/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Evaluate each definite integral.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate each definite integral. $$ \int_{1}^{2} \frac{(x+1)^{2}}{x^{2}} d x $$

Short Answer

Expert verified
The integral evaluates to \( \frac{3}{2} + 2 \ln(2) \).

Step by step solution

01

Simplify the Integrand

Rewrite the expression inside the integral by expanding the numerator before integrating. The original integral is: \( \int_{1}^{2} \frac{(x+1)^2}{x^2} \, dx \).Expanding \( (x + 1)^2 \), we get \( x^2 + 2x + 1 \).Thus, the integrand becomes:\[\int_{1}^{2} \left(\frac{x^2}{x^2} + \frac{2x}{x^2} + \frac{1}{x^2}\right) \, dx = \int_{1}^{2} \left(1 + \frac{2}{x} + \frac{1}{x^2}\right) \, dx\]
02

Integrate Term-by-Term

Now integrate each term separately over the interval \([1, 2]\):\[\int_{1}^{2} \left( 1 \right) \, dx + \int_{1}^{2} \left( \frac{2}{x} \right) \, dx + \int_{1}^{2} \left( \frac{1}{x^2} \right) \, dx\]- \( \int_{1}^{2} 1 \, dx = [x]_{1}^{2} \)- \( \int_{1}^{2} \frac{2}{x} \, dx = 2[\ln|x|]_{1}^{2} \)- \( \int_{1}^{2} \frac{1}{x^2} \, dx = [-\frac{1}{x}]_{1}^{2} \)
03

Evaluate the Integrals

Calculate the definite integrals obtained in Step 2:- For \( [x]_{1}^{2} \), calculate: \( 2 - 1 = 1 \)- For \( 2[\ln|x|]_{1}^{2} \), calculate: \( 2[\ln(2) - \ln(1)] = 2 \ln(2) \)- For \( [-\frac{1}{x}]_{1}^{2} \), calculate: \(-\frac{1}{2} - (-1) = \frac{1}{2} \)Add the results: \( 1 + 2 \ln(2) + \frac{1}{2} \)
04

Sum the Results

Combine all the evaluated parts:\[1 + 2 \ln(2) + \frac{1}{2} = \frac{3}{2} + 2 \ln(2)\]
05

Final Simplified Answer

Thus, the value of the definite integral is \( \frac{3}{2} + 2 \ln(2) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Integration by Parts
Integration by parts is a technique used in calculus to integrate products of functions. It helps when the standard integration methods don’t work straightforwardly. You use the formula:
\[ \int u \, dv = uv - \int v \, du \]
- **Identify the parts:** Often, one function gets differentiated (typically a polynomial) and the other integrated (commonly exponential or trigonometric).- **Calculate components:** Choose which function in your integral should be "u" and which should be "dv."- **Substitution:** Substitute into the formula to simplify and solve the integral.
This method is particularly useful when dealing with products such as polynomial multiplied by a logarithmic or exponential function, though in this problem it wasn't needed since simplification was more appropriate.
Polynomial Expansion
Polynomial expansion simplifies expressions with powers. It's crucial for converting a product of sums into a sum of products.
In the exercise, \( (x + 1)^2 \) is expanded to \( x^2 + 2x + 1 \). This transforms the integrand into terms that are easier to manipulate:- \( \int \frac{x^2}{x^2} \) becomes \( \int 1 \).- \( \int \frac{2x}{x^2} \) becomes \( \int \frac{2}{x} \).- \( \int \frac{1}{x^2} \) remains \( \int \frac{1}{x^2} \).Breaking it down this way simplifies the integration process. Each term fits a well-known integration formula, making computation straightforward. It's a powerful technique because it handles complex expressions by reducing them to simpler, manageable terms.
Logarithmic Integration
Logarithmic integration refers to integrating functions like \( \frac{1}{x} \), where the output involves logarithms.
For the term \( \int \frac{2}{x} \, dx \), the antiderivative is \( 2 \ln|x| \).Some essential points about logarithmic integration are:
  • The integral of \( \frac{1}{x} \) is \( \ln|x| \), and the constant factor "2" remains outside the integral.
  • Always consider the absolute value inside the logarithm to ensure it's defined across all real numbers where the function exists.
Evaluating from 1 to 2 gives terms evaluated as \( 2 \ln(2) \), since \( \ln(1) = 0 \).Logarithmic integrations are common, especially with rational functions, and understanding them helps in calculus and real-world applications.
Simplification of Rational Functions
Simplification of rational functions is a vital skill for integrals, especially when dealing with fractions.
In our problem, initially, we have \( \frac{(x+1)^2}{x^2} \). Using polynomial expansion, it reduces to \( 1 + \frac{2}{x} + \frac{1}{x^2} \). This reformulation allows for easy integration.
  • The simplification helps in focusing on each separate part, especially when dealing with terms like \( \frac{2}{x} \) or \( \frac{1}{x^2} \).
  • You ensure the integral goes faster since each piece uses standard formulas instead of struggling with the original fraction.
By handling rational functions this way, you make complex problems much more approachable and intuitive. Always simplify before integrating.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

On a hot summer afternoon, a city's electricity consumption is \(-3 t^{2}+18 t+10\) units per hour, where \(t\) is the number of hours after noon \((0 \leq t \leq 6) .\) Find the total consumption of electricity between the hours of 1 and \(5 \mathrm{p} \cdot \mathrm{m}\)

Find the derivative of each function. \(\ln \left(x^{3}+6 x\right)\)

Evaluate \(\int_{1}^{1} \frac{x^{43} e^{-17 x}+219 \sqrt[3]{x^{2}}}{\ln \sqrt[29]{6 x^{3}-x^{-11}}-\pi^{3}} d x .\) [Hint: No work necessary.

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). $$ \int(x-2)(x+4)^{5} d x $$

Seat Belts Seat belt use in the United States has risen to \(86 \%,\) but nonusers still risk needless expense and serious injury. The upper curve in the following graph represent an estimate of fatalities per year if seat belts were not used, and the lower curve is a prediction of actual fatalities per year with seat belt use, both in thousands ( \(x\) represents years after 2010). Therefore, the area between the curves represents lives saved by seat belts. Find the area between the curves from 0 to 20 , giving an estimate of the number of lives saved by seat belts during the years \(2010-2030\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.