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 | ||
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&= \frac{1}{2}\ln{|x|}\bigg|_{1}^{9} = \frac{1}{2}\ln{|9|}-\frac{1}{2}\ln{|1|} \\[2ex] | &= \frac{1}{2}\ln{|x|}\bigg|_{1}^{9} | ||
&= \frac{1}{2}\ln{|9|}-\frac{1}{2}\ln{|1|} \\[2ex] | |||
&= \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 = \ln{3} \\[2ex] | &= \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 = \ln{3} \\[2ex] | ||
Revision as of 19:39, 25 August 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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|} \\[2ex] &= \ln{|9^{\frac{1}{2}}|} - \ln{|1^{\frac{1}{2}}|} = \ln{3}-0 = \ln{3} \\[2ex] \end {align} }