5.3 The Fundamental Theorem of Calculus/35: Difference between revisions
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&= \frac{1}{2}\ln{|x|}\bigg|_{1}^{9} \\[2ex] | &= \frac{1}{2}\ln{|x|}\bigg|_{1}^{9} \\[2ex] | ||
&= \frac{1}{2}\ln{|9|}-\frac{1}{2}\ln{|1|} \\[2ex] | &= \frac{1}{2}\ln{|9|}-\frac{1}{2}\ln{|1|} \\[2ex] | ||
&= \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 \\[2ex] | |||
&= \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 = \ln{3} \\[2ex] | &= \ln{3} \\[2ex] | ||
\end {align} | \end {align} | ||
</math> | </math> | ||
Latest revision as of 21:34, 6 September 2022
![{\displaystyle {\begin{aligned}\int _{1}^{9}{\frac {1}{2x}}dx&={\frac {1}{2}}\int _{1}^{9}{\frac {1}{x}}dx\\[2ex]&={\frac {1}{2}}\ln {|x|}{\bigg |}_{1}^{9}\\[2ex]&={\frac {1}{2}}\ln {|9|}-{\frac {1}{2}}\ln {|1|}\\[2ex]&=\ln {|9^{\frac {1}{2}}|}-\ln {|1^{\frac {1}{2}}|}=\ln {3}-0\\[2ex]&=\ln {3}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/d7b1f1a97977e7ce036c3b9ec4add7f5a273a1aa)