how to integrate x^3 multiplied with e^(x^2)?

chhhhelsie

Heres my working

what did I do wrong?

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Reply 1

Sirelis

I might be wrong.... But the intergral of e^(x^2) is e^x2 divided by 2 not 2x :s-smilie:

Reply 2

H0PEL3SS

Split the x^3 so that you have u=x^2 and v'=xe(x^2).

Reply 3

Student20

Are you sure it's not e^2x? I think it's impossible to integrate e^(x^2).

Reply 4

chhhhelsie

Original post by Student20

Are you sure it's not e^2x? I think it's impossible to integrate e^(x^2).

Book doesnt lie.

Reply 5

chhhhelsie

Original post by H0PEL3SS

Split the x^3 so that you have u=x^2 and v'=xe(x^2).

Does it mean what I did is wrong? Can you highlight which part is wrong?

Reply 6

Student20

Hmm I think you need to do u = and dv/dx = the other way around, d(e^(x^2))/dx will give you 2x * e^(x^2) then I guess you got to somehow work your way back with the integral part...

I'll have a go at it tomorrow my mind isn't functioning properly at this severely messed up time

Reply 7

chhhhelsie

Original post by spfl

I think the book probably mean't differentiate rather than integrate.

Edit: Unless it wants you to use integration by parts only once, integrating the x^3 term and differentiating the e^(x^2) term. Which ultimately wouldn't arrive at a final answer, you'd have to just keep doing the same thing over and over just with and increase power on the (first) x term.

That will not result in a completed integral. It will go on forever!

Reply 8

chhhhelsie

Original post by Student20

Ok, thanks.

Reply 9

chhhhelsie

Original post by spfl

Yeah, that's why I think it mean't differentiate rather than integrate. Is there by any chance an answer at the back or something?

Here is the answer.

Reply 10

ClickItBack

Original post by chhhhelsie

Book doesnt lie.

Clearly they want you to use your result from when you differentiated e^x^2 in some way.

So think about it: you know the integral of 2xe^x^2 now, from the above. Can you split the product x^3e^x^2 in some way that lets you use that result when you do IBP?

Nice questions btw.

Posted from TSR Mobile

Reply 11

username1732133

Original post by chhhhelsie

Heres my working

what did I do wrong?

You have good hints from HOPEL3SS and ClickItBack.

Your mistakes were splitting the product in an unhelpful way and trying to integrate

e^{x^2}

.

You can rewrite the integral as

\displaystyle \int x^3 e^{x^2} dx = \int (x^2)\left(x e^{x^2}\right) dx

Reply 12

jman1895

Let

t=e^{x^{2}}

then

dt=2xe^{x^{2}}dx.

Now taking the natural log of

t=e^{x^{2}}

we get

\ln{t}=x^{2}.

Now integrate

\frac{1}{2} t \cdot \ln{t}

by parts and you're set.

(edited 8 years ago)

Reply 13

the1akshay

please tell me this isn't a C4 question?

Reply 14

Notnek

Original post by the1akshay

please tell me this isn't a C4 question?

All the methods involved are C4. And given the hint in part a), this becomes a question that should be doable by many C4 students aiming for a top grade.

Reply 15

einot

Easier with a decent substitution.

Reply 16

cattubato

Let I = integral of x^3 e^(2^x) dx

let u =x^2, du = 2 x dx

then I = 1/2 integral of u e^u du

now let v= e^u, dv = e^u du

now I = 1/2 integral of u dv

= 1/2 ( uv - integral of v du )
=1/2 (u e^u - integral of e^u du)
=1/2 ( u e^u - e^u)
=1/2 (x^2 e^(x^2) - e^(x^2))
=e^(x/2)/2 (x^2 -1) [+C]