[Calculus] It didn't take me long to forget how to Integrate.. - Tilted Forum Project Discussion Community

Artsemis · 10-13-2005, 01:28 AM

It's been about a year since I had Calc. II and I simply cannot integrate this:

((2xe)^-x)^2

Parenthesis added for ease of reading, that's 2xe being raised to the x which is raised to the 2 power.
I was thinking substitution but its still just been too long, and I never got the 'by parts' thing down

Thanks

rsl12 · 10-13-2005, 02:08 AM

you may find the problem easier if you rearrange a little:

(2xe)^-2x dx (I assume there's a dx somewhere in there...)

Artsemis · 10-13-2005, 02:14 AM

Yeah sorry.

That looks much easier but I dont see how you can do that... You're saying that the power -x^2 is the same as the power -2x?

aKula · 10-13-2005, 03:45 AM

No it's not, but the notation: ((2xe)^-x)^2 simplyfies to (2xe)^(-2x) not (2xe)^(-x^2)
i.e. (x^n)^m = x^(nm) != x^(n^m)

I'm not sure but someone once told me that you cannot integrate x^x. Are you sure that the question is possible?
(The derivative of x^x I can do though

).

albania · 10-13-2005, 06:24 AM

you can't X^x can only be done numerically, anyway, are you sure you did the () right. because if that there was no X^x term then it might be possible any way i'm looking at it right now and i don't see a clear way to do it. Search for "mathematica and integrating online" there should be a website which intergrates for you, this way you can at least know if there is a solution.

Lebell · 10-13-2005, 06:38 AM

Christ,

I got straight A's in all my math through diffQ and I don't have a friggin clue.

Course, that was a few years ago...

blizzak · 10-13-2005, 08:09 AM

hmm, this site(http://integrals.wolfram.com/) says:
x^2/8xe^2
I haven't got the cares to check it right now, and I can't find a formula that explains it in my differentials notes, but check it out

Jinn · 10-13-2005, 09:44 AM

It's been three years since I did Calculus, and even longer since I did Algebra -- but I think Integration by Parts DOES solve it. (It's different than Wolfram's Solution, however..)

rsl12 · 10-13-2005, 10:18 AM

That would work if it's 2xe^(-2x), but it looks like it's (2xe)^(-2x).

ps. dunno how blizzak got his answer--i tried that site and came up with no answer. (looks like a neat site though!) I also wonder if this is possible without numerical methods.

stingc · 10-13-2005, 04:46 PM

There's no analytic integral for (2xe)^(-2x). If this is a class, I think the original poster used parentheses incorrectly. 2x e^(-2x) is much more reasonable.

Jinn · 10-14-2005, 07:41 AM

You can't seperate an exponential from its exponent with parentheses..

Quote:

That would work if it's 2xe^(-2x), but it looks like it's (2xe)^(-2x).

What exactly are you E'ing?

rsl12 · 10-14-2005, 08:37 AM

hmmm jinnkai we must follow different schools of notation. looks OK to me. ever hear the expression 'please excuse my dear aunt sally?'

phukraut · 10-14-2005, 03:13 PM

I think we all just ought to wait until the question is restated in a different way by the original poster. Clear it up for us...

Artsemis · 10-15-2005, 08:33 PM

Ack, REALLY sorry! It was my fault the parenthesis are used incorrectly. I should have just left them off completely.

2xe^-x^2

Again, my apologies

Artsemis · 10-15-2005, 08:40 PM

Using substitution I have the integral of:
e^-u du

If I remember correctly, is that just e^-u = e^-x^2?

phukraut · 10-15-2005, 10:10 PM

Letting u=x^2 gives du=2x dx, and hence the integrand is now e^(-u) which integrates to -e^(-u) + C. Note the minus sign in front of exp.

Artsemis · 10-15-2005, 10:23 PM

gotcha! thanks

10-13-2005, 01:28 AM	#1 (permalink)
Artsemis Insane Location: West Virginia	[Calculus] It didn't take me long to forget how to Integrate.. It's been about a year since I had Calc. II and I simply cannot integrate this: ((2xe)^-x)^2 Parenthesis added for ease of reading, that's 2xe being raised to the x which is raised to the 2 power. I was thinking substitution but its still just been too long, and I never got the 'by parts' thing down Thanks __________________ - Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know

10-13-2005, 02:08 AM	#2 (permalink)
rsl12 On the lam Location: northern va	you may find the problem easier if you rearrange a little: (2xe)^-2x dx (I assume there's a dx somewhere in there...) __________________ oh baby oh baby, i like gravy.

10-13-2005, 02:14 AM	#3 (permalink)
Artsemis Insane Location: West Virginia	Yeah sorry. That looks much easier but I dont see how you can do that... You're saying that the power -x^2 is the same as the power -2x? __________________ - Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know

10-13-2005, 06:38 AM	#6 (permalink)
Lebell Cracking the Whip Location: Sexymama's arms...	Christ, I got straight A's in all my math through diffQ and I don't have a friggin clue. Course, that was a few years ago... __________________ "Of all tyrannies, a tyranny exercised for the good of its victims may be the most oppressive. It may be better to live under robber barons than under omnipotent moral busybodies. The robber baron's cruelty may sometimes sleep, his cupidity may at some point be satiated; but those who torment us for our own good will torment us without end, for they do so with the approval of their own conscience." – C. S. Lewis The ONLY sponsors we have are YOU! Please Donate!

10-13-2005, 08:09 AM	#7 (permalink)
blizzak Crazy	hmm, this site(http://integrals.wolfram.com/) says: x^2/8xe^2 I haven't got the cares to check it right now, and I can't find a formula that explains it in my differentials notes, but check it out __________________ Fueled by oxytocin!

10-13-2005, 03:45 AM	#4 (permalink)
aKula Psycho	No it's not, but the notation: ((2xe)^-x)^2 simplyfies to (2xe)^(-2x) not (2xe)^(-x^2) i.e. (x^n)^m = x^(nm) != x^(n^m) I'm not sure but someone once told me that you cannot integrate x^x. Are you sure that the question is possible? (The derivative of x^x I can do though ).

10-13-2005, 06:24 AM	#5 (permalink)
albania Psycho	you can't X^x can only be done numerically, anyway, are you sure you did the () right. because if that there was no X^x term then it might be possible any way i'm looking at it right now and i don't see a clear way to do it. Search for "mathematica and integrating online" there should be a website which intergrates for you, this way you can at least know if there is a solution.

10-13-2005, 09:44 AM	#8 (permalink)
Jinn Lover - Protector - Teacher Location: Seattle, WA	It's been three years since I did Calculus, and even longer since I did Algebra -- but I think Integration by Parts DOES solve it. (It's different than Wolfram's Solution, however..) __________________ "I'm typing on a computer of science, which is being sent by science wires to a little science server where you can access it. I'm not typing on a computer of philosophy or religion or whatever other thing you think can be used to understand the universe because they're a poor substitute in the role of understanding the universe which exists independent from ourselves." - Willravel

10-13-2005, 10:18 AM	#9 (permalink)
rsl12 On the lam Location: northern va	That would work if it's 2xe^(-2x), but it looks like it's (2xe)^(-2x). ps. dunno how blizzak got his answer--i tried that site and came up with no answer. (looks like a neat site though!) I also wonder if this is possible without numerical methods. __________________ oh baby oh baby, i like gravy. Last edited by rsl12; 10-13-2005 at 11:16 AM..

10-13-2005, 04:46 PM	#10 (permalink)
stingc Psycho Location: PA	There's no analytic integral for (2xe)^(-2x). If this is a class, I think the original poster used parentheses incorrectly. 2x e^(-2x) is much more reasonable.

10-14-2005, 08:37 AM	#12 (permalink)
rsl12 On the lam Location: northern va	hmmm jinnkai we must follow different schools of notation. looks OK to me. ever hear the expression 'please excuse my dear aunt sally?' __________________ oh baby oh baby, i like gravy.

10-14-2005, 03:13 PM	#13 (permalink)
phukraut Addict	I think we all just ought to wait until the question is restated in a different way by the original poster. Clear it up for us...

10-15-2005, 08:33 PM	#14 (permalink)
Artsemis Insane Location: West Virginia	Ack, REALLY sorry! It was my fault the parenthesis are used incorrectly. I should have just left them off completely. 2xe^-x^2 Again, my apologies __________________ - Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know

10-15-2005, 08:40 PM	#15 (permalink)
Artsemis Insane Location: West Virginia	Using substitution I have the integral of: e^-u du If I remember correctly, is that just e^-u = e^-x^2? __________________ - Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know

10-15-2005, 10:10 PM	#16 (permalink)
phukraut Addict	Letting u=x^2 gives du=2x dx, and hence the integrand is now e^(-u) which integrates to -e^(-u) + C. Note the minus sign in front of exp. Last edited by phukraut; 10-15-2005 at 10:17 PM..

10-15-2005, 10:23 PM	#17 (permalink)
Artsemis Insane Location: West Virginia	gotcha! thanks __________________ - Artsemis ~~~~~~~~~~~~~~~~~~~~ There are two keys to being the best: 1.) Never tell everything you know