Anti-derivative (Integrate this)

[pakistanitarzan]

e^(t^2 / 2) * t^3

I tried integration by parts but i keep integrating forever without reaching an answer

 

[ashley.1965]

e^(t^2 / 2) * t^3

I tried integration by parts but i keep integrating forever without reaching an answer

Integrating forever is a sure sign that (1) you are on the wrong track, or (2) you need to use induction or iteration.

There is a simpler way - try substitution. Here's how:

1. Let g=t^2
2. This implies that (1/2)dg = t*dt (operator equivalence)
3. Substitute into the integral I = \int e^(t^2/2)*(t^3) dt
4. This simplifies the integral to I = \int e^(g/2) g.dg
5. Now use integration by parts to solve this
6. The result is I = 2*e^(g/2) * (g-2)
7. Use g=t^2 to get the result in terms of the original variable t
8. Final result: I = 2 * e^(t^2/2) *(t^2 - 2)
9. Check: Differentiate this and you recover the original integrand !

Hope this helps !

 

[bronxbull]

sorry guys,all wrong.

 

[pakistanitarzan]

take g=(t^2)/2

dg= tdt

substitute back,

u get I = \int (e^g)*(1/3) dg

thus, I= (e^g)/3 +C

g=(t^2)/2

I= (e^(t^2))/3 + C



There is a mistake in step no.4 dude, the integral should be just dg and not gdg

How did you get 1/3 dg there? isnt that supposed to be equal to t^3 somehow?

 

[bronxbull]

my mistake,i saw thata s t/3.

 

[ashley.1965]

take g=(t^2)/2

dg= tdt

substitute back,

u get I = \int (e^g)*(1/3) dg

thus, I= (e^g)/3 +C

g=(t^2)/2

I= (e^(t^2))/3 + C



There is a mistake in step no.4 dude, the integral should be just dg and not gdg

I just checked my calculation again. Here it is in full:

Looks like I missed a factor of (1/2) in the original calculation for I. But the steps are correct otherwise, as the differential check shows.

Not sure how you can claim that I= (e^(t^2))/3 + C differentiates to the original integrand...

 

[KS]

Calculus = gives me creeps.

 

[pakistanitarzan]

Ok Thank you very much Ashley. It makes much more sense now

 

[A1Kaid]

I just checked my calculation again. Here it is in full:

Looks like I missed a factor of (1/2) in the original calculation for I. But the steps are correct otherwise, as the differential check shows.

Not sure how you can claim that I= (e^(t^2))/3 + C differentiates to the original integrand...

Isn't this similar to U-substitution method?

 

[pakistanitarzan]

Calculus = gives me creeps.

Actualy its differential equation question, this integral is only a small part of the question, now i am on to solving the rest of it

Isn't this similar to U-substitution method?

Its exactly that but Ashley is using other d and dg instead

 

[ashley.1965]

Ok Thank you very much Ashley. It makes much more sense now

Anytime, bro - salaams!

 

[KRAIT]

Calculus = gives me creeps.

Oye, Indians ka naam mat duba.