# Interval of the definite integral

1. Jul 20, 2008

### Nan1teZ

1. The problem statement, all variables and given/known data
Let F(x) = $$\int$$$$^{x}_____________{0}$$ x*e^(t^2) dt for $$x\in[0,1].$$ Find F''(x) for $$x\in(0,1).$$

My only problem is the x, because the interval of the definite integral goes from 0 to x, and x is in the integral, even though the integral is with respect to dt. So I'd just like to know what happens in this case? Is x a constant (as if the interval went from a to b, rather than 0 to x)? Or something different?

Last edited: Jul 20, 2008
2. Jul 20, 2008

### tiny-tim

Hi Nan1teZ!

You can rewrite the integral as F(x) = x*$$\int^{x}_{0}$$ e^(t^2) dt.

Then just use the product rule.

3. Jul 20, 2008

### Nan1teZ

Re: Integral

Okay so i get F''(x) = (2*e^(x^2))(1+x) + x

Is that right?

4. Jul 20, 2008

### tiny-tim

I don't think so …

can you show your detailed working?

5. Jul 20, 2008

### Nan1teZ

Re: Integral

oops just a stupid mistake!!..

okay I got F''(x) = (2*e^(x^2))(1+x^2).

If that's wrong I'm gonna show the detailed working..

6. Jul 20, 2008

### tiny-tim

Woohoo!

7. Jul 20, 2008

Re: Integral

Thanks! :D
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