Problem 90 - 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). $$ \int \frac{x-4}{x-5} d x $$

Step-by-Step Solution

Answer

\((x-5) + \ln|x-5| + C\)

Step 1: Choose Substitution

Choose a substitution to simplify the integral. Let's set \( u = x - 5 \), such that the denominator of the expression can be simplified. Then, \( du = dx \). Since \( x = u + 5 \), replace \( x \) with \( u \).

Step 2: Rewrite the Integral

Rewrite the integral in terms of \( u \):\[ \int \frac{x-4}{x-5} \, dx = \int \frac{u+5-4}{u} \, du = \int \frac{u+1}{u} \, du .\]

Step 3: Simplify the Expression

Separate the fraction and simplify:\[ \int \left(1 + \frac{1}{u}\right) \, du = \int 1 \, du + \int \frac{1}{u} \, du. \]

Step 4: Integrate Simplified Expression

Integrate each term separately:- \( \int 1 \, du = u \).- \( \int \frac{1}{u} \, du = \ln|u| \).Thus, the integral becomes: \( u + \ln|u| + C \).

Step 5: Substitute Back Original Variable

Replace \( u \) with \( x - 5 \) to express the solution in terms of \( x \):\[ (x-5) + \ln|x-5| + C. \]

Key Concepts

Integration Techniques

In calculus, integration techniques help us find the antiderivative or the integral of a given function. One such technique is the substitution method, which is particularly useful when dealing with composite functions. Here’s how it works:

• You choose a substitution variable, typically denoted as \( u \), that transforms part of the integral into a simpler expression.
• Calculate the differential \( du \) in terms of \( dx \) and rewrite the integral in terms of \( u \).
• This new integral is easier to solve, allowing you to apply basic integration rules.
• Finally, substitute back the original variable to obtain the solution in terms of the initial expressions.

By using substitution, you transform a complicated integral into a simpler one, making it easier to handle with basic integration rules. This technique is often used for "definite and indefinite" integrals alike, simplifying otherwise complex problems in integral calculus.

Definite and Indefinite Integrals

In integral calculus, understanding the difference between definite and indefinite integrals is essential. Indefinite integrals represent a family of functions whose derivative is the integrand. They are depicted as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \) and \( C \) is the constant of integration.

• Indefinite integrals provide general solutions without bounds.
• They result in a function with a constant \( C \) since the differentiation of a constant is zero.

On the other hand, definite integrals have limits of integration, giving a specific numerical result that represents the net area under the curve from \( a \) to \( b \). Formally, \( \int_{a}^{b} f(x) \, dx \):

• Calculates the net signed area between the function and the x-axis over the interval \([a, b]\).
• Does not include the constant \( C \) because the limits \( a \) and \( b \) specify a particular range.

Understanding these two types of integrals allows one to apply integral calculus effectively to both theoretical problems and practical applications in fields like physics and engineering.

Integral Calculus

Integral calculus is one of the fundamental branches of calculus focused on the concept of integration. It allows us to calculate areas, volumes, and more by integrating functions. Integral calculus encompasses a variety of methods and techniques to solve integrals, especially when direct integration isn't straightforward. Such techniques include substitution, integration by parts, and partial fraction decomposition.

• The substitution method, as discussed, simplifies integrands through intelligent substitution choices.
• Integration by parts breaks down products of functions by integrating one function and differentiating another.
• Partial fraction decomposition splits complex rational expressions into simpler fractions that are easier to integrate.

The main goal of integral calculus is to find the antiderivative of a given function, thus reversing the process of differentiation. Application of these methods allows mathematicians and scientists to solve complex problems involving continuous change, such as finding the total accumulated value or volume in practical scenarios. Mastery of these techniques imparts a deep understanding of both the geometric and algebraic aspects of calculus.