Lösung 2.2:1a
Aus Online Mathematik Brückenkurs 2
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
K (Solution 2.2:1a moved to Lösung 2.2:1a: Robot: moved page) |
Version vom 10:20, 11. 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
- the integral must be rewritten in terms of the new variable
u ; - the element of integration,
dx , is replaced bydu , according to the formuladu=u ;(x)dx
- the limits of integration are for
x and must be changed to limits of integration for the variableu .
In this case, we will perform the change of variables (3x−1)4
u4
The relation between
![]() ![]() ![]() |
which means that
Furthermore, when
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} |