Integrating (1-x²)f''(x) and (a²-x²)f''(x) without knowing f''(x)

[Falcon]

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

 

[Zurtex]

I can't say I have ever done anything like that. However would it not be fairly easy to say:

$$\int_b^a (1-x^2)f''(x) dx = \int_b^a f''(x) dx - \int_b^a (x^2)f''(x)dx$$

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

And apply by parts on the remaining integral. That's just a guess

 

[HallsofIvy]

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 $(1-x^2)f'(x)\|_a^b+ 2\int_a^bxf'(x)dx$

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

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

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

Of course, we can't evaluate that last integral precisely because we don't know f itself.

 

[Falcon]

ahhh.. thank you both very much for your help!

[bows to HallsofIvy]

very much appreciated :)