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integration

I'm a bit stuck on this integral:

∏∫[(x e^(-x)]^2

As 0 → 2

Thanks for any help.

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Reply 1
Avatar for Fermat
Fermat
8
Ok, you want the integral of [(x e^(-x)]^2 with the limits of from x=0 to x=2.

I=∫ [(x e^(-x)]²
I=∫ x².e^(-2x)

Integrating by parts,

I = -½.x².e^(-2x) + ∫ (x.e^(-2x)
I = -½.x².e^(-2x) + {-½.x.e^(-2x) + ½ ∫ (e^(-2x))}
I = -½.x².e^(-2x) - ½.x.e^(-2x) -¼.e(-2x)

puting in the limits of x=0 and x=2, we get

I = {(-½.4.e^(-4) - ½.2.e^(-4) - ¼.e^(-4)) - (0 - 0 - ¼)}
I = {(-2 -1 - ¼) e^(-4) + ¼}
I = ¼ - 3¼e^(-4)
===========
Reply 2
Avatar for Juwel
Juwel
18
He's using pi-notation, doesn't that indicate a product?
Reply 3
Avatar for GH
GH
OP
13
Amazing, I understand fully now. Thanks very much.
Reply 4
Avatar for GH
GH
OP
13
ZJuwelH
He's using pi-notation, doesn't that indicate a product?
Well, more of a constant. It doesn't really matter.
Reply 5
Avatar for Juwel
Juwel
18
2776
Well, more of a constant. It doesn't really matter.


And the answer is a constant, I see.
Reply 6
Avatar for zazy
zazy
1
Integration by parts
x^2 e^-2x
u= x^2 du/dx= 2x
v= -1/2 e ^-2x dv/dx= e^-2x
x^2(-o.5e^(-2x))-the integral of (-0.5e^(-2x))(2x) dx
Integrate the bold expression by parts:
u= 2x du/dx=2
v= 1/4e^(-2x) dv/dx=(-o.5e^(-2x))
[-0.5 x^2e^(-2x)-0.5xe^(-2x)+1/4e^(-2x)] between 0 and 2
final answer= pi(-11/4e^-4+1/4)
Reply 7
Avatar for GH
GH
OP
13
zazy
Integration by parts
x^2 e^-2x
u= x^2 du/dx= 2x
v= -1/2 e ^-2x dv/dx= e^-2x
x^2(-o.5e^(-2x))-the integral of (-0.5e^(-2x))(2x) dx
Integrate the bold expression by parts:
u= 2x du/dx=2
v= 1/4e^(-2x) dv/dx=(-o.5e^(-2x))
[-0.5 x^2e^(-2x)-0.5xe^(-2x)+1/4e^(-2x)] between 0 and 2
final answer= pi(-11/4e^-4+1/4)
You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...
Reply 8
Avatar for zazy
zazy
1
2776
You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...

Aha, so what's new,clumsy me... :frown:
Reply 9
Avatar for GH
GH
OP
13
I'm stuck on another integration here:

∫(x sin3x)

As 0 → (1/3)pi

Unless I made an error, it doesn't matter whether I make (sin3x) u or dv/dx it seems that I am trapped by ever widening integrals.
Reply 10
Avatar for Juwel
Juwel
18
2776
I'm stuck on another integration here:

∫(x sin3x)

As 0 → (1/3)pi

Unless I made an error, it doesn't matter whether I make (sin3x) u or dv/dx it seems that I am trapped by ever widening integrals.


INT xsin3x

Let u=x, dv/dx = sin3x, then du/dx=1 and v=-cos3x/3 so
INT xsin3x = -xcos3x/3 + INT cos3x/3
= xsin3x/3 + sin3x/9

I think.
Reply 11
Avatar for fishpaste
fishpaste
2776
I'm stuck on another integration here:

∫(x sin3x)

As 0 → (1/3)pi

Unless I made an error, it doesn't matter whether I make (sin3x) u or dv/dx it seems that I am trapped by ever widening integrals.


If you set u = x, and dv/dx = sin3x, it seems to go straight to something you integrate for me. :/
Reply 12
Avatar for zazy
zazy
1
ZJuwelH
INT xsin3x

Let u=x, dv/dx = sin3x, then du/dx=1 and v=-cos3x/3 so
INT xsin3x = -xcos3x/3 + INT cos3x/3
= xsin3x/3 + sin3x/9

I think.

That's what i get...(too lazy to type)
the only differenc ebeing that my final answer is;
-1/3x(cos3x)+1/9sin3x
Reply 13
Avatar for fishpaste
fishpaste
Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?
Reply 14
Avatar for m:)ckel
m:)ckel
14
fishpaste
Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?


yeah, final line should read (i think):

-(xcos3x)/3 + 1/9(sin3x) (between limits.....)
Reply 15
Avatar for GH
GH
OP
13
Cheers people.

Say if you wish to integrate the square of the above function i.e.:

∫(x.sin3x)^2

How would you go about it?
Reply 16
Avatar for Juwel
Juwel
18
2776
Cheers people.

Say if you wish to integrate the square of the above function i.e.:

∫(x.sin3x)^2

How would you go about it?


I'd say:

INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy motherfucker of an integration.
Reply 17
Avatar for Squishy
Squishy
11
ZJuwelH
I'd say:

INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy motherfucker of an integration.


I'd also rewrite it as x²sin²3x, but I'd replace sin²3x with ½-½cos6x instead. Then use integration by parts.
Reply 18
Avatar for Fermat
Fermat
8
ZJuwelH
I'd say:

INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy motherfucker of an integration.

Use the identity,

sin²ø = ½(1-cos2ø)
Reply 19
Avatar for Juwel
Juwel
18
Fermat
Use the identity,

sin²ø = ½(1-cos2ø)


Of course! I was trying to think of it while doing the post but couldn't so I suggested the other substitution. But is mine OK or not?

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