Solution 2.1:2c - Förberedande kurs i matematik 2

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                        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 2.1:2c
			
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- {{NAVCONTENT_START}} + If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as 
- <center> [[Image:2_1_2c.gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\begin{align}
  + \int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
  + &= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt] 
  + &= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + This is a standard integral in which the integrand consists of two terms looking like <math>x^n</math>, where <math>n=1/2</math> and <math>n=-1/2\,</math>, respectively.
  +  
  + We obtain
  +  
  + {{Displayed math||<math>\begin{align}
  + \int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
  + &= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt] 
  + &= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt] 
  + &= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt] 
  + &= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt] 
  + &= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt] 
  + &= 18-6-\frac{16}{3}+4\\[5pt]
  + &= 16-\frac{16}{3}\\[5pt] 
  + &= \frac{16\cdot 3-16}{3}\\[5pt]
  + &= \frac{32}{3}\,\textrm{.}
  + \end{align}</math>}}
Current revision
If we recall that \displaystyle \sqrt{x} = x^{1/2}, the integral can be written as 





\displaystyle \begin{align}
\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt] 
&= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.} 
\end{align}



This is a standard integral in which the integrand consists of two terms looking like \displaystyle x^n, where \displaystyle n=1/2 and \displaystyle n=-1/2\,, respectively.
We obtain





\displaystyle \begin{align}
\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt] 
&= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt] 
&= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt] 
&= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt] 
&= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt] 
&= 18-6-\frac{16}{3}+4\\[5pt]
&= 16-\frac{16}{3}\\[5pt] 
&= \frac{16\cdot 3-16}{3}\\[5pt]
&= \frac{32}{3}\,\textrm{.}
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_2.1:2c"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as
-<center> [[Image:2_1_2c.gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
+\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
+&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]
+&= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.}
+\end{align}</math>}}
+This is a standard integral in which the integrand consists of two terms looking like <math>x^n</math>, where <math>n=1/2</math> and <math>n=-1/2\,</math>, respectively.
+We obtain
+{{Displayed math||<math>\begin{align}
+\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
+&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]
+&= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt]
+&= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt]
+&= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt]
+&= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt]
+&= 18-6-\frac{16}{3}+4\\[5pt]
+&= 16-\frac{16}{3}\\[5pt]
+&= \frac{16\cdot 3-16}{3}\\[5pt]
+&= \frac{32}{3}\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt] 
&= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt] 
&= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt] 
&= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt] 
&= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt] 
&= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt] 
&= 18-6-\frac{16}{3}+4\\[5pt]
&= 16-\frac{16}{3}\\[5pt] 
&= \frac{16\cdot 3-16}{3}\\[5pt]
&= \frac{32}{3}\,\textrm{.}
\end{align}