Intergration by parts.

[Bobby Jones]

How do you solve the question below. Is it by parts?

{ Intergration sign, with limits +2 at top and -1 at bottom} xe^(1-x^2) dx

The e is an exponential.


Is the answer: -0.955

 

[galactus]

Hello.

\(\displaystyle \int_{-1}^{2}xe^{1-x^{2}}dx\)

No, the solution is not -.955.



This can be done with a simple u substitution.

Let \(\displaystyle u=1-x^{2}, \;\ -\frac{du}{2}=xdx\)

 

[soroban]

Hello, Bobby Jones!

How do you solve the question below. .Is it by parts?

. . \(\displaystyle I \;=\;\int^2_{-1} xe^{1-x^2}\,dx\)

You asked about "by parts", but I doubt that you tried it.


There are two choices:

\(\displaystyle \text{We have: }\:\begin{Bmatrix}u &=& x && dv &=& e^{1-x^2}dx \\ du &=& dx && v &=& ?? \end{Bmatrix}\)


\(\displaystyle \text{We have: }\:\begin{Bmatrix}u &=& e^{1-x^2} && dv &=& x\,dx \\ du &=& -2xe^{1-x^2}dx && v &=& \frac{1}{2}x^2 \end{Bmatrix}\)

\(\displaystyle \text{Then: }\:I \;=\;\tfrac{1}{2}x^2e^{1-x^2} + \underbrace{\int x^3e^{1-x^2}dx}_{worse!}\)

Edit: my last line had vanished.
.

 

[Bobby Jones]

Would i take your second lot of values and apply them to this formula:

{Intragrel sign} u.dv = uv - {Intragrel sign} v du

Making

{Intragrel sign} xe^1-x^2 = 1/2.x^2 . e^1-x^2 - {Intragrel sign} 1/2.x^2 . -2xe^1-x^2


={Intragrel sign} xe^1-x^2 = 1/2.x^2 . e^1-x^2 + {Intragrel sign} x^3 . e^1-x^2


Where do i go next?

 

[Bobby Jones]

Soroban,

How did you differentiate e^(1-x^2).

I know if y = e^3x then dy/dx = 3e^3x, bu with the x^2 above the e I'm getting confused.

 

[galactus]

Soroban,

How did you differentiate e^(1-x^2).

I know if y = e^3x then dy/dx = 3e^3x, but with the x^2 above the e I'm getting confused.

You do it the same way as with e^(3x). What's the derivative of 1-x^2?. This is the chain rule.

The chain rule essentially says, "derivative of inside times derivative of outside".

The outside is \(\displaystyle e^{1-x^{2}}\). The inside is \(\displaystyle 1-x^{2}\)

 

[Bobby Jones]

Soroban,

How should I carry on from your edited line, whic h says worse underneath? Should I take u as e^(1-x^2) and dv as x^3

so that:

x^3 . e^(1-x^2) = 1/4. x^4 . e^(1-x^2) + {intregral sign} ..........


I'm getting stuck here, am I on the right path?

 

[soroban]

Hello, Bobby Jones!

How should I carry on from your edited line, which says "worse" underneath?

You don't . . .

Since it's worse than the original integral, we abandon that approch.

Recall that the object of Integration By Parts is to create a simpler integral,
. . one which can be integrated.

 

[pka]

How do you solve the question below. Is it by parts?
{ Intergration sign, with limits +2 at top and -1 at bottom} xe^(1-x^2) dx

Why would anyone even think 'parts' here?

If \(\displaystyle y'=xe^{1-x^2}\) then \(\displaystyle y=\frac{-1}{2}e^{1-x^2}.\)

 

[Bobby Jones]

Pka,

Is that the answer?

I cant see how you've done that. Can you break it done into steps for me please so I can see and understand what you've done, otherwise I'm just getting more stuck.

Thanks

 

[galactus]

Pka,

Is that the answer?

I cant see how you've done that. Can you break it done into steps for me please so I can see and understand what you've done, otherwise I'm just getting more stuck.

Thanks

Did you even look at my post at the beginning of the thread?.

 

[pka]

Pka, Is that the answer?
I cant see how you've done that. Can you break it done into steps for me please so I can see and understand what you've done, otherwise I'm just getting more stuck.

Can you write the derivative of \(\displaystyle \frac{-e^{1-x^2}}{2}~?\).

If not, that may point to your misunderstanding.

 

[Bobby Jones]

Is it

xe^(1-x^2)

 

[Bobby Jones]

Galatus

Going back to your post,if u=1-x^2, I see how you've differenceiated it but why have you left an x on the other side with the dx, instead of under the du with the -1/2.

what happens to the first x in the original equation next to the e^1-x^2, do you not have to seperate them out and intergrate seperately?

e.g.

{ Intergration sign, with limits +2 at top and -1 at bottom} xe^(1-x^2) dx

{ Intergration sign} x . { Intergration sign} e^(1-x^2) dx

= [ (1/2)x^2 . -e^(1-x^2)] +2 at top and -1 at bottom

Then just enter in the values into x.

Am i on the right track or not?

 

[Deleted member 4993]

Galatus

Going back to your post,if u=1-x^2, I see how you've differenceiated it but why have you left an x on the other side with the dx, instead of under the du with the -1/2.

what happens to the first x in the original equation next to the e^1-x^2, do you not have to seperate them out and intergrate seperately?

e.g.

{ Intergration sign, with limits +2 at top and -1 at bottom} xe^(1-x^2) dx

{ Intergration sign} x . { Intergration sign} e^(1-x^2) dx

= [ (1/2)x^2 . -e^(1-x^2)] +2 at top and -1 at bottom

Then just enter in the values into x.

Am i on the right track or not?

Are you in a regular math-class or taking an online course or trying to teach yourself?

For calculus - in my opinion - the last two roads are full of pot-holes and dangerous.