Lösung 2.1:2d
Aus Online Mathematik Brückenkurs 2
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Version vom 10:16, 11. Mär. 2009
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} |
