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V125BC [204]
3 years ago
11

Evaluate ∫ xe2x dx.

Mathematics
2 answers:
mixer [17]3 years ago
6 0
The formula is integral of (udv) = uv - integral of (vdu)
We use integration by parts by letting u = x and dv = (e^2x)dx
Then du = dx, and v = (1/2)(e^2x)
integral of x(e^2x)dx = (1/2)(x)(e^2x) - integral of (1/2)(e^2x)dx
= (1/2)(x)(e^2x) - (1/4)(e^2x) + C
Therefore the correct answer is C.
Helen [10]3 years ago
6 0
The answer is (1/2)xe^(2x) - (1/4)e^(2x) + C
Solution:
Since our given integrand is the product of the functions x and e^(2x), we can use the formula for integration by parts by choosing
     u = xdv/dx = e^(2x)

By differentiating, we get
     du/dx= 1
By integrating dv/dx= e^(2x), we have
     v =∫e^(2x) dx = (1/2)e^(2x)

Then we substitute these values to the integration by parts formula:
     ∫ u(dv/dx) dx = uv −∫ v(du/dx) dx ∫ x e^(2x) dx
                          = (x) (1/2)e^(2x) - ∫ ((1/2) e^(2x)) (1) dx
                          = (1/2)xe^(2x) - (1/2)∫[e^(2x)] dx
                          = (1/2)xe^(2x) - (1/2) (1/2)e^(2x) + C
where c is the constant of integration.

Therefore, our simplified answer is now
<span>     ∫ x e^(2x) dx = (1/2)xe^(2x) - (1/4)e^(2x) + C</span>
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Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

6 0
3 years ago
Read 2 more answers
Someone help me out please!! YOU WILL GET BRAINLIEST
sineoko [7]

nope they are not similar

as in order to say that triangles are similar their sides should be in proportion or angles to be equal but here

here only one angle is equal the other one is not

So they are not similar

8 0
2 years ago
Read 2 more answers
6.) *
labwork [276]

Answer:

the common difference is 5.4 and yes, this is an arithmetic sequence

Step-by-step explanation:

Each new term is derived by adding 5.4 to the previous term:

2.1 + 5.4 = 7.5,

7.5 + 5.4 = 12.9,

and so on,

so the common difference is 5.4 and yes, this is an arithmetic sequence.

8 0
2 years ago
Please please help please please ASAP
NeX [460]

Answer:

15.09

Step-by-step explanation:

with reference angle 28°

perpendicular (p) = 8

base (b) = x

Now

tan 28° = p / b

0.53 = 8 / x

x = 8 / 0.53

x = 15.09

4 0
3 years ago
Barry has an ice tray that makes ice cubes shaped like a cone with a diameter of 2 cm and height of 2 cm. To the nearest cubic c
Lynna [10]
8 because 2^3 or 2 x 2 x 2 = 8
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