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Chapter 8: Problem 52

Evaluate each integral in Exercises \(47-52\) by reducing the improper fraction and using a substitution (if necessary) to reduce it to standard form. $$ \int \frac{2 \theta^{3}-7 \theta^{2}+7 \theta}{2 \theta-5} d \theta $$

Short Answer

Expert verified
The integral is \(\frac{\theta^3}{3} - \frac{\theta^2}{2} + 2\theta + C\).

Step by step solution

01

Polynomial Division

Perform polynomial long division on \( \frac{2\theta^3 - 7\theta^2 + 7\theta}{2\theta - 5}\). The leading term is \(2\theta^2\), multiply \(2\theta - 5\) by \(\theta^2\) to get \(2\theta^3 - 5\theta^2\). Subtract from the original polynomial and continue dividing until the remainder is found. Solutions give \(\theta^2 - \theta + 1\) with a remainder of \(2\theta - 5\).
02

Express Integral as Sum

Write the integral as the sum of two integrals: \(\int (\theta^2 - \theta + 1) \, d\theta + \int \frac{2\theta - 5}{2\theta - 5} \, d\theta\).
03

Integrate First Part

Integrate \(\theta^2 - \theta + 1\) using basic integration rules: \(\int \theta^2 \, d\theta = \frac{\theta^3}{3}\), \(\int \theta \, d\theta = \frac{\theta^2}{2}\), and \(\int 1 \, d\theta = \theta\). This yields \(\frac{\theta^3}{3} - \frac{\theta^2}{2} + \theta\).
04

Integrate Second Part

\(\int \frac{2\theta - 5}{2\theta - 5} \, d\theta = \int 1 \, d\theta = \theta\).
05

Combine Results

Sum the results from Steps 3 and 4: \(\frac{\theta^3}{3} - \frac{\theta^2}{2} + 2\theta + C\), where \(C\) is the constant of integration.

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Key Concepts

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

Improper Fractions
In the context of integration, improper fractions appear when the degree of the polynomial in the numerator is greater than or equal to that in the denominator. This can complicate the process of finding integrals.
By reducing the fraction to a simpler form, we facilitate the integration process. The first step to simplify an improper fraction is usually to perform polynomial division.
This helps rewrite the original expression into a form that can be broken down into more manageable parts.
  • For instance, in the integral provided, the numerator has a degree of 3 and the denominator a degree of 1.
  • Thus, it's necessary to perform polynomial division to simplify the fraction into a proper fraction or someone easier to integrate.
Simplifying these expressions can often involve several arithmetic operations, but it ultimately leads to a solution that's simpler to solve using integration techniques.
Polynomial Division
Polynomial division is much like long division but applied to polynomials. It's instrumental in simplifying improper fractions before integration.
Here's a breakdown of how this applies:
  • First, divide the leading terms of the numerator by the leading term of the denominator.
  • Multiply the entire polynomial denominator by this quotient term and subtract it from the original numerator polynomial to find the new dividend.
  • Continue this process until the degree of the remainder is lower than that of the divisor.
In the example given, the division of \(2\theta^3 - 7\theta^2 + 7\theta\) by \(2\theta - 5\) resulted in \(\theta^2 - \theta + 1\) with a remainder \(2\theta - 5\).
This simplifies the integration process by providing terms that can be directly integrated.
Basic Integration
Once the fraction is reduced via polynomial division, we turn to basic integration techniques to find the primitive function. Basic integration involves applying standard integration formulas to each term of the polynomial separately.
In this context, consider these steps:
  • The integral of \(\theta^n\) is \(\frac{\theta^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.
  • Applying this rule, for \(\theta^2\), \(\theta\), and the constant term, their integrals become \(\frac{\theta^3}{3}\), \(\frac{\theta^2}{2}\), and \(\theta\) respectively.
Basic integration helps to individually integrate these components, illustrating how core integration rules facilitate solving complex expressions when neatly reduced.
Definite and Indefinite Integrals
When integrating a function, we often encounter definite or indefinite integrals.
Definite integrals have specified limits, giving a numerical value as a result, while indefinite integrals involve a constant of integration and produce a family of functions.
In the solution given:
  • Since no limits of integration are specified, this is an indefinite integral.
  • The result is \(\frac{\theta^3}{3} - \frac{\theta^2}{2} + 2\theta + C\), where \(C\) represents the constant of integration.
Indefinite integrals are crucial in representing the general solution to an integration problem, encompassing all potential solutions before specifying conditions like initial values or boundaries that turn them into definite integrals.

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Most popular questions from this chapter

Consider the integral \(\int_{-1}^{1} \sin \left(x^{2}\right) d x\) a. Find \(f^{\prime \prime}\) for \(f(x)=\sin \left(x^{2}\right)\) b. Graph \(y=f^{\prime \prime}(x)\) in the viewing window \([-1,1]\) by \([-3,3]\) c. Explain why the graph in part (b) suggests that \(\left|f^{\prime \prime}(x)\right| \leq 3\) for \(-1 \leq x \leq 1 .\) d. Show that the error estimate for the Trapezoidal Rule in this case becomes $$ \left|E_{T}\right| \leq \frac{(\Delta x)^{2}}{2} $$ e. Show that the Trapezoidal Rule error will be less than or equal to 0.01 in magnitude if \(\Delta x \leq 0.1 .\) f. How large must \(n\) be for \(\Delta x \leq 0.1 ?\)

Evaluate the integrals in Exercises \(1-28\). $$ \int_{0}^{3 / 2} \frac{d x}{\sqrt{9-x^{2}}} $$

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int \frac{d x}{1+x^{2}} $$

In Exercises \(29-36\) , use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$ \int_{1 / 12}^{1 / 4} \frac{2 d t}{\sqrt{t}+4 t \sqrt{t}} $$

As we mentioned at the beginning of the section, the definite integrals of many continuous functions cannot be evaluated with the Fundamental Theorem of Calculus because their antiderivatives lack elementary formulas. Numerical integration offers a practical way to estimate the values of these so-called nonelementary integrals. If your calculator or computer has a numerical integration routine, try it on the integrals in Exercises \(39-42\) . $$ \int_{0}^{\pi / 2} \frac{\sin x}{x} d x $$
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