Lösung 2.1:3c

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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If we multiply the factors in the integrand together and use the power laws,
If we multiply the factors in the integrand together and use the power laws,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\int e^{2x}\bigl(e^x+1\bigr)\,dx
\int e^{2x}\bigl(e^x+1\bigr)\,dx
&= \int\bigl(e^{2x}e^{x} + e^{2x}\bigr)\,dx\\[5pt]
&= \int\bigl(e^{2x}e^{x} + e^{2x}\bigr)\,dx\\[5pt]
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<math>a</math> is a constant. The indefinite integral is therefore
<math>a</math> is a constant. The indefinite integral is therefore
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{{Displayed math||<math>\int \bigl(e^{3x}+e^{2x}\bigr)\,dx = \frac{e^{3x}}{3} + \frac{e^{2x}}{2} + C\,,</math>}}
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{{Abgesetzte Formel||<math>\int \bigl(e^{3x}+e^{2x}\bigr)\,dx = \frac{e^{3x}}{3} + \frac{e^{2x}}{2} + C\,,</math>}}
where <math>C</math> is an arbitrary constant.
where <math>C</math> is an arbitrary constant.

Version vom 12:59, 10. Mär. 2009

If we multiply the factors in the integrand together and use the power laws,

\displaystyle \begin{align}

\int e^{2x}\bigl(e^x+1\bigr)\,dx &= \int\bigl(e^{2x}e^{x} + e^{2x}\bigr)\,dx\\[5pt] &= \int\bigl(e^{2x+x} + e^{2x}\bigr)\,dx\\[5pt] &= \int{\bigl(e^{3x} + e^{2x}\bigr)}\,dx\,, \end{align}

we obtain a standard integral with two terms of the type \displaystyle e^{ax}, where \displaystyle a is a constant. The indefinite integral is therefore

\displaystyle \int \bigl(e^{3x}+e^{2x}\bigr)\,dx = \frac{e^{3x}}{3} + \frac{e^{2x}}{2} + C\,,

where \displaystyle C is an arbitrary constant.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:3c“