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2 years ago

Be granted 10 points plus brainliest

Mathematics

1 answer:

BartSMP [9]2 years ago

5 0

Answer:

8 square feet

Step-by-step explanation:

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Dafna11 [192]

Answer:

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

General Formulas and Concepts:
Calculus

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:

Step 1: Define

Identify given.

$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution/u-solve.

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

Step 3: Integrate Pt. 2

[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]:
$\displaystyle \int {\frac{x}{\sqrt{4 - x^2}}} \, dx = -\sqrt{u} + C$
[u] 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 find 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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