5.3 The Fundamental Theorem of Calculus/35: Difference between revisions
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\int_{1}^{9}\frac{1}{2x}dx = \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx | \int_{1}^{9}\frac{1}{2x}dx = \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx | ||
&= \frac{1}{2}\ln{|x|}\bigg|_{1}^{9} = \frac{1}{2}\ln{|9|}-\frac{1}{2}\ln{|1|} \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 = \ln{3} | &= \frac{1}{2}\ln{|x|}\bigg|_{1}^{9} = \frac{1}{2}\ln{|9|}-\frac{1}{2}\ln{|1|} = \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 = \ln{3} | ||
\end {align} | \end {align} | ||
</math> | </math> | ||
Revision as of 19:32, 25 August 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{9}{\frac {1}{2x}}dx={\frac {1}{2}}\int _{1}^{9}{\frac {1}{x}}dx&={\frac {1}{2}}\ln {|x|}{\bigg |}_{1}^{9}={\frac {1}{2}}\ln {|9|}-{\frac {1}{2}}\ln {|1|}=\ln {|9^{\frac {1}{2}}|}-\ln {|1^{\frac {1}{2}}|}=\ln {3}-0=\ln {3}\end{aligned}}}