Interval of the definite integral

Nan1teZ
Messages
11
Reaction score
0

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?
 
Last edited:
Physics news on Phys.org
Nan1teZ said:

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! :smile:

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

Then just use the product rule. :smile:
 


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

Is that right?
 
Nan1teZ said:
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? :smile:
 


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..
 
Nan1teZ said:
oops just a stupid mistake!..

okay I got F''(x) = (2*e^(x2))(1+x2).

:biggrin: Woohoo! :biggrin:
 


Thanks! :D
 
Back
Top