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January 16th, 2016, 11:27 PM   #1
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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 05:01 AM.
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January 17th, 2016, 12:30 AM   #2
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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.
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January 17th, 2016, 12:35 AM   #3
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Oh sorry, this is the question:

Find the Taylor polynomial of order 1 for the function $\displaystyle f$ about the point $\displaystyle x_{0}$.
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January 17th, 2016, 04:18 AM   #4
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$P(x)=f(4)+f'(4) \cdot (x-4) = 0 + \dfrac{5}{8e}(x-4)$
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January 17th, 2016, 05:03 AM   #5
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Yes! Now i understand it. Thank you.
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