# integration

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I'm a bit stuck on this integral:

∏∫[(x e^(-x)]^2

As 0 → 2

Thanks for any help.

∏∫[(x e^(-x)]^2

As 0 → 2

Thanks for any help.

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#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)

===========

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)

===========

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(Original post by

He's using pi-notation, doesn't that indicate a product?

**ZJuwelH**)He's using pi-notation, doesn't that indicate a product?

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#6

(Original post by

Well, more of a constant. It doesn't really matter.

**2776**)Well, more of a constant. It doesn't really matter.

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#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))-

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)

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)

0

(Original post by

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))-

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)

**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)

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#9

(Original post by

You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...

**2776**)You've gone wrong numerically I think. Its (-13/4)pi...not (-11/4)pi...

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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.

∫(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.

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#11

(Original post by

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.

**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.

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.

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#12

**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.

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#13

(Original post by

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.

**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.

the only differenc ebeing that my final answer is;

-1/3x(cos3x)+1/9sin3x

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#14

Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?

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#15

(Original post by

Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?

**fishpaste**)Just the final line, am I being silly, or did you copy down the sin term instead of the cos term?

-(xcos3x)/3 + 1/9(sin3x) (between limits.....)

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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?

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

∫(x.sin3x)^2

How would you go about it?

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#17

(Original post by

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?

**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?

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

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#18

(Original post by

I'd say:

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

**ZJuwelH**)I'd say:

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

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#19

**ZJuwelH**)

I'd say:

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

sin²ø = ½(1-cos2ø)

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#20

(Original post by

Use the identity,

sin²ø = ½(1-cos2ø)

**Fermat**)Use the identity,

sin²ø = ½(1-cos2ø)

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