Aus Online Mathematik Brückenkurs 2

Version vom 14:40, 16. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.1:5a moved to Solution 2.1:5a: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:29, 18. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	(HINT: multiply the top and bottom by the conjugate of the denominator.)
-	<center> [[Image:2_1_5a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	If we multiply top and bottom of the fraction by the conjugate expression,
-	{{NAVCONTENT_START}}	+	<math>\sqrt{x+9}+\sqrt{x}</math>, then the conjugate rule gives that denominator's root is squared away:
-	<center> [[Image:2_1_5a-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
		+	<math>\begin{align}
		+	& \frac{1}{\sqrt{x+9}-\sqrt{x}}=\frac{1}{\sqrt{x+9}-\sqrt{x}}\centerdot \frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}} \\
		+	& =\frac{\sqrt{x+9}+\sqrt{x}}{\left( \sqrt{x+9} \right)^{2}-\left( \sqrt{x} \right)^{2}} \\
		+	& =\frac{\sqrt{x+9}+\sqrt{x}}{x+9-x} \\
		+	& =\frac{\sqrt{x+9}+\sqrt{x}}{9} \\
		+	\end{align}</math>
		+
		+
		+	Thus,
		+
		+
		+	<math>\int{\frac{\,dx}{\sqrt{x+9}-\sqrt{x}}}=\frac{1}{9}\int{\left( \sqrt{x+9}+\sqrt{x} \right)}\,dx</math>
		+
		+
		+	If we write the square roots in power form,
		+
		+
		+	<math>\frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx</math>,
		+
		+	we see that we have a standard integral and can write down the primitive functions directly:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx=\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right)+C \\
		+	& =\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right)+C \\
		+	& =\frac{1}{9}\left( \frac{2}{3}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{3}x^{\frac{3}{2}} \right)+C \\
		+	& =\frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \\
		+	& \\
		+	\end{align}</math>,
		+
		+	where C is an arbitrary constant.
		+
		+	This can also be written with square roots as
		+
		+
		+	<math>\frac{2}{27}\left( x+9 \right)\sqrt{x+9}+\frac{2}{27}x\sqrt{x}+C</math>
		+
		+
		+	To be completely certain that we have everything correctly, we differentiate the answer and see if we get back the integrand:
		+
		+
		+	<math>\begin{align}
		+	& \frac{d}{dx}\left( \frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \right) \\
		+	& =\frac{2}{27}\centerdot \frac{3}{2}\left( x+9 \right)^{\frac{3}{2}-1}+\frac{2}{27}\centerdot \frac{3}{2}x^{\frac{3}{2}-1}+0 \\
		+	& =\frac{1}{9}\left( x+9 \right)^{\frac{1}{2}}+\frac{1}{9}x^{\frac{1}{2}} \\
		+	\end{align}</math>

---

Version vom 13:29, 18. Okt. 2008

(HINT: multiply the top and bottom by the conjugate of the denominator.)

If we multiply top and bottom of the fraction by the conjugate expression, \displaystyle \sqrt{x+9}+\sqrt{x}, then the conjugate rule gives that denominator's root is squared away:

\displaystyle \begin{align} & \frac{1}{\sqrt{x+9}-\sqrt{x}}=\frac{1}{\sqrt{x+9}-\sqrt{x}}\centerdot \frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}} \\ & =\frac{\sqrt{x+9}+\sqrt{x}}{\left( \sqrt{x+9} \right)^{2}-\left( \sqrt{x} \right)^{2}} \\ & =\frac{\sqrt{x+9}+\sqrt{x}}{x+9-x} \\ & =\frac{\sqrt{x+9}+\sqrt{x}}{9} \\ \end{align}

Thus,

\displaystyle \int{\frac{\,dx}{\sqrt{x+9}-\sqrt{x}}}=\frac{1}{9}\int{\left( \sqrt{x+9}+\sqrt{x} \right)}\,dx

If we write the square roots in power form,

\displaystyle \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx,

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

\displaystyle \begin{align} & \frac{1}{9}\int{\left( \left( x+9 \right)^{\frac{1}{2}}+x^{\frac{1}{2}} \right)}\,dx=\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right)+C \\ & =\frac{1}{9}\left( \frac{\left( x+9 \right)^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right)+C \\ & =\frac{1}{9}\left( \frac{2}{3}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{3}x^{\frac{3}{2}} \right)+C \\ & =\frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \\ & \\ \end{align},

where C is an arbitrary constant.

This can also be written with square roots as

\displaystyle \frac{2}{27}\left( x+9 \right)\sqrt{x+9}+\frac{2}{27}x\sqrt{x}+C

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

\displaystyle \begin{align} & \frac{d}{dx}\left( \frac{2}{27}\left( x+9 \right)^{\frac{3}{2}}+\frac{2}{27}x^{\frac{3}{2}}+C \right) \\ & =\frac{2}{27}\centerdot \frac{3}{2}\left( x+9 \right)^{\frac{3}{2}-1}+\frac{2}{27}\centerdot \frac{3}{2}x^{\frac{3}{2}-1}+0 \\ & =\frac{1}{9}\left( x+9 \right)^{\frac{1}{2}}+\frac{1}{9}x^{\frac{1}{2}} \\ \end{align}