Lösung 2.2:1c
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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With the given variable substitution, <math>u=x^3</math>, we obtain | With the given variable substitution, <math>u=x^3</math>, we obtain | ||
| - | {{ | + | {{Abgesetzte Formel||<math>du = \bigl(x^3\bigr)'\,dx = 3x^2\,dx</math>}} |
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>, | 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>, | ||
| - | {{ | + | {{Abgesetzte Formel||<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 | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int e^{x^3}x^2\,dx = \frac{1}{3}e^{x^3} + C\,,</math>}} |
where <math>C</math> is an arbitrary constant. | where <math>C</math> is an arbitrary constant. | ||
Version vom 13:00, 10. Mär. 2009
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.
