Integrating Difficult Integrals: (1-x^2)(f ''(x)) & (a^2-x^2)(f ''(x))

In summary, Chris said that if he couldn't get the integral to work, perhaps he should try using the IBP formula.
  • #1
Falcon
20
0
Here's a pretty difficult integral that our prof threw at us a couple days ago... left the whole class a little dumbfounded. I wish I had symbols to use (such as the ones that a lot of the pros on this site use)... but I'm sure you'll be able to understand.

Integrate (from a to b) the following (1-x^2)(f ''(x)) dx



He went on to say that if we couldn't get that, perhaps we should look at this one:


Integrate (from -a to a) the following (a^2-x^2)(f ''(x))
(he suggested IBP)


I think what threw most people off was how he knew what f''(x) was without identifying it directly.

Thanks for everyones help in advance!

-Chris
 
Physics news on Phys.org
  • #2
I can't say I have ever done anything like that. However would it not be fairly easy to say:

[tex]\int_b^a (1-x^2)f''(x) dx = \int_b^a f''(x) dx - \int_b^a (x^2)f''(x)dx[/tex]

[tex]\left[ f'(x) \right]_b^a - \int_b^a (x^2)f''(x)dx [/tex]

And apply by parts on the remaining integral. That's just a guess :confused:
 
  • #3
There's no reason to separate the (1- x2).

Let u= 1-x2 and dv= f"(x)dx
Then du= -2xdx and v= f '(x)

so the integral becomes [itex](1-x^2)f'(x)\|_a^b+ 2\int_a^bxf'(x)dx[/itex]

To do that second integral let u= x, dv= f'(x)dx so
du= dx and v= f(x). The second integral becomes [itex]xf(x)\|_a^b- 2\int_a^bf(x)dx[/itex].

The entire integral is [itex](1-b^2)f'(b)-(1-a^2)f'(a)+ 2bf(b)-2af(a)- 2\int_a^bf(x)dx[/itex]

The "advantage" of using a2-x2 instead of 1- x2 is that the "(1-a2) term is 0 and we have only [itex](a^2-b^2)f'(b)+ 2bf(b)-2af(a)- 2\int_a^bf(x)dx[/itex].

Of course, we can't evaluate that last integral precisely because we don't know f itself.
 
Last edited by a moderator:
  • #4
ahhh.. thank you both very much for your help!

[bows to HallsofIvy]

very much appreciated :)
 

1. How do I find the antiderivative of (1-x^2)(f ''(x))?

To find the antiderivative of (1-x^2)(f ''(x)), you can use the substitution method or integration by parts. First, substitute u = 1-x^2, then use the power rule to find the antiderivative. Alternatively, you can use integration by parts by letting u = (1-x^2) and dv = f ''(x) dx.

2. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and represents the area under a curve within those limits. An indefinite integral does not have specific limits and represents the antiderivative of a function.

3. Can I use integration by parts for (a^2-x^2)(f ''(x))?

Yes, you can use integration by parts for (a^2-x^2)(f ''(x)). Let u = a^2-x^2 and dv = f ''(x) dx.

4. Is there a general formula for integrating difficult integrals?

Unfortunately, there is no general formula for integrating difficult integrals. It often requires a combination of techniques such as substitution, integration by parts, trigonometric substitutions, and partial fractions.

5. Can I use a calculator to integrate (1-x^2)(f ''(x)) & (a^2-x^2)(f ''(x))?

While some calculators have integration capabilities, they may not be able to solve more complex integrals like (1-x^2)(f ''(x)) & (a^2-x^2)(f ''(x)). It is best to use hand methods or specialized software for these types of integrals.

Similar threads

Replies
4
Views
1K
Replies
16
Views
2K
Replies
8
Views
1K
Replies
3
Views
1K
Replies
1
Views
2K
Replies
1
Views
926
Replies
4
Views
639
Replies
8
Views
1K
Replies
2
Views
2K
Back
Top