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IRINA_888 [86]
3 years ago
11

Evaluate ∫ xe2x dx. 1 2 3x A./xe +C 6 B.1/xe2x-1/ xe2x+C 22 C.1/xe2x-1/ e2x+C 24 1 2 1 4x D./x-/e +C 28

Mathematics
1 answer:
Dafna1 [17]3 years ago
7 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 = x
     dv/dx = e^(2x)

By differentiating u, 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, 
     ∫ x e^(2x) dx = (1/2)xe^(2x) - (1/4)e^(2x) + C
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between 0 and 1: divide 1 by 2 = 0.5

between 1 and 1/2: divide 1/2 by 2 (0.25) and multiply that by 3 = 0.75

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Hope this helps. Pls give brainliest!

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1360 concert tickets were sold for a total of $13,328. If students paid $8 and non students paid $12, how many student tickets w
lyudmila [28]

748 student tickets were sold.

Step-by-step explanation:

Given,

Number of tickets sold = 1360

Revenue generated = $13328

Cost of each student ticket = $8

Cost of each non student ticket = $12

Let,

x be the number of student tickets sold

y be the number of non student tickets sold

According to given statement;

x+y=1360        Eqn 1

8x+12y=13328    Eqn 2

Multiplying Eqn 1 by 12

12(x+y=1360)\\12x+12y=16320\ \ \ Eqn\ 3

Subtracting Eqn 2 from Eqn 3

(12x+12y)-(8x+12y)=16320-13328\\12x+12y-8x-12y=2992\\4x=2992\\

Dividing both sides by 4

\frac{4x}{4}=\frac{2992}{4}\\x=748

748 student tickets were sold.

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/774670
  • brainly.com/question/763150

#LearnwithBrainly

5 0
3 years ago
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