Aus Online Mathematik Brückenkurs 2

If we rewrite \displaystyle \sqrt{x} as \displaystyle x^{{1}/{2}\;}, the integrand can then be simplified using the power laws:

\displaystyle \int\limits_{1}^{4}{\frac{\sqrt{x}}{x^{2}}}\,dx=\int\limits_{1}^{4}{\frac{x^{\frac{1}{2}}}{x^{2}}}\,dx=\int\limits_{1}^{4}{x^{\frac{1}{2}-2}}\,dx=\int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx

We can now use the fact that a primitive function for \displaystyle x^{n} is \displaystyle \frac{x^{n+1}}{n+1} and calculate the integral's value:

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