Lösung 2.1:2c

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If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as
If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]
&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]
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We obtain
We obtain
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]
&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]

Version vom 12:58, 10. Mär. 2009

If we recall that \displaystyle \sqrt{x} = x^{1/2}, the integral can be written as

\displaystyle \begin{align}

\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx &= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt] &= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.} \end{align}

This is a standard integral in which the integrand consists of two terms looking like \displaystyle x^n, where \displaystyle n=1/2 and \displaystyle n=-1/2\,, respectively.

We obtain

\displaystyle \begin{align}

\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx &= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt] &= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt] &= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt] &= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt] &= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt] &= 18-6-\frac{16}{3}+4\\[5pt] &= 16-\frac{16}{3}\\[5pt] &= \frac{16\cdot 3-16}{3}\\[5pt] &= \frac{32}{3}\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:2c“