/*! 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 51 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 51

Evaluate the following integrals. $$\int_{-1}^{0} \frac{x}{x^{2}+2 x+2} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral $\int_{-1}^{0} \frac{x}{x^{2}+2x+2}dx$. Answer: The definite integral evaluates to $\frac{1}{2} \ln 2 - \frac{\pi}{4}$.

Step by step solution

01

Completing the square in the denominator

In order to simplify the given integral, we need to complete the square in the denominator of the integrand. We can rewrite the square part by factoring out the common term and applying the completing the square technique as follows: $$\frac{x}{x^2 + 2x + 2} = \frac{x}{( x^2 + 2x) + 2} = \frac{x}{((x+1)^2 -1^2) + 2} = \frac{x}{(x+1)^2 -1 + 2} = \frac{x}{(x+1)^2 + 1}$$
02

Apply substitution method

Let us make the substitution: $$u = x+1 \Rightarrow x = u-1$$ and $$d u = d x$$ Now, we'll also update the limits of integration: When \(x=-1, u=0\) and When \(x=0, u=1\) So the new integral becomes, $$\int_{u=0}^{1} \frac{(u-1)}{u^{2}+1} d u$$
03

Finding the antiderivative

To find the antiderivative of our integrand, let's separate the fraction: $$\int_{u=0}^{1} \frac{u-1}{u^2+1}du = \int_{u=0}^{1}\left(\frac{u}{u^2+1}-\frac{1}{u^2+1}\right)du$$ Now integrate the separated terms: $$\int_{u=0}^{1}\frac{u}{u^2+1}du - \int_{u=0}^{1}\frac{1}{u^2+1}du = \frac{1}{2} \ln\left|u^2+1\right| - \arctan(u)+ C$$
04

Apply the Fundamental Theorem of Calculus

Now we can evaluate the definite integral using the fundamental theorem of calculus (FTC) by finding the difference of the anti-derivative at the limits of integration: $$\left[\frac{1}{2} \ln\left|u^2+1\right| - \arctan(u)\right]_{u=0}^{1} = \left(\frac{1}{2} \ln 2 - \arctan (1)\right) - \left(\frac{1}{2} \ln 1 - \arctan (0)\right)$$ Since \(\ln 1 = 0\) and \(\arctan 0 = 0\), $$= \frac{1}{2} \ln 2 - \arctan (1) = \frac{1}{2} \ln 2 - \frac{\pi}{4}$$ The final answer is: $$\int_{-1}^{0} \frac{x}{x^{2}+2x+2}dx = \boxed{\frac{1}{2} \ln 2 - \frac{\pi}{4}}$$

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 91影视!

Key Concepts

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

Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square form. This technique is particularly useful in calculus, as it simplifies expressions and helps solve integrals. Let's break down the original quadratic expression in the integral:
  • We start with the expression in the denominator: \(x^2 + 2x + 2\).
  • To complete the square, we focus on the quadratic and linear terms, \(x^2 + 2x\).
  • Add and subtract \((\frac{2}{2})^2 = 1\) inside the expression, rewriting it as \((x+1)^2 - 1 + 2\).
  • This simplifies to \((x+1)^2 + 1\).
This new form makes it easier to rewrite and integrate the function in subsequent steps. Completing the square transforms the quadratic into a more manageable form, setting the foundation for further simplification.
Substitution Method
The substitution method is a technique for solving integrals by changing the variable to simplify the expression. It is especially helpful when dealing with composite functions, like in our integral. Here's a closer look at how it works:
  • First, identify a substitution that simplifies the expression, such as \(u = x + 1\).
  • With this substitution, the integral changes from terms of \(x\) to terms of \(u\). In this example, differentiate: \(du = dx\).
  • Substitute both the function and the limits of integration for the new variable. When \(x = -1\), \(u = 0\) and when \(x = 0\), \(u = 1\).
The revised integral becomes \(\int_{0}^{1} \frac{u-1}{u^2+1} du\), which is simpler and more straightforward to solve. Substitution aids in transforming the integral into a form that highlights its antiderivative.
Antiderivative
An antiderivative is a function whose derivative gives back the original function. Finding antiderivatives is essential for evaluating integrals. It involves identifying a function that differentiates to the given integrand. In our scenario:
  • We start by separating the expression: \(\int \frac{u}{u^2+1}du - \int \frac{1}{u^2+1}du\).
  • The first part, \(\int \frac{u}{u^2+1}du\), can be rewritten using a simple substitution if needed, and it often directly simplifies to a logarithmic form involving \(\ln|u^2+1|\).
  • The second part, \(\int \frac{1}{u^2+1}du\), is a standard form that integrates to \(\arctan(u)\).
Combining these, the antiderivative of the integrand is \(\frac{1}{2}\ln|u^2+1| - \arctan(u)\). This function represents the collection of all possible integrals of the initial function with respect to \(u\).
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) connects differentiation and integration, providing a method to evaluate definite integrals. Here's how it applies:
  • According to the FTC, the integral of a function over an interval can be found by evaluating its antiderivative at the endpoints.
  • In our case, we take the antiderivative \(\frac{1}{2}\ln|u^2+1| - \arctan(u)\) and evaluate it between the limits \(u=0\) and \(u=1\).
  • This results in \(\left(\frac{1}{2} \ln 2 - \arctan(1)\right) - \left(\frac{1}{2} \ln 1 - \arctan(0)\right)\).
Since \(\ln 1 = 0\) and \(\arctan(0) = 0\), the expression simplifies to \(\frac{1}{2} \ln 2 - \frac{\pi}{4}\). This final value represents the area under the curve of the integrand from \(-1\) to \(0\). The Fundamental Theorem of Calculus provides the rigorous framework needed to evaluate these calculations.

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

The following integrals may require more than one table look-up. Evaluate the integrals using a table of integrals, and then check your answer with a computer algebra system. $$\int x \sin ^{-1} 2 x d x$$

It can be shown that $$\begin{array}{l}\int_{0}^{\pi / 2} \sin ^{n} x d x=\int_{0}^{\pi / 2} \cos ^{n} x d x= \\\\\quad\left\\{\begin{array}{ll}\frac{1 \cdot 3 \cdot 5 \cdot \cdots(n-1)}{2 \cdot 4 \cdot 6 \cdots n} \cdot \frac{\pi}{2} & \text { if } n \geq 2 \text { is an eveninteger } \\\\\frac{2 \cdot 4 \cdot 6 \cdots(n-1)}{3 \cdot 5 \cdot 7 \cdots n} & \text { if } n \geq 3 \text { is an odd integer. }\end{array}\right.\end{array}$$ a. Use a computer algebra system to confirm this result for \(n=2,3,4,\) and 5 b. Evaluate the integrals with \(n=10\) and confirm the result. c. Using graphing and/or symbolic computation, determine whether the values of the integrals increase or decrease as \(n\) increases.

Evaluate the following integrals. $$\int_{1}^{\sqrt[3]{2}} y^{8} e^{y^{3}} d y$$

A remarkable integral \(1 \mathrm{t}\) is a fact that \(\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{m} x}=\frac{\pi}{4},\) for all real numbers \(m\). a. Graph the integrand for \(m=-2,-3 / 2,-1,-1 / 2,0,1 / 2,1\) 3/2, and 2, and explain geometrically how the area under the curve on the interval \([0, \pi / 2]\) remains constant as \(m\) varies. b. Use a computer algebra system to confirm that the integral is constant for all \(m\).

Trapezoid Rule and Simpson's Rule Consider the following integrals and the given values of \(n .\) a. Find the Trapezoid Rule approximations to the integral using \(n\) and \(2 n\) subintervals. b. Find the Simpson's Rule approximation to the integral using \(2 n\) subintervals. It is easiest to obtain Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(8 .\) c. Compute the absolute errors in the Trapezoid Rule and Simpson's Rule with \(2 n\) subintervals. $$\int_{1}^{e} \frac{d x}{x} ; n=50$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation
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.