Lösung 2.2:1a
Aus Online Mathematik Brückenkurs 2
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The relation between <math>dx</math> and <math>du</math> reads | The relation between <math>dx</math> and <math>du</math> reads | ||
| - | {{ | + | {{Abgesetzte Formel||<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}} |
which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>. | which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>. | ||
| Zeile 20: | Zeile 20: | ||
One usually writes the whole substitution of variables as | One usually writes the whole substitution of variables as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align} |
u &= 3x-1\\[5pt] | u &= 3x-1\\[5pt] | ||
du &= 3\,dx | du &= 3\,dx | ||
| Zeile 27: | Zeile 27: | ||
Sometimes, we are more brief and hide the details, | Sometimes, we are more brief and hide the details, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}} |
After the substitution of variables, we have a standard integral which is easy to compute. | After the substitution of variables, we have a standard integral which is easy to compute. | ||
| Zeile 33: | Zeile 33: | ||
In summary, the whole calculation is, | In summary, the whole calculation is, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
\int\limits_1^2 \frac{dx}{(3x-1)^4} | \int\limits_1^2 \frac{dx}{(3x-1)^4} | ||
&= \left\{\begin{align} | &= \left\{\begin{align} | ||
Version vom 13:00, 10. Mär. 2009
A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.
When we carry out a substitution of variables \displaystyle u=u(x), there are three things which are affected in the integral:
- the integral must be rewritten in terms of the new variable \displaystyle u;
- the element of integration, \displaystyle dx, is replaced by \displaystyle du, according to the formula \displaystyle du=u'(x)\,dx;
- the limits of integration are for \displaystyle x and must be changed to limits of integration for the variable \displaystyle u.
In this case, we will perform the change of variables \displaystyle u=3x-1, mainly because the integrand \displaystyle 1/(3x-1)^4 will then be replaced by \displaystyle 1/u^4.
The relation between \displaystyle dx and \displaystyle du reads
| \displaystyle du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,, |
which means that \displaystyle dx is replaced by \displaystyle \tfrac{1}{3}\,du.
Furthermore, when \displaystyle x=1 in the lower limit of integration, the corresponding u-value becomes \displaystyle u=3\cdot 1-1=2, and when \displaystyle x=2, we obtain the u-value \displaystyle u=3\cdot 2-1=5\,.
One usually writes the whole substitution of variables as
| \displaystyle \int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}
u &= 3x-1\\[5pt] du &= 3\,dx \end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.} |
Sometimes, we are more brief and hide the details,
| \displaystyle \int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.} |
After the substitution of variables, we have a standard integral which is easy to compute.
In summary, the whole calculation is,
| \displaystyle \begin{align}
\int\limits_1^2 \frac{dx}{(3x-1)^4} &= \left\{\begin{align} u &= 3x-1\\[5pt] du &= 3\,dx \end{align}\right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4} = \frac{1}{3}\int\limits_2^5 u^{-4}\,du\\[5pt] &= \frac{1}{3}\Bigl[\ \frac{u^{-4+1}}{-4+1}\ \Bigr]_2^5 = -\frac{1}{9}\Bigl[\ \frac{1}{u^3}\ \Bigr]_2^5 = -\frac{1}{9}\Bigl(\frac{1}{5^3} - \frac{1}{2^3} \Bigr) = -\frac{1}{9}\cdot\frac{2^3-5^3}{2^3\cdot 5^3}\\[5pt] &= \frac{117}{3^2\cdot 2^3\cdot 5^3} = \frac{3^2\cdot 13}{3^2\cdot 2^3\cdot 5^3} = \frac{13}{2^3\cdot 5^3} = \frac{13}{1000}\,\textrm{.} \end{align} |
