All categories

3 years ago

Which of the equations or

Mathematics

1 answer:

Oksi-84 [34.3K]3 years ago

3 0

Answer:

Step-by-step explanation:

Have a great day ^^ :)

You might be interested in

Dai has 3 meters of fabric that he wants to cut into 5 equal-sized pieces.

Mariulka [41]

Answer:each piece has 1/5 of 3 meters of fabric.

Step-by-step explanation:

7 0

3 years ago

VERY URGENT PLZ HELP

Mamont248 [21]

Answer:

Step-by-step explanation:

(a-4)²-a-4

a²-8a+16-a-4

a²-9a+12

8 0

3 years ago

What is the range of the absolute value function shown in the graph?

Alex787 [66]

Answer:

Range : [-6, ∞)

Step-by-step explanation:

Domain of any function on a graph is represented by the x-values (input values).

Similarly, Range of function is represented by the y-values or output values of the function on a graph.

Therefore, domain of the given absolute function will be (-∞, ∞) Or set of all real numbers.

Range of the function → [-6, ∞) Or {y | y ≥ -6}

3 0

3 years ago

Solve for x: -4(3x - 2) = 6x +2

lawyer [7]

Answer:

x = 1/3

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%5Crm%5Cint_%7B0%7D%5E%7B%20%5Cinfty%20%7D%20%5Cleft%28%20%20%5Cfrac%7B%2

mylen [45]

Recall the geometric sum,

$\displaystyle \sum_{k=0}^{n-1} x^k = \frac{1-x^k}{1-x}$

It follows that

$1 - x + x^2 - x^3 + \cdots + x^{2020} = \dfrac{1 + x^{2021}}{1 + x}$

So, we can rewrite the integral as

$\displaystyle \int_0^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Split up the integral at x = 1, and consider the latter integral,

$\displaystyle \int_1^\infty \frac{x^2 + 1}{x^4 + x^2 + 1} \frac{\ln(1 + x^{2021}) - \ln(1 + x)}{\ln(x)} \, dx$

Substitute $x\to\frac1x$ to get

$\displaystyle \int_0^1 \frac{\frac1{x^2} + 1}{\frac1{x^4} + \frac1{x^2} + 1} \frac{\ln\left(1 + \frac1{x^{2021}}\right) - \ln\left(1 + \frac1x\right)}{\ln\left(\frac1x\right)} \, \frac{dx}{x^2}$

Rewrite the logarithms to expand the integral as

$\displaystyle - \int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2021}+1) - \ln(x^{2021}) - \ln(x+1) + \ln(x)}{\ln(x)} \, dx$

Grouping together terms in the numerator, we can write

$\displaystyle -\int_0^1 \frac{1+x^2}{1+x^2+x^4} \frac{\ln(x^{2020}+1)-\ln(x+1)}{\ln(x)} \, dx + 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

and the first term here will vanish with the other integral from the earlier split. So the original integral reduces to

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx$

Substituting $x\to\frac1x$ again shows this integral is the same over (0, 1) as it is over (1, ∞), and since the integrand is even, we ultimately have

$\displaystyle \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \frac{\ln(1-x+\cdots+x^{2020})}{\ln(x)} \, dx = 2020 \int_0^1 \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 1010 \int_0^\infty \frac{1+x^2}{1+x^2+x^4} \, dx \\\\ = 505 \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx$

We can neatly handle the remaining integral with complex residues. Consider the contour integral

$\displaystyle \int_\gamma \frac{1+z^2}{1+z^2+z^4} \, dz$

where γ is a semicircle with radius R centered at the origin, such that Im(z) ≥ 0, and the diameter corresponds to the interval [-R, R]. It's easy to show the integral over the semicircular arc vanishes as R → ∞. By the residue theorem,

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4}\, dx = 2\pi i \sum_\zeta \mathrm{Res}\left(\frac{1+z^2}{1+z^2+z^4}, z=\zeta\right)$

where $\zeta$ denotes the roots of $1+z^2+z^4$ that lie in the interior of γ; these are $\zeta=\pm\frac12+\frac{i\sqrt3}2$ . Compute the residues there, and we find

$\displaystyle \int_{-\infty}^\infty \frac{1+x^2}{1+x^2+x^4} \, dx = \frac{2\pi}{\sqrt3}$

and so the original integral's value is

$505 \times \dfrac{2\pi}{\sqrt3} = \boxed{\dfrac{1010\pi}{\sqrt3}}$

8 0

2 years ago