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 November 29th, 2015, 06:58 PM #1 Newbie   Joined: Nov 2015 From: Philippines Posts: 1 Thanks: 0 Application of First Fundamental Theorem of Calculus Find the second derivative of f($x$)=$\displaystyle \int_{-\infty}^{x^2/2} e^{-(x^2+1)t^2} dt$ My solution: f($x$)=$\displaystyle \int_{-\infty}^0 e^{-(x^2+1)t^2}dt + \int_0^{x^2/2} e^{-(x^2+1)t^2}dt$ I may be able to find a closed form for the first integral $\displaystyle \int_{-\infty}^0 e^{-(x^2+1)t^2}dt$ by transforming into polar coordinates. My concern is the second integral: h($x$) = $\displaystyle \int_0^{x^2/2} e^{-(x^2+1)t^2}dt$ How do I apply Chain rule to get h'($x$)? Last edited by skipjack; November 29th, 2015 at 11:59 PM. Reason: to use latex November 30th, 2015, 12:05 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 I doubt that the final answer can be given in closed form. I suggest that you make a substitution so that the original integral can be written in terms of the error function, erf(x). November 30th, 2015, 02:00 PM #3 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 There is no point in splitting the integral. If you apply the basic definition of the derivative, $\displaystyle f'(x)=\lim_{h ->0}\frac{f(x+h)-f(x)}{h}$, you will be able to work it out by brute force. For your problem let $\displaystyle f(x)=\int_{-\infty}^{g(x)}p(x,t)dt$ $\displaystyle f'(x)=p(x,g(x))g'(x)+\int_{-\infty}^{g(x)}\frac{\partial p(x,t)}{\partial x}dt$ I'll let you work out the second derivative. Tags application, calculus, fundamental, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mr Davis 97 Calculus 1 April 16th, 2015 04:06 PM MadSoulz Real Analysis 2 April 15th, 2014 03:19 PM layd33foxx Calculus 3 December 12th, 2011 07:32 PM riotsandravess Calculus 3 November 25th, 2010 12:44 PM mrguitar Calculus 3 December 9th, 2007 01:22 PM

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