Solving Integral Question: f(x) = \int^{x+1}_{x} e^{-t^2} dt

[daniel_i_l]

Homework Statement

$$f(x) = \int^{x+1}_{x} e^{-t^2} dt$$
1) Where does f have a derivative?
2) Prove that f(x)>0 for all x in R.
3) Find the segments of R where f goes up and down and the extream points.

Homework Equations

The Attempt at a Solution

1) The integral of a continues function has a derivative everywhere.

2) $$e^{-t^2}$$ has a miniumum (m) in [x,x+1] and so
$$m(x+1 -x) = m <= \int^{x+1}_{x} e^{-t^2} dt$$
But e^x > 0 for all x and so m>0.

3)If $$F(x) = \int^{x}_{0} e^{-t^2} dt$$
Then f(x) = -F(x) + F(x+1) and so
$$f'(x) =-e^{-x^2} + e^{-(x+1)^2} = e^{-x^2} ( e^{-(2x+1)} -1)$$
And so f'(x)=0 only when x= -1/2. For all x>-1/2 the f'(x) < 0 and the function goes down, and for all x<-1/2 f'(x)>0 and the function goes up.

Is that right?
Thanks.

 

[Dick]

I don't see any problems.