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Chapter 1: Problem 262

In the following exercises, find the antiderivative using the indicated substitution. $$ \int(x-1)^{5} d x ; u=x-1 $$

Short Answer

Expert verified
\( \frac{(x-1)^6}{6} + C \)

Step by step solution

01

Substitution

We start by using the substitution provided in the exercise. Let \( u = x - 1 \). Then, differentiate both sides to find \( du \). We have \( du = dx \).
02

Transform the Integral

Substitute \( u = x - 1 \) and \( du = dx \) into the integral. The integral \( \int (x-1)^{5} \, dx \) becomes \( \int u^5 \, du \).
03

Find the Antiderivative

Now that we have transformed the integral, we find the antiderivative of \( \int u^5 \, du \). The antiderivative of \( u^5 \) is \( \frac{u^6}{6} + C \), where \( C \) is the constant of integration.
04

Substitute Back

Finally, substitute \( u = x - 1 \) back into the antiderivative \( \frac{u^6}{6} + C \). The antiderivative becomes \( \frac{(x-1)^6}{6} + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful tool in calculus, especially when dealing with integrals. It simplifies the process of finding antiderivatives by turning a complex expression into a simpler one. In this exercise, the substitution method is used to evaluate the integral \( \int (x-1)^5 \, dx \).
  • We choose a substitution to simplify the expression inside the integral. Here, the substitution is \( u = x - 1 \).
  • Differentiate the substitution to connect the new variable \( u \) back to the original variable \( x \). Thus, \( du = dx \).
Using this method, the original problem is transformed into a much easier form, \( \int u^5 \, du \), which is simpler to solve. By substituting back after finding the antiderivative, you return to the terms of the original variable. This technique is akin to changing the perspective, making complex problems more manageable.
Differentiation
Differentiation plays a crucial role when utilizing the substitution method. It is the process through which we find how much a function changes as its input changes. In this exercise, differentiation connects our substituted variable \( u \) with the original variable \( x \):
  • When we differentiate the substitution \( u = x - 1 \), it gives \( du = dx \).
This differentiation ensures that the differential \( du \) matches with \( dx \) in our integral, allowing a seamless transition from \( dx \) to \( du \). This step may seem minor, but it's critical to ensure that the substitution is correctly applied. Without it, the integral in terms of \( u \) would not correspond correctly to the original integral in terms of \( x \).
Integral Transformation
Integral transformation is the core idea of substitution. We transform the original integral into a more tractable form using the substituted variable. This process involves replacing part of the integral expression so that calculations become simpler:
  • Upon substitution, the integral \( \int (x-1)^5 \, dx \) is transformed into \( \int u^5 \, du \).
  • This new integral, \( \int u^5 \, du \), is easier to work with because the expression inside is a simple power function.
The essence of this transformation is to reduce the complexity inherent in the integration problem by strategically altering the variable and the bounds of integration, when necessary. The goal is always to reach an expression where finding the antiderivative is straightforward. This step leverages known formulas or simpler integral solutions, ultimately easing the path to finding a solution.
Constant of Integration
In integration, the constant of integration \( C \) is vital. It represents an infinite number of solutions that satisfy an antiderivative. When you integrate a function, you determine one particular form of its antiderivative. However, since derivatives of constants are zero, any constant added to an antiderivative will still satisfy the derivative relation:
  • In the exercise, after integrating \( u^5 \), we arrive at \( \frac{u^6}{6} + C \).
  • This constant \( C \) is an integral part of indefinite integrals, ensuring the result includes all possible antiderivatives.
Always include \( C \) in your final answer when determining the antiderivative of a function. Its presence acknowledges that the solution is part of a family of infinitely many solutions, each differing by a constant. Remember, every indefinite integral comes with this hidden scope of flexibility given by \( C \).

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