# Integration should blow up ?

1. Dec 7, 2008

### touqra

Using Maple 8, I integrate this:

$$\int_0^1\frac{1}{\sqrt(p(1-p))} dp$$

and I get $$\pi$$
but, this function goes to infinity at p = 0 and 1.
How can this be possible ?

2. Dec 7, 2008

### Hurkyl

Staff Emeritus
Why wouldn't it be?

3. Dec 7, 2008

### touqra

The denominator has sqrt[p*(1-p)], which will give infinity on both p = 0 and 1. And so, the area under this curve is infinity too, right ?

4. Dec 7, 2008

### Hurkyl

Staff Emeritus
Can you think of a reason why that should be true, though? If you're convinced the area should be infinite, then try to work out a proof of it -- such exercises are often really good at clearing up misunderstandings.

Incidentally, it might help to play with simpler functions, e.g.

$$\int_0^1 \frac{1}{\sqrt{x}} \, dx$$

or

$$\int_1^\infty \frac{1}{y^2} \, dy$$