I don't know why $$\int_{-3}^{-2} \frac{dx} x = \ln \frac{2}{3}.$$ How can i solve this to get that answer?
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1Because the anti derivative of 1/x in ln(x) and then evaluate. – user258250 Aug 24 '15 at 17:28
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3We have that $\int \frac{dx}{x}=\ln(|x|)+C$, where $C$ is an arbitrary constant. – Scounged Aug 24 '15 at 17:28
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3it is $$\ln(|-2|)-\ln(|-3|)$$ – Dr. Sonnhard Graubner Aug 24 '15 at 17:30
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If you are bothered by the negative limits of integration, make a change of variable $u = -x$ and $du = -dx$ and be careful to change the limits of integration accordingly. – hardmath Aug 24 '15 at 21:23
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$$ I = \int_{-3}^{-2}\frac{dx}{x} = \left[\ln |x|\vphantom{\frac11}\right]_{-3}^{-2} = \ln|-2|-\ln|-3| = \ln(2)-\ln(3) = \ln\left(\frac{2}{3}\right)$$
juantheron
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Notice,
\begin{align}\int_{-3}^{-2}\frac{dx}{x}&=\left[\ln|x| \vphantom{\frac 1 1} \right]_{-3}^{-2}\\ &=\left[\ln|-2|-\ln|-3| \vphantom{\frac 1 1} \right]\end{align}
$$=\ln\left(\frac{2}{3}\right)$$
Harish Chandra Rajpoot
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