Question

In Problems 1-54, perform the indicated integrations. $$ \int(x-2)^{5} d x $$

Short answer

\(\int(x-2)^5\, dx = \frac{(x-2)^6}{6} + C\)

Step-by-step solution

Identify the Integration Technique

The expression \((x-2)^5\) is a polynomial raised to a power. To integrate this, we can apply the power rule for integration.

Apply the Power Rule for Integration

The power rule for integration states that for \(\int u^n\,du\), the result is \(\frac{u^{n+1}}{n+1} + C\), where \(n eq -1\). Here, \(u = x-2\) and \(n = 5\).

Write the Substitute Expression for Integration

Since \(u = x-2\), \(du = dx\). Therefore, the integral \(\int (x-2)^5 \, dx\) becomes \(\int u^5 \, du\).

Integrate Using the Power Rule

Perform the integration: \(\int u^5 \, du = \frac{u^{5+1}}{5+1} + C\), which simplifies to \(\frac{u^6}{6} + C\).

Substitute Back to Original Variable

Replace \(u\) with \(x-2\) in the integrated expression: \(\frac{(x-2)^6}{6} + C\).

Simplify the Expression

The final integrated expression, with simplification, is \(\frac{(x-2)^6}{6} + C\).

Key concepts

Power Rule for Integration

The Power Rule for Integration is one of the most fundamental techniques in calculus. It is applied when you're integrating functions that can be expressed in the form of a variable raised to an exponent. This rule significantly simplifies the process of finding antiderivatives for power functions.
To use the power rule, remember the formula:

• For an integral of the form \( \int u^n \, du \), the antiderivative is \( \frac{u^{n+1}}{n+1} + C \), where \( n eq -1 \).

In our problem, the expression \((x-2)^5\) is a perfect candidate for this rule. By identifying \( u = x-2 \) and \( n = 5 \), we can directly apply the power rule and integrate with ease.
It’s crucial to notice that this rule is not applicable directly if \( n = -1 \). In that case, the integral would involve the natural logarithm, a concept used in another set of integration techniques.

Polynomial Functions

Polynomial functions, like the one in our problem \((x-2)^5\), consist of variables raised to whole number exponents. These functions can be as simple as linear expressions, such as \(x\), or more complex, like high-degree polynomials.

• Polynomials are expressions made up of constants, coefficients, and variables combined using addition, subtraction, and multiplication.
• They are defined by their degree, which is the highest power of the variable in the expression.

In calculus, polynomials are particularly straightforward to integrate. This is because each term in a polynomial can be integrated separately by applying the power rule, as shown in our original exercise.
Understanding polynomial functions is key to mastering calculus since they form the basis for many real-world applications, such as modeling curves and trends in economics and the natural sciences.

Definite and Indefinite Integrals

In calculus, integrals can be categorized into definite and indefinite integrals, each serving different purposes.
Indefinite integrals, like the one in this exercise, represent a family of functions. These are expressed generally as \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant of integration. The primary aim is to find the antiderivative of the function.
Definite integrals, on the other hand, calculate the net area under the curve between two points. They have limits of integration, specified as \( \int_{a}^{b} f(x) \, dx \), and yield a numerical value rather than a function.

• Understanding the distinction between these two types is central to solving a variety of integration problems.
• Indefinite integrals are used for finding general solutions, while definite integrals are essential in physical applications, like computing areas and volumes.

In our problem, once the indefinite integral was found using the power rule, we ended up with the antiderivative plus the integration constant \(C\).

Calculus Education

Calculus is a branch of mathematics that deals with rates of change and quantities. In a calculus education, learning integration techniques is essential. Knowledge of these techniques allows students to solve complex problems involving areas, volumes, work, and more.
Comprehending topics such as the power rule, polynomial functions, and the roles of definite and indefinite integrals is critical for students aspiring to excel in calculus.

• These foundational concepts enable students to tackle a wide range of problems not only in mathematics but also in fields like physics and engineering.
• Effective calculus education involves practicing a variety of problems to better grasp integration methods and their real-world applications.

By mastering the rudiments of integration, students lay the groundwork for more advanced topics, such as differential equations and multivariable calculus, which further expand their analytical skills and problem-solving capabilities.