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yanalaym [24]
3 years ago
11

Solve the Anti derivative.​

Mathematics
1 answer:
Alex Ar [27]3 years ago
7 0

Answer:

\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C

General Formulas and Concepts:

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

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

Trig Integration:                                                                                                           \displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx<u />

<u />

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

  1. [Integral] Factor fraction denominator:                                                         \displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx
  2. [Integral] Integration Property - Multiplied Constant:                                   \displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx

<u>Step 3: Identify Variables</u>

<em>Set up u-substitution for the arctan trig integration.</em>

\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx

<u>Step 4: Integrate Pt. 2</u>

  1. [Integral] Substitute u-du:                                                                               \displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du
  2. [Integral] Trig Integration:                                                                               \displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C
  3. [Integral] Simplify:                                                                                           \displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C
  4. [integral] Multiply:                                                                                           \displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C
  5. [Integral] Back-Substitute:                                                                             \displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

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I got 23 u might want to check it tho
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3 years ago
An open box is made from an 8 by ten-inch rectangular piece of cardboard by cutting squares from each corner and folding up the
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Please find the attachment.

Let x represent the side length of the squares.

We have been given that an open box is made from an 8 by ten-inch rectangular piece of cardboard by cutting squares from each corner and folding up the sides. We are asked to find the volume of the box.

The side of box will be 8-x-x=8-2x and 10-x-x=10-2x.

The height of the box will be x.

The volume of box will be area of base times height.

\text{Volume of box}=(8-2x)(10-2x)\cdot x

Now we will use FOIL to simplify our expression.

\text{Volume of box}=(80-16x-20x+4x^2)\cdot x

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Now we will distribute x.

\text{Volume of box}=80x-36x^2+4x^3

V(x)=80x-36x^2+4x^3

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3 years ago
<img src="https://tex.z-dn.net/?f=%20%20log_%7B2%7D%28x%29%20%20%3D%20%20-%204" id="TexFormula1" title=" log_{2}(x) = - 4" al
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Step-by-step explanation:

Since the inverse of a Logarithm is an exponential function, we know that the final solution has to involve an exponential function somewhere in it.

1. log B(2) {x} = -4  || given

2. x = 2 ^ -4  || Logarithm rule that allows you to move the base of the logarithm to the base of the exponent on the other side. For example, if you had log B(5) {x} = 3, the base of 5 would move over to the other side and it would be raised to 3; x = 5^3.

3. x = (1) / (2^4) || Simplify. Use the negative exponent rule. This rule always leaves a numerator of 1, and a denominator of your exponent. In this case, it will be 2 ^ -4, so you will do 2^4 which is 16 and you will put that over 1. Resulting in your final answer of x = 1/16

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