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) ??

Answer:

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}

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