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 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: 2,820 Thanks: 1464 $P(x)=f(4)+f'(4) \cdot (x-4) = 0 + \dfrac{5}{8e}(x-4)$ 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.

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