# Thread: Express F(x) as a definite integral

1. ## Express F(x) as a definite integral

Hi, my question is:

The function F(x) satisfies F'(x)=e^(-x^2) and F(0)=0. Express F(x) as a definite integral.

I'm really stuck on this question, so any help and hints would be greatly appreciated!

2. ## Re: Express F(x) as a definite integral

Originally Posted by sakuraxkisu
The function F(x) satisfies F'(x)=e^(-x^2) and F(0)=0. Express F(x) as a definite integral.
$\displaystyle F(x) = \int_0^x {{e^{ - {t^2}}}dt}~?~?$

3. ## Re: Express F(x) as a definite integral

What confused me about this question is that, when I try to compute the integral, I get:

F(x)= [-e^(-t^2)/2t] upper limit x and lower limit 0

When I put in the limits, I get the definite integral being:

(-e^(x^2)/2x)-(-e^(0)/0)=(-e^(x^2)/2x)-(-1/0)

What confuses me is the second part of this definite integral, as it is divided by zero and can't be computed. For this reason I'm stuck on this question. Is there some other method, or is this the correct way to approach this question? Thank you

4. ## Re: Express F(x) as a definite integral

Originally Posted by sakuraxkisu
What confused me about this question is that, when I try to compute the integral, I get:
F(x)= [-e^(-t^2)/2t] upper limit x and lower limit 0
It is impossible to compute $\displaystyle \int {{e^{ - {t^2}}}dt}$ because there is no elementary anti-derivative.

However, the integral I gave has the properties: $\displaystyle F(0) = 0~\& ~F'(x) = {e^{ - {x^2}}}$.

5. ## Re: Express F(x) as a definite integral

So for this question, am I right in saying that I need to state what the definite integral is (ie the integral you gave), but I don't actually have to compute it? Thank you

6. ## Re: Express F(x) as a definite integral

Originally Posted by sakuraxkisu
So for this question, am I right in saying that I need to state what the definite integral is (ie the integral you gave), but I don't actually have to compute it?
You can't compute using elementary functions.
You were not told to compute it.