D/dx[int(0,x) e^(-t^2) dt] : two methods, two answers

[nomadreid]

There are two methods to take d/dx [ ∫_t=0^x exp(-t^2) dt].

First method: using the relationship of integral and antiderivative, one gets
(exp(-t^2) , from t = 0 to x, so exp(-x^2) - 1.

Second method: the integral is (1/2)sqrt(pi)*erf(t) from 0 to x, which is (1/2)sqrt(pi)*erf(x), and the derivative of this is exp(-x^2).

So, which answer is correct, and what is wrong with the other method?

(Wolfram alpha favors the second answer, but I have another source that favors the first answer.)

 

[jgens]

First method: using the relationship of integral and antiderivative, one gets (exp(-t^2) , from t = 0 to x, so exp(-x^2) - 1.

The second answer is right. Can you show how you got this so we can help point out your mistake?

 

[nomadreid]

The second answer is right. Can you show how you got this so we can help point out your mistake?

Thank you, gladly. I am using the idea that d(∫_a^bf(t)dt)/dx = f(x)|_a^b; here a=0 , b = x, and f(t)= exp(-t^2).

 

[jgens]

Thank you, gladly. I am using the idea that d(∫_a^bf(t)dt)/dx = f(x)|_a^b; here a=0 , b = x, and f(t)= exp(-t^2).

The formula you are using is not correct. The correct formula is given here: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#First_part

 

[nomadreid]

Ah, I see my mistake. Thank you very much, gjens.