# integration

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#1
I'm a bit stuck on this integral:

∫[(x e^(-x)]^2

As 0 → 2

Thanks for any help.
0
16 years ago
#2
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)
===========
0
16 years ago
#3
He's using pi-notation, doesn't that indicate a product?
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#4
Amazing, I understand fully now. Thanks very much.
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#5
(Original post by ZJuwelH)
He's using pi-notation, doesn't that indicate a product?
Well, more of a constant. It doesn't really matter.
0
16 years ago
#6
(Original post by 2776)
Well, more of a constant. It doesn't really matter.
And the answer is a constant, I see.
0
16 years ago
#7
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
0
#8
(Original post by 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
You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...
0
16 years ago
#9
(Original post by 2776)
You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...
Aha, so what's new,clumsy me...
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#10
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.
0
16 years ago
#11
(Original post by 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.
0
16 years ago
#12
(Original post by 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. :/
0
16 years ago
#13
(Original post by 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
0
16 years ago
#14
Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?
0
16 years ago
#15
(Original post by 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.....)
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#16
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?
0
16 years ago
#17
(Original post by 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 mother****er of an integration.
0
16 years ago
#18
(Original post by ZJuwelH)
I'd say:

INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy mother****er of an integration.
I'd also rewrite it as x²sin²3x, but I'd replace sin²3x with ½-½cos6x instead. Then use integration by parts.
0
16 years ago
#19
(Original post by ZJuwelH)
I'd say:

INT xsin3x = INT x²sin²3x, substitute sin²3x for 2sin3xcos3x and get one messy mother****er of an integration.
Use the identity,

sin²ø = ½(1-cos2ø)
0
16 years ago
#20
(Original post by 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?
0
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