Find the absolute value of |-93.1| and |3-8|

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

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If a patient dosage decrease by 25 milligram per day ,how much change will it be in a week?

timofeeve [1]

After one week it will be a total decrease of 175 milligrams

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3 years ago

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A used motorcycle is on sale for $3,600. Erik makes an offer equal to 3/4 of this price. How much does Erik offer for the motorc

lesantik [10]

3/4 of 3600..." of " means multiply
3/4 * 3600 =
10800/4 =
2700 <==

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3 years ago

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Find the nth term of the arithmetic sequences a1=5,d=6,n=11

ki77a [65]

here's the solution,

n = 11
a = 5 ( a = first term )
d = 6 ( d = common difference )

we know,

=》

$nth \: \: term \: = a \: + (n - 1) \times d$

=》

$11th \: \: term = 5 + (11 - 1) \times 6$

=》

$11th \: \: term = 5 + (10 \times 6)$

=》

$11th \: \: term = 5 + 60$

=》

$11th \: \: term = 65$

nth term ( 11th term ) = 65

3 0

3 years ago

Graph y= x−2. PLZZ HELP

nataly862011 [7]

Explanation: To graph the equation y = x - 2, the idea is simple.

We pick three values for x, plug them into the

equation, and find the corresponding values for y.

In order to do this in an organized way, we set up a chart.

Down the left side we make a column for our x's, down the right side we make a column for our y's, and in the middle we make a column for the side of the equation that contains the x, in this case, x - 2.

So reading the chart from left to right, we have our x or what we plug into the equation, then we have the x - 2 or what happens to the x when we plug it into the equation, and finally we have our y or what we end up with after plugging x into the equation.

So our first step in filling out the chart is choosing

the x's that we want to plug into the equation.

When graphing lines, I always choose three x's,

a positive number, zero, and a negative number.

The easiest numbers to pick are usually 1, 0, and -1.

So for 1, we plug a 1 into the equation for x

and we have y = (1) - 2 or -1

For 0 we plug a 0 into the equation for x

and we have y = (0) - 2 or -2.

For -1 we plug a -1 into the equation for x

and we have y = (-1) - 2 or -3.

So our points are (1, -1), (0, -2), and (-1, -3).

I looked at the x and y column to find those points.

Since our domain is all real numbers, we can connect these points based on the pattern that they are forming on the graph which is a line.

5 0

3 years ago

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