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Lösung 2.1:2d

Aus Online Mathematik Brückenkurs 2

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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,
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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{{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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{{Abgesetzte Formel||<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>}}
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,
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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{{Abgesetzte Formel||<math>\begin{align}
\int\limits_1^4 x^{-3/2}\,dx
\int\limits_1^4 x^{-3/2}\,dx
&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]
&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]

Version vom 12:58, 10. Mär. 2009

If we rewrite x  as x12 , the integrand can then be simplified using the power laws,

41x2xdx=41x2x12dx=41x12−2dx=41x−32dx. 

We can now use the fact that a primitive function for xn is xn+1(n+1) and calculate the integral's value,

41x−32dx= x−32+1−32+1 41= −12x−12 41= −21x12 41= −2x 41=−24−−21=−22+2=1.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:2d“