# Thread: Integral Problem

1. ## Integral Problem

In this problem $F(1)$ where $F'(x)=e^{-t^2}$

And $F(0) = 2$

find the value of the function so I wind up with this

$F(x) = 2 + \int e^{-t^2}$

plug in one for x but how do I anti differentiate this

$\int e^{-t^2}$

2. Hello The Power!

Originally Posted by The Power
In this problem $F(1)$ where $F'(x)=e^{-t^2}$
Huh? Is it $F'(t) = e^{-t^2}$ ? Because I think it should be

Originally Posted by The Power
And $F(0) = 2$

find the value of the function so I wind up with this

$F(x) = 2 + \int e^{-t^2}$
It would be better, if you first calculate

$F'(t) = e^{-t^2}$

=> $F(t) = \int e^{-t^2} dt + c$

and then you substitute t = 0, so you solve F(0) = 2 to solve for c.

Originally Posted by The Power
plug in one for x but how do I anti differentiate this

$\int e^{-t^2}$
Do you know integration by substitution?

Use the substitution z = t^2... That would work.

Yours
Rapha

3. Thanks Rapha!
Im currently learning how to do substitution is it quite tricky because at first I made u = -t^2 and do i not know when to solve for du = some derivative dx completely for dx if you can make any sense of that

In this case C = 2 so its

2 + integral

If this is correct

4. Originally Posted by The Power
In this problem $F(1)$ where $F'(x)=e^{-t^2}$

And $F(0) = 2$

find the value of the function so I wind up with this

$F(x) = 2 + \int e^{-t^2}$

plug in one for x but how do I anti differentiate this

$\int e^{-t^2}$
$\int{e^{-t^2}\,dt}$ does not have a solution from elementary functions.

5. So how do I obtain a value for F(1) somehow the book retrieved the answer numerically of 2.747

6. Originally Posted by The Power
So how do I obtain a value for F(1) somehow the book retrieved the answer numerically of 2.747
As said earlier, a closed form solution using elementary functions cannot be obtained. The book would have used technology to get an approximate value.

7. I wish the book would state this but sadly they do not I should just skip through those type of problems because calculators are not allowed on test