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Hoochie [10]
3 years ago
14

Determine the area of the given region under the curve (1/x^4) [1,2]

Mathematics
1 answer:
Nikolay [14]3 years ago
6 0

Answer:

\displaystyle \frac{7}{24}

General Formulas and Concepts:

<u>Algebra I</u>

  • Exponential Rule [Rewrite]: \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

[Area] Limits of Riemann's Sums - Integrals

Integration Rule [Reverse Power Rule]:                                                                    \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                          \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle f(x) = \frac{1}{x^4} \\ \ [1, 2]<u />

<u />

<u>Step 2: Find Area</u>

  1. [Integral] Set up area:                                                                                    \displaystyle \int\limits^2_1 {\frac{1}{x^4}} \, dx
  2. [Integral] Rewrite:                                                                                            \displaystyle \int\limits^2_1 {x^{-4}} \, dx
  3. [Integral] Reverse Power Rule:                                                                      \displaystyle \frac{-1}{3x^3} \bigg| \limits^2_1
  4. [Area] Fundamental Theorem of Calculus:                                                   \displaystyle \frac{7}{24}

Topic: Calculus

Unit: Basic Integration/Riemann Sums

Book: College Calculus 10e

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Yes it's totally correct
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Estimate the square root to the nearest integer and tenth square root of 8
Sever21 [200]

Answer:

√

8

≈

3

Explanation:

Note that:

2

2

=

4

<

8

<

9

=

3

2

Hence the (positive) square root of  

8

is somewhere between  

2

and  

3

. Since  

8

is much closer to  

9

=

3

2

than  

4

=

2

2

, we can deduce that the closest integer to the square root is  

3

.

We can use this proximity of the square root of  

8

to  

3

to derive an efficient method for finding approximations.

Consider a quadratic with zeros  

3

+

√

8

and  

3

−

√

8

:

(

x

−

3

−

√

8

)

(

x

−

3

+

√

8

)

=

(

x

−

3

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2

−

8

=

x

2

−

6

x

+

1

From this quadratic, we can define a sequence of integers recursively as follows:

⎧

⎪

⎨

⎪

⎩

a

0

=

0

a

1

=

1

a

n

+

2

=

6

a

n

+

1

−

a

n

The first few terms are:

0

,

1

,

6

,

35

,

204

,

1189

,

6930

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...

The ratio between successive terms will tend very quickly towards  

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+

√

8

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So:

√

8

≈

6930

1189

−

3

=

3363

1189

≈

2.828427

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