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Lösung 2.1:5a

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 2: Zeile 2:
<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{\sqrt{x+9}-\sqrt{x}}
\frac{1}{\sqrt{x+9}-\sqrt{x}}
&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
Zeile 12: Zeile 12:
Thus,
Thus,
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{{Displayed math||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}</math>}}
If we write the square roots in power form,
If we write the square roots in power form,
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{{Displayed math||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
+
{{Abgesetzte Formel||<math>\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,</math>}}
we see that we have a standard integral and can write down the primitive functions directly,
we see that we have a standard integral and can write down the primitive functions directly,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
\frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx
&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]
&= \frac{1}{9}\biggl(\frac{(x+9)^{1/2+1}}{\tfrac{1}{2}+1} + \frac{x^{1/2+1}}{\tfrac{1}{2}+1} \biggr)+C\\[5pt]
Zeile 32: Zeile 32:
This can also be written with square roots as
This can also be written with square roots as
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{{Displayed math||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}</math>}}
Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
\frac{d}{dx}\Bigl( \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2} + C \Bigr)
&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]
&= \frac{2}{27}\cdot \frac{3}{2}(x+9)^{3/2-1} + \frac{2}{27}\cdot\frac{3}{2} x^{3/2-1} + 0\\[5pt]
&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
&= \frac{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 13:00, 10. Mär. 2009

If we multiply top and bottom of the fraction by the conjugate expression x+9+x  then the formula for the difference of two squares gives that denominator's root is squared away,

1x+9−x=1x+9−xx+9+xx+9+x=x+9+xx+92−x2=x+9−xx+9+x=9x+9+x.

Thus,

dxx+9−x=91x+9+xdx. 

If we write the square roots in power form,

91(x+9)12+x12dx 

we see that we have a standard integral and can write down the primitive functions directly,

91(x+9)12+x12dx=9121+1(x+9)12+1+21+1x12+1+C=9132(x+9)32+32x32+C=9132(x+9)32+32x32+C=227(x+9)32+227x32+C

where C is an arbitrary constant.

This can also be written with square roots as

227(x+9)x+9+227xx+C. 


Note: To be completely certain that we have done everything correctly, we differentiate the answer and see if we get back the integrand,

ddx227(x+9)32+227x32+C=22723(x+9)32−1+22723x32−1+0=91(x+9)12+91x12.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:5a“