Lösung 2.1:2d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
If we rewrite <math>\sqrt{x}</math> as <math>x^{1/2}</math>, the integrand can then be simplified using the power laws,
If we rewrite <math>\sqrt{x}</math> as <math>x^{1/2}</math>, the integrand can then be simplified using the power laws,
-
{{Displayed math||<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}
We can now use the fact that a primitive function for <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> and calculate the integral's value,
We can now use the fact that a primitive function for <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> and calculate the integral's value,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_1^4 x^{-3/2}\,dx
\int\limits_1^4 x^{-3/2}\,dx
&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]
&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]

Version vom 12:58, 10. Mär. 2009

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^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}

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

\displaystyle \begin{align}

\int\limits_1^4 x^{-3/2}\,dx &= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt] &= \Bigl[\ \frac{x^{-1/2}}{-1/2}\ \Bigr]_1^4\\[5pt] &= \Bigl[\ -2\frac{1}{x^{1/2}}\ \Bigr]_1^4\\[5pt] &= \Bigl[\ -\frac{2}{\sqrt{x}}\ \Bigr]_1^4\\[5pt] &= -\frac{2}{\sqrt{4}} - \Bigl(-\frac{2}{\sqrt{1}}\Bigr)\\[5pt] &= -\frac{2}{2}+2\\[5pt] &= 1\,\textrm{.} \end{align}