/*! 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 15 Find each integral by using the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 15

Find each integral by using the integral table on the inside back cover. $$ \int \sqrt{x^{2}-4} d x $$

Short Answer

Expert verified
\( \int \sqrt{x^2 - 4} \, dx = \frac{x}{2} \sqrt{x^2 - 4} - 2 \ln|x + \sqrt{x^2 - 4}| + C \)

Step by step solution

01

Identify the Integral Form

Recognize that the integral \( \int \sqrt{x^2 - 4} \, dx \) resembles an integral in the form \( \int \sqrt{x^2 - a^2} \, dx \), where \( a = 2 \).
02

Use the Integral Table Formula

According to standard integral tables, the integral \( \int \sqrt{x^2 - a^2} \, dx \) can be expressed as \( \frac{x}{2} \sqrt{x^2 - a^2} - \frac{a^2}{2} \ln|x + \sqrt{x^2 - a^2}| + C \). Substitute \( a = 2 \) into this formula.
03

Substitute the Value of a

For \( a = 2 \), substitute into the formula: \( \int \sqrt{x^2 - 4} \, dx = \frac{x}{2} \sqrt{x^2 - 4} - \frac{4}{2} \ln|x + \sqrt{x^2 - 4}| + C \).
04

Simplify the Expression

Simplify the expression to: \( \frac{x}{2} \sqrt{x^2 - 4} - 2 \ln|x + \sqrt{x^2 - 4}| + C \).
05

Verify the Solution

Ensure that all substitutions and transformations align with the standard tables and simplify correctly. The final integral solution matches the known form.

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.

Integral Tables
Imagine having a list where you can quickly check solutions to many integrals without calculating them from scratch. That's exactly what integral tables are! They compile a catalogue of solved integrals, organized in a way that allows you to find solutions by matching your integral expressions to known forms.
Think of it like a dictionary, but for integrals. These tables are particularly handy when dealing with complex integrals that do not have an obvious solution through simple algebraic manipulation.
  • To use them, simply match the integral's form with a corresponding format in the table.
  • Most tables will have common algebraic forms, trigonometric integrals, and even exponential ones.
  • Once you find the right form, apply it directly to your problem.
It's important to note that these tables typically include both definite and indefinite forms, but you'll need to know what kind of integral you're working with to choose the right formula.
Definite Integrals
Definite integrals represent the area under the curve of a function between two points on a graph. Imagine the curve as a hill, and the integral calculates the total space that the hill covers from start to end.
Just like determining how much paint you'd need to cover a specific area, definite integrals help you find exact values, which is unlike indefinite integrals that give a general formula.
  • In the process, you replace the function鈥檚 variable with the numeric boundaries provided in the integral.
  • Evaluate the antiderivative at the upper limit, and subtract the value at the lower limit.
  • This operation gives you the specific value for that integral within the set limits.
Since definite integrals provide specific results, they are invaluable in real-world applications such as physics and engineering, where exact values are crucial.
Indefinite Integrals
An indefinite integral is like a treasure map that leads you to a family of functions, rather than just a single solution. When you solve an indefinite integral, you don't get a specific number. Instead, you get a function plus a constant (usually denoted as +C).
This makes them quite different from definite integrals that have specific upper and lower limits. Indefinite integrals tell you about the general form of the antiderivative of the function.
  • They don鈥檛 have boundaries, implying the solution is a general form.
  • The constant C is crucial because it represents any constant multiple that could be added to the function.
  • Finding the indefinite integral is essentially the reverse process of differentiation.
When you solve an indefinite integral, you're essentially unwrapping the layers of a function to see its broad potential shapes. This openness is what makes them so flexible and important for further calculus operations.

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

Think of the slope field for the differential equation \(\frac{d y}{d x}=\frac{6 x}{y^{2}} .\) What is the sign of the slope in quadrant I (where \(x\) and \(y\) are both positive)? What is the sign of the slope in each of the other three quadrants? Check your answers by looking at the slope field on page 463 .

In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T, p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. ( \(K\) and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself.) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=K I_{0}-\frac{K}{R} p$$ Find the general solution of this differential equation. \(\left(I_{0}\right.\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example 7.] d. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0}\). e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right)\). Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0)\). Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$

A medical examiner called to the scene of a murder will usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If \(y(t)\) is the temperature (in degrees Fahrenheit) of the body thours after the murder, and if the room temperature is \(70^{\circ}\), then \(y\) satisfies $$\begin{aligned} y^{\prime} &=-0.32(y-70) \\ y(0) &\left.=98.6 \text { (body temperature initially } 98.6^{\circ}\right) \end{aligned}$$ a. Solve this differential equation and initial condition. b. Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the body when it was discovered was \(80^{\circ}\).

Your company has developed a new product, and your marketing department has predicted how it will sell. Let \(y(t)\) be the (monthly) sales of the product after \(t\) months. a. Write a differential equation that says that the rate of growth of the sales will be four times the one-half power of the sales. b. Write an initial condition that says that at time \(t=0\) sales were 10,000 . c. Solve this differential equation and initial value. d. Use your solution to predict the sales at time \(t=12\) months.

A cell receives nutrients through its surface, and its surface area is proportional to the two-thirds power of its weight. Therefore, if \(w(t)\) is the cell's weight at time \(t\), then \(w(t)\) satisfies \(w^{\prime}=a w^{2 / 3}\), where \(a\) is a positive constant. Solve this differential equation with the initial condition \(w(0)=1\) (initial weight 1 unit).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Geometry

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.