Lösung 2.1:2d

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If we rewrite
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If we rewrite <math>\sqrt{x}</math> as <math>x^{1/2}</math>, the integrand can then be simplified using the power laws,
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<math>\sqrt{x}</math>
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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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{{Displayed math||<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}
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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 <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> and calculate the integral's value,
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{{Displayed math||<math>\begin{align}
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We can now use the fact that a primitive function for
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\int\limits_1^4 x^{-3/2}\,dx
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<math>x^{n}</math>
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&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]
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is
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&= \Bigl[\ \frac{x^{-1/2}}{-1/2}\ \Bigr]_1^4\\[5pt]
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<math>\frac{x^{n+1}}{n+1}</math>
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&= \Bigl[\ -2\frac{1}{x^{1/2}}\ \Bigr]_1^4\\[5pt]
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and calculate the integral's value:
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&= \Bigl[\ -\frac{2}{\sqrt{x}}\ \Bigr]_1^4\\[5pt]
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&= -\frac{2}{\sqrt{4}} - \Bigl(-\frac{2}{\sqrt{1}}\Bigr)\\[5pt]
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&= -\frac{2}{2}+2\\[5pt]
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<math>\begin{align}
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&= 1\,\textrm{.}
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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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\end{align}</math>}}
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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>
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Version vom 13:09, 21. 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^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}

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

\displaystyle \begin{align}

\int\limits_1^4 x^{-3/2}\,dx &= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt] &= \Bigl[\ \frac{x^{-1/2}}{-1/2}\ \Bigr]_1^4\\[5pt] &= \Bigl[\ -2\frac{1}{x^{1/2}}\ \Bigr]_1^4\\[5pt] &= \Bigl[\ -\frac{2}{\sqrt{x}}\ \Bigr]_1^4\\[5pt] &= -\frac{2}{\sqrt{4}} - \Bigl(-\frac{2}{\sqrt{1}}\Bigr)\\[5pt] &= -\frac{2}{2}+2\\[5pt] &= 1\,\textrm{.} \end{align}

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