## calculus question

Posted: May 12, 2013 in Mathematics

Question:

If $f(x) = \displaystyle \int_x^{x+1} e^{t^2} - e^{x^2} \;dt$

What is $f'(x)$ ??

$f(x) = \displaystyle \int_x^{x+1} e^{t^2} - e^{x^2} \;dt$

$= \displaystyle\int_x^{x+1}e^{t^2} \; dt - e^{x^2}\int_x^{x+1}\;dt$

$= \displaystyle\int_x^{x+1}e^{t^2} \;dt - e^{x^2}$

By the part one of Fundamental Theorem of Calculus,

$f'(x) = e^{(x+1)^2} - e^{x^2} - 2xe^{x^2}$