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Integral of Exponential Help (1 Viewer)

jyu

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f3nr15 said:
Find



This one in particular took "exponentially" longer than usual than all other questions.
Using du = 2xex dx when u = x2ex does not yield the right answer. :(
du = (2x+x2)ex dx when u = x2ex
 

LoneShadow

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Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + Integral[x^2*Exp(x)dx]

let dv = Exp(x) dx => v = Exp(x)
let u = x^2 => du=2xdx

Then using Integration by parts: Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + x^2*Exp(x) - Integral[2x*Exp(x)dx] = x^2*Exp(x) + C

ooopse. I just noticesd this is in 2 unit maths section. Use jyu's method.
 
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jyu said:
du = (2x+x2)ex dx when u = x2ex
I must have differentiated u improperly.
After finding du/dx then ...

= ∫ 2xex+x2ex dx
= ∫ (2x+x2)ex dx
= ∫ du
= ∫ 1 du
= u + C
= x2ex + C

Right ? Because it is really the working out I'm after.

LoneShadow said:
Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + Integral[x^2*Exp(x)dx]

let dv = Exp(x) dx => v = Exp(x)
let u = x^2 => du=2xdx

Then using Integration by parts: Integral[x*(2+x)*Exp(x)dx] = Integral[2x*Exp(x)dx] + x^2*Exp(x) - Integral[2x*Exp(x)dx] = x^2*Exp(x) + C

ooopse. I just noticesd this is in 2 unit maths section. Use jyu's method.
Is using 2 substitutions the Extension 2 method ?
I've formatted your working out so I can comprehend it more easily.

∫[x(2+x)exdx] = ∫[2xexdx] + ∫[x2exdx]

let dv = ex dx => v = ex
let u = x2 => du = 2x dx

Then using Integration by parts:
∫[x(2+x)exdx]
= ∫[2xexdx] + x2ex - ∫[2xexdx] = x2ex + C


jyu's one is a little more simpler to workout but thanks anyway.
 

LoneShadow

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It's called "integration by parts". Pretty useful thing.

edit: and yes it's ext2. Very easy to learn though.
 

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