Help me please!!!!!!

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

Unit: Integration

5 0

2 years ago

NEED HELP PLEASE!!!!

Masja [62]

Answer:

Yes

Step-by-step explanation:

$\sqrt{\frac{2x}{9} }={\frac{\sqrt{2x} }\sqrt{} {9} } =\frac{\sqrt{2x} }{3}=\frac{1}{3}\sqrt{2x}$

3 0

2 years ago

PLEASE HELP ME PLEASE!!

Harman [31]

Answer:

hypotenuse (h)=?

perpendicular (p)=3

base(b)=2

using pythagoras theorem

h²=p²+b²

h²=3²+2²

h²=9+4

h²=13

h=√13

hypotenuse (h)=12
perpendicular (p)=10
base(b)=?

using pythagoras theorem

b²=h²-p²

b²=12²-10²

b²=144-100

b=√44

b=2√11

6 0

2 years ago

Solve please and SHOW UR WORK

Ksenya-84 [330]

Answer:

1.3x

Step-by-step explanation:

X=1

1+0.3=1.3 (don't forget to add 'X')

So, the answer is 1.3x

hope this helps!

8 0

3 years ago

The image shows a ray of light bending at the boundary of a glass block. According to a law of optics, = n, where n is a constan

Salsk061 [2.6K]

The law of optics applicable in this scenario is the Snell's Law which states that the ratio of the index of refraction of two media is equal to the ratio of the sine of the angle of reflection and sine of the angle of refraction. In equation form:
n₁ / n₂ = sin b / sin a

8 0

3 years ago

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