Lösung 2.2:1a

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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.
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.
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When we carry out a substitution of variables
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When we carry out a substitution of variables <math>u=u(x)</math>, there are three things which are affected in the integral:
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<math>u=u\left( x \right)</math>, there are three things which are affected in the integral:
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1. the integral must be rewritten in terms of the new variable
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# the integral must be rewritten in terms of the new variable <math>u</math>;
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<math>u</math>;
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# the element of integration, <math>dx</math>, is replaced by <math>du</math>, according to the formula <math>du=u'(x)\,dx</math>;
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2. the element of integration,
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# the limits of integration are for <math>x</math> and must be changed to limits of integration for the variable <math>u</math>.
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<math>dx</math>, is replaced by
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<math>du</math>, according to the formula
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<math>du={u}'\left( x \right)dx</math>;
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3. the limits of integration are for
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<math>x\text{ }</math>
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and must be changed to limits of integration for the variable
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<math>u</math>.
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In this case, we will perform the change of variables
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In this case, we will perform the change of variables <math>u=3x-1</math>, mainly because the integrand <math>1/(3x-1)^4</math> will then be replaced by <math>1/u^4</math>.
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<math>u=3x-1</math>, mainly because the integrand
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<math>\frac{1}{\left( 3x-1 \right)^{4}}</math>
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will then be replaced by
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<math>\frac{1}{u^{4}}</math>.
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The relation between
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The relation between <math>dx</math> and <math>du</math> reads
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<math>dx</math>
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and
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<math>du</math>
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reads
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{{Displayed math||<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}
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<math>du={u}'\left( x \right)\,dx=\left( 3x-1 \right)^{\prime }\,dx=3\,dx</math>,
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which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>.
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which means that
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<math>dx</math>
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is replaced by
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<math>\frac{1}{3}\,du</math>.
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Furthermore, when
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<math>x=\text{1}</math>
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in the lower limit of integration, the corresponding
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<math>u</math>
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-value becomes
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<math>u=3\centerdot 1-1=2</math>, and when
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<math>x=2</math>, we obtain the u-value
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<math>u=3\centerdot 2-1=5</math>
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 +
Furthermore, when <math>x=1</math> in the lower limit of integration, the corresponding ''u''-value becomes <math>u=3\cdot 1-1=2</math>, and when
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<math>x=2</math>, we obtain the ''u''-value <math>u=3\cdot 2-1=5\,</math>.
One usually writes the whole substitution of variables as
One usually writes the whole substitution of variables as
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{{Displayed math||<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}
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u &= 3x-1\\[5pt]
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du &= 3\,dx
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\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}
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<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}
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Sometimes, we are more brief and hide the details,
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u=3x-1 \\
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du=3\,dx \\
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\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}</math>
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Sometimes, we are more brief and hide the details:
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<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>
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 +
{{Displayed math||<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.
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In summary, the whole calculation is:
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In summary, the whole calculation is,
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}
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\int\limits_1^2 \frac{dx}{(3x-1)^4}
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u=3x-1 \\
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&= \left\{\begin{align}
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du=3\,dx \\
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u &= 3x-1\\[5pt]
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\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}} \\
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du &= 3\,dx
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& =\frac{1}{3}\int\limits_{2}^{5}{\,u^{-4}du}=\left[ \frac{u^{-4+1}}{-4+1} \right]_{2}^{5} \\
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\end{align}\right\}
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& =-\frac{1}{9}\left[ \frac{1}{u^{3}} \right]_{2}^{5}=-\frac{1}{9}\left( \frac{1}{5^{3}}-\frac{1}{2^{3}} \right) \\
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= \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}
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& =-\frac{1}{9}\centerdot \frac{2^{3}-5^{3}}{2^{3}\centerdot 5^{3}}=\frac{117}{3^{2}\centerdot 2^{3}\centerdot 5^{3}} \\
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= \frac{1}{3}\int\limits_2^5 u^{-4}\,du\\[5pt]
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& =\frac{3^{2}\centerdot 13}{3^{2}\centerdot 2^{3}\centerdot 5^{3}}=\frac{13}{2^{3}\centerdot 5^{3}}=\frac{13}{1000} \\
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&= \frac{1}{3}\Bigl[\ \frac{u^{-4+1}}{-4+1}\ \Bigr]_2^5
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\end{align}</math>.
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= -\frac{1}{9}\Bigl[\ \frac{1}{u^3}\ \Bigr]_2^5
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= -\frac{1}{9}\Bigl(\frac{1}{5^3} - \frac{1}{2^3} \Bigr)
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= -\frac{1}{9}\cdot\frac{2^3-5^3}{2^3\cdot 5^3}\\[5pt]
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&= \frac{117}{3^2\cdot 2^3\cdot 5^3}
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= \frac{3^2\cdot 13}{3^2\cdot 2^3\cdot 5^3}
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= \frac{13}{2^3\cdot 5^3}
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= \frac{13}{1000}\,\textrm{.}
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\end{align}</math>}}

Version vom 12:31, 28. 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(x), 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'(x)\,dx;
  3. 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}

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