Solution 2.1:5a - Förberedande kurs i matematik 2

-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
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+			Solution 2.1:5a
			
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- {{NAVCONTENT_START}} + If we multiply top and bottom of the fraction by the conjugate expression 
- <center> [[Image:2_1_5a-1(2).gif]] </center> + <math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
- {{NAVCONTENT_STOP}} +  
- {{NAVCONTENT_START}} + {{Displayed math||<math>\begin{align}
- <center> [[Image:2_1_5a-2(2).gif]] </center> + \frac{1}{\sqrt{x+9}-\sqrt{x}}
- {{NAVCONTENT_STOP}} + &= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
  + &= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
  + &= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
  + &= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + Thus,
  +  
  + {{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>}}
  +  
  + If we write the square roots in power form,
  +  
  + {{Displayed math||<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,
  +  
  + {{Displayed math||<math>\begin{align}
  + \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}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
  + &= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
  + &= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
  + \end{align}</math>}}
  +  
  + where C is an arbitrary constant.
  +  
  + This can also be written with square roots as
  +  
  + {{Displayed math||<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,
  +  
  + {{Displayed math||<math>\begin{align}
  + \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{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
  + \end{align}</math>}}
Current revision
If we multiply top and bottom of the fraction by the conjugate expression 
\displaystyle \sqrt{x+9}+\sqrt{x} then the formula for the difference of two squares 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}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
\end{align}



Thus,





\displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}


If we write the square roots in power form,





\displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,


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





\displaystyle \begin{align}
\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}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
\end{align}



where C is an arbitrary constant.
This can also be written with square roots as





\displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}




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





\displaystyle \begin{align}
\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{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
\end{align}





Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-ImperialCollege/index.php/Solution_2.1:5a"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+If we multiply top and bottom of the fraction by the conjugate expression
-<center> [[Image:2_1_5a-1(2).gif]] </center>
+<math>\sqrt{x+9}+\sqrt{x}</math> then the formula for the difference of two squares gives that denominator's root is squared away,
-{{NAVCONTENT_STOP}}
-{{NAVCONTENT_START}}
+{{Displayed math||<math>\begin{align}
-<center> [[Image:2_1_5a-2(2).gif]] </center>
+\frac{1}{\sqrt{x+9}-\sqrt{x}}
-{{NAVCONTENT_STOP}}
+&= \frac{1}{\sqrt{x+9}-\sqrt{x}}\cdot\frac{\sqrt{x+9}+\sqrt{x}}{\sqrt{x+9}+\sqrt{x}}\\[5pt]
+&= \frac{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt]
+&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt]
+&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.}
+\end{align}</math>}}
+Thus,
+{{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>}}
+If we write the square roots in power form,
+{{Displayed math||<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,
+{{Displayed math||<math>\begin{align}
+\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}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt]
+&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt]
+&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
+\end{align}</math>}}
+where C is an arbitrary constant.
+This can also be written with square roots as
+{{Displayed math||<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,
+{{Displayed math||<math>\begin{align}
+\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{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
\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{\sqrt{x+9}+\sqrt{x}}{\bigl(\sqrt{x+9}\,\bigr)^2 - \bigl(\sqrt{x}\,\bigr)^2}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{x+9-x}\\[5pt] 
&= \frac{\sqrt{x+9}+\sqrt{x}}{9}\,\textrm{.} 
\end{align}
 \displaystyle \int \frac{dx}{\sqrt{x+9}-\sqrt{x}} = \frac{1}{9}\int\bigl(\sqrt{x+9}+\sqrt{x}\,\bigr)\,dx\,\textrm{.}
 \displaystyle \frac{1}{9}\int\bigl((x+9)^{1/2} + x^{1/2}\bigr)\,dx\,,
 \displaystyle \begin{align}
\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}\Bigl(\frac{(x+9)^{3/2}}{3/2} + \frac{x^{3/2}}{3/2} \Bigr)+C\\[5pt] 
&= \frac{1}{9}\Bigl(\frac{2}{3}(x+9)^{3/2} + \frac{2}{3}x^{3/2} \Bigr)+C\\[5pt] 
&= \frac{2}{27}(x+9)^{3/2} + \frac{2}{27}x^{3/2}+C\,,
\end{align}
 \displaystyle \frac{2}{27}(x+9)\sqrt{x+9} + \frac{2}{27}x\sqrt{x} + C\,\textrm{.}
 \displaystyle \begin{align}
\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{1}{9}(x+9)^{1/2} + \frac{1}{9}x^{1/2}\,\textrm{.} 
\end{align}