Solution 2.2:2c - 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 2.2:2c
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_2c.gif </center> {{NAVCONTENT_STOP}})
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- {{NAVCONTENT_START}} + If we focus on the integrand, then the substitution <math>u=3x+1</math> seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate. There is also no risk involved in using a linear substitution such as <math>u=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,
- <center> [[Bild:2_2_2c.gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}
  +  
  + which does not cause any problems.
  +  
  + We obtain
  +  
  + {{Displayed math||<math>\begin{align}
  + \int\limits_0^5 \sqrt{3x+1}\,dx
  + &= \left\{\begin{align}
  + u &= 3x+1\\[5pt]
  + du &= 3\,dx
  + \end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt] 
  + &= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt] 
  + &= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt] 
  + &= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.} 
  + \end{align}</math>}}
Current revision
If we focus on the integrand, then the substitution \displaystyle u=3x+1 seems suitable, since we then get \displaystyle \sqrt{u} which we can integrate. There is also no risk involved in using a linear substitution such as \displaystyle u=3x+1, because the relation between \displaystyle dx and \displaystyle du will be a constant factor,





\displaystyle du = (3x+1)'\,dx = 3\,dx\,,


which does not cause any problems.
We obtain





\displaystyle \begin{align}
\int\limits_0^5 \sqrt{3x+1}\,dx
&= \left\{\begin{align}
u &= 3x+1\\[5pt]
du &= 3\,dx
\end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt] 
&= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt] 
&= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt] 
&= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.} 
\end{align}




Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_2.2:2c"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+If we focus on the integrand, then the substitution <math>u=3x+1</math> seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate. There is also no risk involved in using a linear substitution such as <math>u=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,
-<center> [[Bild:2_2_2c.gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}
+which does not cause any problems.
+We obtain
+{{Displayed math||<math>\begin{align}
+\int\limits_0^5 \sqrt{3x+1}\,dx
+&= \left\{\begin{align}
+u &= 3x+1\\[5pt]
+du &= 3\,dx
+\end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt]
+&= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt]
+&= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt]
+&= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.}
+\end{align}</math>}}
 \displaystyle du = (3x+1)'\,dx = 3\,dx\,,
 \displaystyle \begin{align}
\int\limits_0^5 \sqrt{3x+1}\,dx
&= \left\{\begin{align}
u &= 3x+1\\[5pt]
du &= 3\,dx
\end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt] 
&= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt] 
&= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt] 
&= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.} 
\end{align}