Lösung 2.2:1c

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With the given variable substitution,
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With the given variable substitution, <math>u=x^3</math>, we obtain
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<math>u=x^{3}</math>
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we obtain
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{{Displayed math||<math>du = \bigl(x^3\bigr)'\,dx = 3x^2\,dx</math>}}
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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 <math>x^2</math> as a factor, we can bundle it together with <math>dx</math> and replace the combination with <math>\tfrac{1}{3}\,du</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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{{Displayed math||<math>\int e^{x^3}x^2\,dx = \bigl\{\,u=x^3\,\bigr\} = \int e^u\tfrac{1}{3}\,du = \frac{1}{3}e^u + C\,\textrm{.}</math>}}
Thus, the answer is
Thus, the answer is
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{{Displayed math||<math>\int e^{x^3}x^2\,dx = \frac{1}{3}e^{x^3} + C\,,</math>}}
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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 <math>C</math> is an arbitrary constant.
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where
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<math>C</math>
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is an arbitrary constant.
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Version vom 12:48, 28. Okt. 2008

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

\displaystyle du = \bigl(x^3\bigr)'\,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 \tfrac{1}{3}\,du,

\displaystyle \int e^{x^3}x^2\,dx = \bigl\{\,u=x^3\,\bigr\} = \int e^u\tfrac{1}{3}\,du = \frac{1}{3}e^u + C\,\textrm{.}

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“