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3 years ago

5

This year Jack will be 2 years from being twice as old as your sister Jen. The sum of Jack's age and three times Jen's age is 66

. How old is jen?

1 answer:

FinnZ [79.3K]3 years ago

3 0

The age of jenn is 13.6 years old.

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

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]:

f(x) = cxⁿ
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>.

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

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

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

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

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

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

Unit: Integration

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