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Chapter 5: Problem 68

(a) Evaluate the integral \(\int(5 x-1)^{2} d x\) by two methods: first square and integrate, then let \(u=5 x-1\) (b) Explain why the two apparently different answers obtained in part (a) are really equivalent.

Short Answer

Expert verified
Both methods yield the same integral value up to a constant of integration.

Step by step solution

01

Expand the Expression

First, let's square the expression inside the integral: \((5x-1)^2 = (5x)^2 - 2 \times 5x \times 1 + 1^2 = 25x^2 - 10x + 1\). So the integral becomes \(\int (25x^2 - 10x + 1) dx\).
02

Integrate Term by Term

Now, integrate each term separately: - The integral of \(25x^2\) is \(\frac{25}{3}x^3\) - The integral of \(-10x\) is \(-5x^2\) - The integral of \(1\) is \(x\).So the integrated result is \(\frac{25}{3}x^3 - 5x^2 + x + C\), where \(C\) is the constant of integration.
03

Use Substitution Method

Now, use the substitution method. Let \(u = 5x - 1\), then \(du = 5dx\) or \(dx = \frac{1}{5}du\).Rewrite the integral: \(\int (5x-1)^2 dx = \int u^2 \cdot \frac{1}{5} du\), which simplifies to \(\frac{1}{5}\int u^2 du\).
04

Integrate Using Substitution

Integrate \(\int u^2 du\) to get \(\frac{u^3}{3} + C\).Thus, \(\frac{1}{5} \cdot \frac{u^3}{3} + C = \frac{u^3}{15} + C\).Substitute back \(u = 5x - 1\) to get \(\frac{(5x-1)^3}{15} + C\).
05

Simplify and Compare Results

Expand \(\frac{(5x-1)^3}{15}\) and compare with the result from the first method.Upon expansion, \((5x-1)^3 = 125x^3 - 75x^2 + 15x - 1\).So, \(\frac{(5x-1)^3}{15} = \frac{125}{15}x^3 - 5x^2 + x - \frac{1}{15}\).This simplifies to \(\frac{25}{3}x^3 - 5x^2 + x - \frac{1}{15} + C\), which is equivalent to the first result when you adjust the constants \(C\).

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Key Concepts

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

Integration by Substitution
Integration by substitution is a helpful method in calculus to simplify difficult integrals. It is similar to the reverse of the chain rule from differentiation. Essentially, you choose a substitution that will convert the integral into an easier one. Here's how to understand it better:
  • Identify the part of the integral that can be substituted. This is usually a function and its derivative.
  • Choose a substitution like \( u = g(x) \), where \( g(x) \) is inside the complex part of the integral. Compute the derivative \( \, du = g'(x) \, dx \).
  • Rewrite the integral in terms of \( u \) and \( du \). This should simplify the integration process.
  • Perform the integration as a simpler \( u \) function, and then substitute back the original variable at the end.
This method is a wonderful tool to transform tough integrals into an easier form. In the given problem, substituting \( u = 5x - 1 \) transformed the integral into a form that is straightforward to integrate.
Definite Integrals
Definite integrals are used to calculate the area under a curve between two points. They provide a numeric value that represents an accumulated quantity, such as area or total change. Understanding definite integrals involves:
  • Setting boundaries on the integral, which are the limits of integration \([a, b]\).
  • Computing the antiderivative or integral of the function within these bounds.
  • Substituting the upper and lower bounds into the antiderivative and subtracting to find the result.
This results in a specific number rather than a function with an arbitrary constant \( C \). Even though the presented problem involves indefinite integrals, understanding definite integrals is critical when specific limits are involved, completing the link between antiderivatives and area under curves.
Indefinite Integrals
Indefinite integrals refer to the integration of a function without specific limits. They yield a general expression that includes an arbitrary constant, \( C \). Here's how indefinite integrals work:
  • They are represented as \( \int f(x) \, dx \), indicating the antiderivative of \( f(x) \).
  • The result is a family of functions differing by constant \( C \), since integration is the reverse operation of differentiation.
  • This constant is necessary because different functions have the same derivative.
In the solved exercise, both methods (expanding and substitution) provided indefinite integrals of the same function. These techniques ultimately yielded equivalent results despite different appearances, due to the presence of the constant \( C \). Always remember: indefinite integrals are about finding a general solution to the integral.

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