Question

Tabular integration Consider the integral \(\int f(x) g(x) d x,\) where \(f\) can be differentiated repeatedly and \(g\) can be integrated repeatedly. Let \(G_{k}\) represent the result of calculating \(k\) indefinite integrals of \(g,\) where the constants of integration are omitted. a. Show that integration by parts, when applied to \(\int f(x) g(x) d x\) with the choices \(u=f(x)\) and \(d v=g(x) d x,\) leads to \(\int f(x) g(x) d x=f(x) G_{1}(x)-\int f^{\prime}(x) G_{1}(x) d x .\) This formula can be remembered by utilizing the following table, where a right arrow represents a product of functions on the right side of the integration by parts formula, and a left arrow represents the integral of a product of functions (also appearing on the right side of the formula). Explain the significance of the signs associated with the arrows. (Table cannot copy) b. Perform integration by parts again on \(\int f^{\prime}(x) G_{1}(x) d x\) (from part (a)) with the choices \(u=f^{\prime}(x)\) and \(d v=G_{1}(x) d x\) to show that \(\int f(x) g(x) d x=f(x) G_{1}(x)-f^{\prime}(x) G_{2}(x)+\) \(\int f^{\prime \prime}(x) G_{2}(x) d x .\) Explain the connection between this integral formula and the following table, paying close attention to the signs attached to the arrows. (Table cannot copy) c. Continue the pattern established in parts (a) and (b) and integrate by parts a third time. Write the integral formula that results from three applications of integration by parts, and construct the associated tabular integration table (include signs of the arrows). d. The tabular integration table from part (c) is easily extended to allow for as many steps as necessary in the process of integration by parts. Evaluate \(\int x^{2} e^{x / 2} d x\) by constructing an appropriate table, and explain why the process terminates after four rows of the table have been filled in. e. Use tabular integration to evaluate \(\int x^{3} \cos x d x .\) How many rows of the table are necessary? Why? f. Explain why tabular integration is particularly suited to integrals of the form \(\int p_{n}(x) g(x) d x,\) where \(p_{n}\) is a polynomial of degree \(n>0\) (and where, as before, we assume \(g\) is easily integrated as many times as necessary).

Short answer

#Answer# For the first integral, we have evaluated \(\int x^2 e^{x/2} dx = 2x^2 e^{x/2} - 8xe^{x/2} + 16e^{x/2} + C\). For the second integral, we have evaluated \(\int x^3 \cos x dx = x^3\sin x + 3x^2\cos x - 6x\sin x + 6\cos x + C\). Tabular integration is particularly suitable for integrals of the form \(\int p_n(x) g(x) dx\) because it takes advantage of the termination when differentiating a polynomial, and if the function g(x) is easily integrable, the method simplifies the process, leading to a quicker result.

Step-by-step solution

Perform integration by parts with chosen u and dv

We choose \(u = f(x)\) and \(dv = g(x)dx\). From the integration by parts formula, we have: $$ \int f(x)g(x)dx = u \cdot v - \int v \cdot du = f(x) \cdot G_1(x) - \int f'(x)G_1(x)dx $$ Here, \(G_1(x)\) is the first antiderivative of \(g(x)\). The right arrow represents the product of the functions, and the left arrow represents the integral of the product of functions. #b. Perform integration by parts again on the integral from part (a). Explain the connection to the table method.#

Perform integration by parts for the second time

We choose \(u = f'(x)\) and \(dv = G_1(x)dx\), and apply the integration by parts formula again: $$ \int f'(x)G_1(x)dx = u \cdot v - \int v \cdot du = f'(x)G_2(x) -\int f''(x)G_2(x)dx $$ Substituting this back into the formula from part (a), we have: $$ \int f(x)g(x)dx = f(x)G_1(x)-[f'(x)G_2(x) - \int f''(x)G_2(x)dx] = f(x)G_1(x) - f'(x)G_2(x) + \int f''(x)G_2(x)dx $$ Here, \(G_2(x)\) is the second antiderivative of \(g(x)\). The alternating signs from the table are due to the integration by parts formula involving subtraction. #c. Continue the pattern established in parts (a) and (b) and integrate by parts a third time. Write the integral formula that results from three applications of integration by parts, and construct the associated tabular integration table (include signs of the arrows).#

Perform integration by parts for the third time

Applying the integration by parts formula again with \(u = f''(x)\) and \(dv = G_2(x)dx\): $$ \int f''(x)G_2(x)dx = f''(x)G_3(x) - \int f^{(3)}(x)G_3(x)dx $$ Substituting this back into the formula obtained in part (b), we have: $$ \int f(x)g(x)dx = f(x)G_1(x) - f'(x)G_2(x) + f''(x)G_3(x) - \int f^{(3)}(x)G_3(x)dx $$ Here, \(G_3(x)\) is the third antiderivative of \(g(x)\). The tabular integration table has alternating signs from the left and alternating derivatives and antiderivatives on the right. The "+" and "-" signs in the table help to keep track of the signs as we apply integration by parts multiple times. #d. Evaluate \(\int x^2 e^{x/2} dx\) by constructing an appropriate table. Explain the process and termination.#

Apply tabular integration to the given integral

We will create a table with derivatives of \(x^2\) in the first column, and antiderivatives of \(e^{x/2}\) in the second column: $$ \begin{array}{c|c} x^2 & e^{x/2} \\ -\!\!\to & + \\ 2x & 2e^{x/2} \\ -\!\!\to & - \\ 2 & 4e^{x/2} \\ -\!\!\to & + \\ 0 & 8e^{x/2} \\ \end{array} $$ The integral is the sum of products across the rows, remembering to include the signs from the arrows. Since the derivative of \(2\) is \(0\), the process terminates after four rows. $$ \int x^2 e^{x/2} dx = x^2(2e^{x/2}) - 2x(4e^{x/2}) + 2(8e^{x/2}) = 2x^2 e^{x/2} - 8xe^{x/2} + 16e^{x/2} + C $$ #e. Use tabular integration to evaluate \(\int x^3 \cos x dx .\) How many rows of the table are necessary? Why?#

Apply tabular integration to the second integral

We will create a table with derivatives of \(x^3\) in the first column and antiderivatives of \(\cos x\) in the second column: $$ \begin{array}{c|c} x^3 & \sin x \\ -\!\!\to & + \\ 3x^2 & -\cos x \\ -\!\!\to & - \\ 6x & -\sin x \\ -\!\!\to & + \\ 6 & \cos x \\ -\!\!\to & - \\ 0 & \sin x \\ \end{array} $$ The integral is the sum of products across the rows, with the signs from the arrows. Since the derivative of \(6\) is \(0\), the process terminates after four rows. $$ \int x^3 \cos x dx = x^3(\sin x) - 3x^2 (-\cos x) - 6x (-\sin x) + 6 (\cos x) = x^3\sin x + 3x^2\cos x - 6x\sin x + 6\cos x + C $$ #f. Explain why tabular integration is particularly suitable for integrals of the form \(\int p_n(x) g(x) dx.\)#

Explain the suitability of the tabular integration method

Tabular integration is particularly suitable for integrals of the form \(\int p_n(x)g(x)dx\) where \(p_n(x)\) is a polynomial of degree \(n\) and \(g(x)\) is easily integrable, because: 1. Differentiating \(p_n(x)\) will eventually lead to a constant, and further differentiation will result in zero, ensuring that the process will terminate after a finite number of steps. 2. If \(g(x)\) can be easily integrated repeatedly, we can easily complete the antiderivative column in the table. These two factors combined make tabular integration a simple and effective technique for integrating products of polynomials and easily integrable functions.

Key concepts

Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and is formally given by the formula:
\[\begin{equation} \int u\text{ } dv = uv - \int v\text{ } du. \end{equation}\]
To apply this rule, one must choose which part of the integral will be 'u' (to be differentiated) and which part will be 'dv' (to be integrated). This choice can greatly affect the simplicity of the resulting integral. Ideally, you select 'u' such that its derivative simplifies the integrand when substituted back into the equation, and select 'dv' such that its antiderivative is easily computed.
When tabular integration is employed, this process is systematized, allowing for repeated applications of integration by parts. This is particularly helpful when dealing with the integral of a product where one function can be repeatedly differentiated to zero, such as polynomials.

Antiderivatives

An antiderivative of a function is another function whose derivative is the original function. It is also known as an indefinite integral and represents the reversed process of differentiation. If 'F(x)' is the antiderivative of 'f(x),' then
\[\begin{equation} F'(x) = f(x). \end{equation}\]
In the context of tabular integration, antiderivatives play an important role as they are generated in one column of the table and paired with the successive derivatives of the other function to calculate the integral in a stepwise fashion.

Indefinite Integrals

Indefinite integrals are integrals without limits, representing a family of functions with an added constant 'C'—since the derivative of a constant is zero. The indefinite integral of a function 'f(x)' is symbolically represented as
\[\begin{equation} \int f(x)dx. \end{equation}\]
It is important to note that the constant of integration 'C' must be included in the solution. In tabular integration, the process will yield a series of indefinite integrals, each representing a stage where integration by parts has been applied. The sum of these stages, taking into account the alternating signs, gives the final integrated function plus 'C'.

Polynomial Functions

Polynomial functions are algebraic expressions consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables. Their general form is
\[\begin{equation} p_n(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0, \end{equation}\]
where 'n' is the degree of the polynomial, and 'a_n', 'a_{n-1}', ..., 'a_1', 'a_0' are constants. Polynomial functions possess certain properties that are useful in integration, such as the ability to differentiate them down to zero. This terminating property makes them ideal candidates for tabular integration, as they ensure a finite series of integration by parts.

Application in Tabular Integration

In tabular integration, rows are created for each derivative of the polynomial until reaching zero, allowing the procedure to conclude in a finite number of steps. Combined with a function that is easily integrable, this creates a streamlined pathway to finding the integral using the table method.