Lösung 2.1:2d

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{{NAVCONTENT_START}}
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If we rewrite
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<center> [[Image:2_1_2d.gif]] </center>
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<math>\sqrt{x}</math>
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{{NAVCONTENT_STOP}}
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as
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<math>x^{{1}/{2}\;}</math>, the integrand can then be simplified using the power laws:
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<math>\int\limits_{1}^{4}{\frac{\sqrt{x}}{x^{2}}}\,dx=\int\limits_{1}^{4}{\frac{x^{\frac{1}{2}}}{x^{2}}}\,dx=\int\limits_{1}^{4}{x^{\frac{1}{2}-2}}\,dx=\int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx</math>
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We can now use the fact that a primitive function for
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<math>x^{n}</math>
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is
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<math>\frac{x^{n+1}}{n+1}</math>
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and calculate the integral's value:
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<math>\begin{align}
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& \int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx=\left[ \frac{x^{-\frac{3}{2}+1}}{^{-\frac{3}{2}+1}} \right]_{1}^{4} \\
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& =\left[ \frac{x^{-\frac{1}{2}}}{^{-\frac{1}{2}}} \right]_{1}^{4}=\left[ -2\frac{1}{x^{{1}/{2}\;}} \right]_{1}^{4} \\
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& =\left[ -\frac{2}{\sqrt{x}} \right]_{1}^{4}=-\frac{2}{\sqrt{4}}-\left( -\frac{2}{\sqrt{1}} \right) \\
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& =-\frac{2}{2}+2=1 \\
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\end{align}</math>

Version vom 13:27, 17. Okt. 2008

If we rewrite \displaystyle \sqrt{x} as \displaystyle x^{{1}/{2}\;}, the integrand can then be simplified using the power laws:


\displaystyle \int\limits_{1}^{4}{\frac{\sqrt{x}}{x^{2}}}\,dx=\int\limits_{1}^{4}{\frac{x^{\frac{1}{2}}}{x^{2}}}\,dx=\int\limits_{1}^{4}{x^{\frac{1}{2}-2}}\,dx=\int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx


We can now use the fact that a primitive function for \displaystyle x^{n} is \displaystyle \frac{x^{n+1}}{n+1} and calculate the integral's value:


\displaystyle \begin{align} & \int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx=\left[ \frac{x^{-\frac{3}{2}+1}}{^{-\frac{3}{2}+1}} \right]_{1}^{4} \\ & =\left[ \frac{x^{-\frac{1}{2}}}{^{-\frac{1}{2}}} \right]_{1}^{4}=\left[ -2\frac{1}{x^{{1}/{2}\;}} \right]_{1}^{4} \\ & =\left[ -\frac{2}{\sqrt{x}} \right]_{1}^{4}=-\frac{2}{\sqrt{4}}-\left( -\frac{2}{\sqrt{1}} \right) \\ & =-\frac{2}{2}+2=1 \\ \end{align}

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