Question

integral from 1 to infinity of x^2*(e^-(2x))

Answer 1

Explain with steps please

Answer 2

1 to inf hell where'd you get that

Answer 3

textbook

Answer 4

I think you misunderstood the question. It is to the power of

Answer 5

\[\int\limits_{1}^{\infty} x^{2e^{2x}} \]right ?

Answer 6

typo -2x not 2x .....

Answer 7

\[\int\limits_{1}^{\infty} x ^{2} e ^{-2x}dx\]

Answer 8

This is the correct one

Answer 9

Power is -2x

Answer 10

I dont know why the (-) sign does not show up

Answer 11

or is it \[\int\limits_{1}^{\infty}x^2 \times e^{-2x}\] oh okay i got it ...its okay its a minus only ...i got it wring so many times lol

Answer 12

you know by parts did you try it with that

Answer 13

yes i tried that. I dont get the answer. I said let u= x^2, du=2xdx, dv=e^-2x and v=-1/2e^-2x

Answer 14

if i try by parts then it will have me


but how would you integrate it afterwards ...

Answer 15

ignore the first sentence

Answer 16

wait lemme gets some help ..

Answer 17

see maya when i tried by parts then i get x^2 ..as one term ...but if i put inft it leads to undefined

Answer 18

check your question again

Answer 19

I get that too. But 1/inf = 0

Answer 20

see after repeated by parts i get

\[-\frac{x^2*e^{-2x}}{2} - \frac{x*e^{-2x}}{2} + \frac{e^{-2x}}{4}\] Now none of the expression gives a defined limit ...I could be wrong ..and I wish I'm wrong otherwise your question must have some errors

Answer 21

Textbook answer is 5/(4*e^2)

Answer 22

what ? i must be wrong the : (

Answer 23

I'm slowly working it by parts but I'm a wee bit rusty on these.

Answer 24

maya what if it is 2x*e^(x^2)

Answer 25

nono Im sure that is the question.

Answer 26

http://www.wolframalpha.com/input/?i=+x^2*%28e^-%282x%29%29
I got the answers right

Answer 27

trust me there is some problem with question

Answer 28

I got the same answer as Maya. Give me a few minutes to type this beastie out.....

Answer 29

These things were a beast in my Calc 2 class. They still are now :)

Answer 30

hey there is no problem with your question
http://www.wolframalpha.com/input/?i=+integrate+x^2*%28e^-%282x%29%29+from+1+to+infinity+

sorry for blaming your question

Answer 31

\[\int\limits_{1}^{\infty}(x^{2}e^{-2x})dx\]
Intergration by parts.
\[Let u=x^{2}\]
\[Let dv = e^{-2x}\]
\[du=2xdx\]
\[v=(-1/2)e^{-2x}\]
Now we have:
\[[(-1/2)x^{2}e^{-2x}]^{\infty}_{1} -\int\limits_{1}^{\infty}(2x(-1/2)e^{-2x})dx\]
Solving the first part we get:
\[(0-(1/2)e^{-2})\]

Answer 32

-1 e^-2x(2x^2 + 2x + 1)
---x ---------
4

maybe infinity is not considered .....maybe

if i put 1 then i get

e^-2(5)
---
4

Answer 33

which is your answer

Answer 34

you have to ask someone else about it

Answer 35

maya i got it here i'm posting the solution

Answer 36

We now need to work the second part as another integration of parts:
\[Let u = 2x\]
\[du = 2dx\]
\[Let dv = (-1/2)e^{-2x}\]
\[v=(1/4)e^{-2x}\]
Putting into the integration we get:
\[-[[(1/2)xe^{-2x}]^{\infty}_{1}-\int\limits_{1}^{\infty}((1/2)e^{-2x})dx]\]
Solving the left we get:
\[-[(0-(1/2)e^{-2x})-\int\limits_{1}^{\infty}((1/2)e^{-2x})dx]\]
Integrating the right we get:
\[-[(1/2)e^{-2x} - [(1/4)e^{-2x}]^{\infty}_{1}\]
Evaluating the right we get:
\[-[(1/2)e^{-2x} - [0-(1/4)e^{-2}]]\]
Putting it all back together you get:
\[(1/2)e^{-2}+(1/2)e^{-2}+(1/4)e^{-2}=(5/4)e^{-2}=(5/(4e^{2}))\]

Answer 37

now when we have to put infinity then

we know integral is -1/4*e^-2x*(2x^2 + 2x +1)

\[\lim_{x \rightarrow \infty } - \frac{e^{-2x}(2x^2 + 2x +1)}{4}\]

Answer 38

which gives zero

Answer 39

now all you've got is the second limit which is one so you get 5/4e^2

Answer 40

BTW I made a typo in my first post. It should read:
\[(0-(-1/2)e^{-2})\]

Answer 41

That was a beast. I sure don't miss Calc 2. Nothing like page full solutions where one minus sign in the wrong spot is a wrong answer.

Answer 42

Lol Osirisis

Answer 43

Thank you :) Sorry I just got back to this problem again.

Answer 44

Source: http://www.acalculator.com/math-calculators.html