Aus Online Mathematik Brückenkurs 2

If we recall that \displaystyle \sqrt{x}=x^{{1}/{2}\;}, the integral can be written as

\displaystyle \begin{align} & \int\limits_{4}^{9}{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}\,dx=\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-\frac{1}{x^{{1}/{2}\;}} \right)}\,dx \\ & =\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx \\ \end{align}

This is a standard integral in which the integrand consists of two terms looking like \displaystyle x^{n}, where \displaystyle n=\frac{1}{2} and \displaystyle n=-\frac{1}{2}.

We obtain

\displaystyle \begin{align} & \int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx=\left[ \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right]_{4}^{9} \\ & =\left[ \frac{x^{1+\frac{1}{2}}}{\frac{3}{2}}-\frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{4}^{9} \\ & =\left[ \frac{2}{3}x\sqrt{x}-2\sqrt{x} \right]_{4}^{9} \\ & =\frac{2}{3}\centerdot 9\centerdot \sqrt{9}-2\sqrt{9}-\left( \frac{2}{3}\centerdot 4\centerdot \sqrt{4}-2\sqrt{4} \right) \\ & =\frac{2}{3}93-2\centerdot 3-\left( \frac{2}{3}\centerdot 4\centerdot 2-2\centerdot 2 \right) \\ & =18-6-\frac{16}{3}+4=16-\frac{16}{3} \\ & =\frac{16\centerdot 3-16}{3}=\frac{32}{3} \\ \end{align}