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Karo-lina-s [1.5K]
3 years ago
5

What is different between solving inequalities and absolute value inequalities?

Mathematics
2 answers:
notka56 [123]3 years ago
6 0

Answer:

Solving Inequalities are different because in Absolute Value Inequalities the values always have to be positive and in Solving Inequalities doesn't always have to be positive.

Step-by-step explanation:

Ivenika [448]3 years ago
5 0

Answer:

The absolute number of a number a is written as

|a|

And represents the distance between a and 0 on a number line.

An absolute value equation is an equation that contains an absolute value expression. The equation

|x|=a

Has two solutions x = a and x = -a because both numbers are at the distance a from 0.

To solve an absolute value equation as

|x+7|=14

You begin by making it into two separate equations and then solving them separately.

x+7=14

x+7−7=14−7

x=7

or

x+7=−14

x+7−7=−14−7

x=−21

An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative.

The inequality

|x|<2

Represents the distance between x and 0 that is less than 2

Whereas the inequality

|x|>2

Represents the distance between x and 0 that is greater than 2

You can write an absolute value inequality as a compound inequality.

−2<x<2

This holds true for all absolute value inequalities.

|ax+b|<c,wherec>0

=−c<ax+b<c

|ax+b|>c,wherec>0

=ax+b<−corax+b>c

You can replace > above with ≥ and < with ≤.

When solving an absolute value inequality it's necessary to first isolate the absolute value expression on one side of the inequality before solving the inequality.

Step-by-step explanation:

Hope this helps :)

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Step-by-step explanation:

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Help evaluating the indefinite integral
Dafna11 [192]

Answer:

\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }

General Formulas and Concepts:
<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle (cu)' = cu'

Derivative Property [Addition/Subtraction]:
\displaystyle (u + v)' = u' + v'
Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:
\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Property [Multiplied Constant]:
\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

Integration Methods: U-Substitution and U-Solve

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

<em />\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution/u-solve</em>.

  1. Set <em>u</em>:
    \displaystyle u = 4 - x^2
  2. [<em>u</em>] Differentiate [Derivative Rules and Properties]:
    \displaystyle du = -2x \ dx
  3. [<em>du</em>] Rewrite [U-Solve]:
    \displaystyle dx = \frac{-1}{2x} \ du

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Apply U-Solve:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-x}{2x\sqrt{u}}} \, du
  2. [Integrand] Simplify:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \int {\frac{-1}{2\sqrt{u}}} \, du
  3. [Integral] Rewrite [Integration Property - Multiplied Constant]:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \frac{-1}{2} \int {\frac{1}{\sqrt{u}}} \, du
  4. [Integral] Apply Integration Rule [Reverse Power Rule]:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = -\sqrt{u} + C
  5. [<em>u</em>] Back-substitute:
    \displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = \boxed{ -\sqrt{4 - x^2} + C }

∴ we have used u-solve (u-substitution) to <em>find</em> the indefinite integral.

---

Learn more about integration: brainly.com/question/27746495

Learn more about Calculus: brainly.com/question/27746485

---

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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2 years ago
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