Lösung 2.2:1a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.2:1a moved to Solution 2.2:1a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
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.
-
<center> [[Image:2_2_1a-1(3).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
When we carry out a substitution of variables
-
{{NAVCONTENT_START}}
+
<math>u=u\left( x \right)</math>, there are three things which are affected in the integral:
-
<center> [[Image:2_2_1a-2(3).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
1. the integral must be rewritten in terms of the new variable
-
{{NAVCONTENT_START}}
+
<math>u</math>;
-
<center> [[Image:2_2_1a-3(3).gif]] </center>
+
2. the element of integration,
-
{{NAVCONTENT_STOP}}
+
<math>dx</math>, is replaced by
 +
<math>du</math>, according to the formula
 +
<math>du={u}'\left( x \right)dx</math>;
 +
3. the limits of integration are for
 +
<math>x\text{ }</math>
 +
and must be changed to limits of integration for the variable
 +
<math>u</math>.
 +
 
 +
In this case, we will perform the change of variables
 +
<math>u=3x-1</math>, mainly because the integrand
 +
<math>\frac{1}{\left( 3x-1 \right)^{4}}</math>
 +
will then be replaced by
 +
<math>\frac{1}{u^{4}}</math>.
 +
 
 +
The relation between
 +
<math>dx</math>
 +
and
 +
<math>du</math>
 +
reads
 +
 
 +
 
 +
<math>du={u}'\left( x \right)\,dx=\left( 3x-1 \right)^{\prime }\,dx=3\,dx</math>,
 +
 
 +
which means that
 +
<math>dx</math>
 +
is replaced by
 +
<math>\frac{1}{3}\,du</math>.
 +
 
 +
Furthermore, when
 +
<math>x=\text{1}</math>
 +
in the lower limit of integration, the corresponding
 +
<math>u</math>
 +
-value becomes
 +
<math>u=3\centerdot 1-1=2</math>, and when
 +
<math>x=2</math>, we obtain the u-value
 +
<math>u=3\centerdot 2-1=5</math>
 +
 
 +
 
 +
One usually writes the whole substitution of variables as
 +
 
 +
 
 +
<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}
 +
u=3x-1 \\
 +
du=3\,dx \\
 +
\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}</math>
 +
 
 +
 
 +
Sometimes, we are more brief and hide the details:
 +
 
 +
 
 +
<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ u=3x-1 \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}</math>
 +
 
 +
 
 +
After the substitution of variables, we have a standard integral which is easy to compute.
 +
 
 +
In summary, the whole calculation is:
 +
 
 +
 
 +
<math>\begin{align}
 +
& \int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}
 +
u=3x-1 \\
 +
du=3\,dx \\
 +
\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}} \\
 +
& =\frac{1}{3}\int\limits_{2}^{5}{\,u^{-4}du}=\left[ \frac{u^{-4+1}}{-4+1} \right]_{2}^{5} \\
 +
& =-\frac{1}{9}\left[ \frac{1}{u^{3}} \right]_{2}^{5}=-\frac{1}{9}\left( \frac{1}{5^{3}}-\frac{1}{2^{3}} \right) \\
 +
& =-\frac{1}{9}\centerdot \frac{2^{3}-5^{3}}{2^{3}\centerdot 5^{3}}=\frac{117}{3^{2}\centerdot 2^{3}\centerdot 5^{3}} \\
 +
& =\frac{3^{2}\centerdot 13}{3^{2}\centerdot 2^{3}\centerdot 5^{3}}=\frac{13}{2^{3}\centerdot 5^{3}}=\frac{13}{1000} \\
 +
\end{align}</math>.

Version vom 09:21, 19. Okt. 2008

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\left( x \right), there are three things which are affected in the integral:

1. the integral must be rewritten in terms of the new variable \displaystyle u; 2. the element of integration, \displaystyle dx, is replaced by \displaystyle du, according to the formula \displaystyle du={u}'\left( x \right)dx; 3. the limits of integration are for \displaystyle x\text{ } 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 \frac{1}{\left( 3x-1 \right)^{4}} will then be replaced by \displaystyle \frac{1}{u^{4}}.

The relation between \displaystyle dx and \displaystyle du reads


\displaystyle du={u}'\left( x \right)\,dx=\left( 3x-1 \right)^{\prime }\,dx=3\,dx,

which means that \displaystyle dx is replaced by \displaystyle \frac{1}{3}\,du.

Furthermore, when \displaystyle x=\text{1} in the lower limit of integration, the corresponding \displaystyle u -value becomes \displaystyle u=3\centerdot 1-1=2, and when \displaystyle x=2, we obtain the u-value \displaystyle u=3\centerdot 2-1=5


One usually writes the whole substitution of variables as


\displaystyle \int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix} u=3x-1 \\ du=3\,dx \\ \end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}


Sometimes, we are more brief and hide the details:


\displaystyle \int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ u=3x-1 \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}


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}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix} u=3x-1 \\ du=3\,dx \\ \end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}} \\ & =\frac{1}{3}\int\limits_{2}^{5}{\,u^{-4}du}=\left[ \frac{u^{-4+1}}{-4+1} \right]_{2}^{5} \\ & =-\frac{1}{9}\left[ \frac{1}{u^{3}} \right]_{2}^{5}=-\frac{1}{9}\left( \frac{1}{5^{3}}-\frac{1}{2^{3}} \right) \\ & =-\frac{1}{9}\centerdot \frac{2^{3}-5^{3}}{2^{3}\centerdot 5^{3}}=\frac{117}{3^{2}\centerdot 2^{3}\centerdot 5^{3}} \\ & =\frac{3^{2}\centerdot 13}{3^{2}\centerdot 2^{3}\centerdot 5^{3}}=\frac{13}{2^{3}\centerdot 5^{3}}=\frac{13}{1000} \\ \end{align}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:1a“