Interval of the definite integral

[Nan1teZ]

Homework Statement

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?

 

[tiny-tim]

Homework Statement

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?

Hi Nan1teZ!

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

Then just use the product rule.

 

[Nan1teZ]

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

Is that right?

 

[tiny-tim]

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

Is that right?

I don't think so …

can you show your detailed working?

 

[Nan1teZ]

oops just a stupid mistake!..

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

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

 

[tiny-tim]

oops just a stupid mistake!..

okay I got F''(x) = (2*e^(x²))(1+x²).

Woohoo! ​

 

[Nan1teZ]

Thanks! :D