Problem 87 - Applied Calculus

Question

The substitution method can be used to find integrals that do not fit our formulas. For example, observe how we find the following integral using the substitution \(u=x+4\) which implies that \(x=u-4\) and so \(d x=d u\). $$ \begin{aligned} \int(x-2)(x+4)^{8} d x &=\int(u-4-2) u^{8} d u \\ &=\int(u-6) u^{8} d u \\ &=\int\left(u^{9}-6 u^{8}\right) d u \\ &=\frac{1}{10} u^{10}-\frac{2}{3} u^{9}+C \\ &=\frac{1}{10}(x+4)^{10}-\frac{2}{3}(x+4)^{9}+C \end{aligned} $$ It is often best to choose \(u\) to be the quantity that is raised to a power. The following integrals may be found as explained on the left (as well as by the methods of Section 6.1). $$ \int x(x-2)^{6} d x $$

Step-by-Step Solution

Answer

\( \frac{1}{8}(x-2)^8 + \frac{2}{7}(x-2)^7 + C \)

Step 1: Choose the substitution

For the integral \( \int x(x-2)^{6} \ d x \), a suitable substitution is to let \( u = x - 2 \). This implies that \( du = dx \) and \( x = u + 2 \).

Step 2: Substitute in terms of u

Substitute \( x = u+2 \) and \( dx = du \) into the integral. This results in: \( \int (u+2)u^{6} \ du \).

Step 3: Expand and integrate

Expand the expression as \( \int (u^7 + 2u^6) \ du \). Now integrate each term separately: \( \int u^7 \ du = \frac{1}{8}u^8 \) and \( \int 2u^6 \ du = \frac{2}{7}u^7 \). Combining gives \( \frac{1}{8}u^8 + \frac{2}{7}u^7 + C \).

Step 4: Substitute back to x

Replace \( u \) with \( x - 2 \) to express the antiderivative in terms of \( x \). The final expression is \( \frac{1}{8}(x-2)^8 + \frac{2}{7}(x-2)^7 + C \).

Key Concepts

Antiderivative

The antiderivative is also known as the indefinite integral. It is essentially the reverse operation of differentiation. When we talk about finding the antiderivative, we mean determining a function whose derivative is the original function we started with.
For example, if we take a function, say \( f(x) = x^2 \), the derivative would be \( f'(x) = 2x \). The antiderivative of \( 2x \) is \( x^2 \), but since there can be multiple functions with the same derivative, we add a constant \( C \), so it becomes \( x^2 + C \).

• Think of antiderivatives like finding the potential original state of a function before it was differentiated.
• They include a constant \( C \), because differentiation of constants results in zero, and we lose that information when going backward.

In integration by substitution, finding the antiderivative is one of the key steps since it involves redefining the variable and then integrating the function in terms of this new variable. This helps simplify complex expressions and make integration manageable.

Integral Calculus

Integral calculus is the branch of mathematics that deals with integrals. It is concerned with the accumulation of quantities and the spaces under and between curves. There are two main types of integrals in calculus: definite and indefinite integrals.
An indefinite integral, like the one discussed in the exercise, provides a general form of the antiderivative, while a definite integral gives the actual accumulated value over a specific interval.

• Indefinite integrals are expressed as functions plus a constant \( C \) because they represent a family of functions.
• Definite integrals give a numerical value and don't include the constant \( C \).

The goal of integral calculus is to take a derivative and go back to the original function (or a family of them). Substitution is a technique used within integral calculus to make solving integrals easier, essentially modifying the function into something simpler to integrate. This method is powerful for dealing with complex expressions that aren't easily integrated using standard formulas.

Change of Variables

The change of variables is a method used in integral calculus to simplify the process of integration. By substituting a new variable into the integral, we can reshape complicated expressions into ones that are easier to integrate.
This process typically involves switching from the original variable \( x \) to a new variable \( u \), like in the example from the exercise where \( u = x - 2 \).

• This technique often makes use of the chain rule from derivatives, but in reverse.
• One must also change the differential; if \( u = x - 2 \), then \( du = dx \).

The change of variables not only simplifies complex integrals but also helps avoid errors that might arise from skipping steps when manually expanding expressions. It's a foundational technique that's widely applicable for solving integrals involving polynomials, trigonometric functions, and many other forms. By using a substitution, integrals can be evaluated more easily, enabling a more rounded understanding of how functions behave in calculus.