derivatives are:- \(p'(x) = 3x^2 - 2\)- \(p''(x) = 6x\)- \(p'''(x) = 6\)

Identify Antiderivatives for Part (a)

Since \(g(x) = e^x\), the antiderivatives are:- \(G_1(x) = e^x\)- \(G_2(x) = e^x\)- \(G_3(x) = e^x\)- \(G_4(x) = e^x\)

Apply Repeated Integration by Parts for Part (a)

Using the formula, \(\int p(x) g(x) \, dx\) becomes:\[\begin{align*} & (x^3 - 2x)e^x - (3x^2 - 2)e^x + 6xe^x - 6e^x + C & = x^3 e^x - 2xe^x - 3x^2 e^x + 2e^x + 6xe^x - 6e^x + C. \end{align*}\] Combining like terms, the result is \(x^3 e^x - 3x^2 e^x + 4xe^x - 4e^x + C\).

Identify the Polynomial Terms for Part (b)

We have \(p(x) = x^2 - 3x + 1\). The degree of the polynomial is 2. The derivatives are:- \(p'(x) = 2x - 3\)- \(p''(x) = 2\)

Identify Antiderivatives for Part (b)

Since \(g(x) = \sin x\), the antiderivatives are:- \(G_1(x) = -\cos x\)- \(G_2(x) = -\sin x\)- \(G_3(x) = \cos x\)

Apply Repeated Integration by Parts for Part (b)

Using the formula, \(\int p(x) g(x) \, dx\) becomes:\[\begin{align*}& (x^2 - 3x + 1)(-\cos x) - (2x - 3)(-\sin x) + 2\cos x + C & = -(x^2 \cos x) + 3x \cos x - \cos x + 2x \sin x - 3\sin x + 2\cos x + C \end{align*}\] Combining like terms, the result is \(-x^2 \cos x + 3x \cos x + 2x \sin x + \cos x - 3\sin x + C\).

Key concepts

Antiderivatives

Antiderivatives are a fundamental concept in calculus. They refer to the reverse process of differentiation. If a function, when differentiated, gives another function, then the original function is called an antiderivative of the latter. For example, if the derivative of function \( F(x) \) is \( f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \). The notation for antiderivatives usually involves the integral symbol \( \int \) and indicates that you are trying to find a function whose derivative is a given function.

Antiderivatives are not unique. Every function has an infinite number of antiderivatives, each differing by a constant. For example, the antiderivatives of \( f(x) = 2x \) are \( F(x) = x^2 + C \), where \( C \) is any constant. This constant is often denoted as \( +C \) in the solution.

In the context of the given problem, the process of finding successive antiderivatives is crucial to applying the method of integration by parts. For instance, in part (a) when \( g(x) = e^x \), all successive antiderivatives \( G_n(x) \) remain \( e^x \) because the derivative of \( e^x \) is \( e^x \) again.

Polynomial Integration

Polynomial integration is an important concept in calculus that involves integrating functions that have terms formed from powers of \( x \). The integration of polynomials can be straightforward due to the power rule. This rule states that:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]where \( n eq -1 \). This formula forms the basis of polynomial integration and is applied to each term of the polynomial separately.

The exercise example involves applying integration by parts to polynomials. The given polynomial for part (a), \( p(x) = x^3 - 2x \), is broken down by differentiating it sequentially as needed. It is crucial to note that decreasing the degree of each term through differentiation is essential in using integration by parts effectively. Through this method, the integration of this polynomial term combined with an exponential function \( e^x \) is tackled.

Integration by parts is a bit more complex than straightforward polynomial integration. It involves choosing parts of the integrand as \( u \) (where \( u \) is differentiated) and \( dv \) (where \( dv \) is integrated) in a strategic manner to simplify the problem.

Trigonometric Integration

Trigonometric integration deals with integrating functions that involve trigonometric functions like \( \sin x \), \( \cos x \), etc. These types of integrals require specific techniques or identities, due to the periodic and oscillating nature of trigonometric functions.

In the given example for part (b), the function to be integrated is \( (x^2 - 3x + 1) \sin x \). For this, integration by parts is used. Appropriate selection of \( u \) and \( dv \) is key: select \( u \) as a polynomial and \( dv \) as the part involving the trigonometric function.

• \( u = x^2 - 3x + 1 \)
• \( dv = \sin x \, dx \)

By applying integration by parts repeatedly, different trigonometric identities and derivatives simplify the integration process. In each step, you integrate the trigonometric part and differentiate the polynomial part, leading to a manageable expression. Successive antiderivatives of \( \sin x \) involve shifting between \( \sin x \) and \(-\cos x \), and then results are combined to form the solution.

These approaches show how integration by parts intertwines with trigonometric and polynomial integrations to solve complex integrals effectively.