 My Math Forum Integral: Fundamental theorem of calculus

 Calculus Calculus Math Forum

 January 17th, 2016, 12:27 AM #1 Newbie   Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0 Integral: Fundamental theorem of calculus Hi, I have to solve this integral with the Fundamental theorem of calculus: $\displaystyle f(x)=\int_{\frac{2}{\sqrt{x}}}^{\frac{x^2}{16}} \frac{e^{-u}}{u}du$, where $\displaystyle x_{0}=4$ With the help of a similar example, I have done as following, but I am not sure whether I have done right: $\displaystyle \int_{\frac{2}{\sqrt{x}}}^{C}\left(\frac{e^{-u}}{u}\right)du + \int_{C}^{\frac{x^2}{16}}\left(\frac{e^{-u}}{u}\right)du = -\int_{C}^{\frac{2}{\sqrt{x}}}\left(\frac{e^{-u}}{u}\right)du+\int_{C}^{\frac{x^2}{16}} \left(\frac{e^{-u}}{u}\right)du$ $\displaystyle F'(x)=-\frac{e^{\frac{-2}{\sqrt{x}}}}{\frac{2}{\sqrt{x}}}\left ( \frac{-1}{x^{\frac{3}{2}}} \right )+\frac{e^{-\frac{x^{2}}{16}}}{\frac{x^{2}}{16}}\left ( \frac{x}{8} \right )=\frac{e^{\frac{-2}{\sqrt{x}}}+4e^{-\frac{x^{2}}{16}}}{2x}$ Since: $\displaystyle x_{0}=4$ Answer: $\displaystyle \frac{e^{-1}+4e^{-1}}{8}$ Well, I have skipped some steps here when I calculated $\displaystyle F'(x)$ because of too many codings. But is my answer right? If not, can I get some hints? Thank you so much. Last edited by skipjack; January 17th, 2016 at 06:01 AM. January 17th, 2016, 01:30 AM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Your question seems to be unrelated to the solution. Is the following question correct? If $\displaystyle f(x) = \int^{x^2/16}_{2/\sqrt{x}}\dfrac{e^{-u}}{u}\,du$, find $f'(4)$. If that's the correct question, then your working is spot on. January 17th, 2016, 01:35 AM #3 Newbie   Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0 Oh sorry, this is the question: Find the Taylor polynomial of order 1 for the function $\displaystyle f$ about the point $\displaystyle x_{0}$. January 17th, 2016, 05:18 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 $P(x)=f(4)+f'(4) \cdot (x-4) = 0 + \dfrac{5}{8e}(x-4)$ Thanks from Johanovegas January 17th, 2016, 06:03 AM #5 Newbie   Joined: Jan 2016 From: sweden Posts: 11 Thanks: 0 Yes! Now i understand it. Thank you. Tags calculus, fundamental, integral, 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 Jhenrique Calculus 6 April 12th, 2014 09:34 PM layd33foxx Calculus 3 December 12th, 2011 08:32 PM Aurica Calculus 1 June 14th, 2009 09:04 AM Aurica Calculus 1 June 10th, 2009 06:39 PM mrguitar Calculus 3 December 9th, 2007 02:22 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top       