Aus Online Mathematik Brückenkurs 2

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}.