Lösung 2.2:1c

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K (Lösning 2.2:1c moved to Solution 2.2:1c: Robot: moved page)
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{{NAVCONTENT_START}}
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With the given variable substitution,
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<center> [[Image:2_2_1c).gif]] </center>
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<math>u=x^{3}</math>
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{{NAVCONTENT_STOP}}
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we obtain
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<math>du=\left( x^{3} \right)^{\prime }\,dx=3x^{2}\,dx</math>
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and because the integral contains
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<math>x^{2}</math>
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as a factor, we can bundle it together with
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<math>dx</math>
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and replace the combination with
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<math>\frac{1}{3}\,du</math>,
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<math>\int{e^{x^{3}}x^{2}\,dx=\left\{ u=x^{3} \right\}}=\int{e^{u}}\frac{1}{3}\,du=\frac{1}{3}e^{u}+C</math>
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Thus, the answer is
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<math>\int{e^{x^{3}}x^{2}\,dx=}\frac{1}{3}e^{x^{3}}+C</math>
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where
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<math>C</math>
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is an arbitrary constant.

Version vom 09:55, 19. Okt. 2008

With the given variable substitution, \displaystyle u=x^{3} we obtain


\displaystyle du=\left( x^{3} \right)^{\prime }\,dx=3x^{2}\,dx


and because the integral contains \displaystyle x^{2} as a factor, we can bundle it together with \displaystyle dx and replace the combination with \displaystyle \frac{1}{3}\,du,


\displaystyle \int{e^{x^{3}}x^{2}\,dx=\left\{ u=x^{3} \right\}}=\int{e^{u}}\frac{1}{3}\,du=\frac{1}{3}e^{u}+C


Thus, the answer is


\displaystyle \int{e^{x^{3}}x^{2}\,dx=}\frac{1}{3}e^{x^{3}}+C


where \displaystyle C is an arbitrary constant.

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