need help with definite integral

**gracy** · November 25th 2006, 05:44 AM

find the value of integral(-e to -e^2) (3/x) dx
(a) -3
(b)-e
(c)e^3
(d)1/3
(3) none

**Soroban** · November 25th 2006, 06:30 AM

Hello, Gracy!

This is a trick question ... on a number of levels.

Find the value of: . $\int^{\text{-}e^2}_{\text{-}e}\frac{3}{x}\,dx$

$(a)\;-3\qquad(b)\;-e\qquad(c)\;e^3\qquad(d)\;\frac{1}{3}\qquad(e) \text{ none of these}$

We have: . $3\int^{\text{-}e^2}_{\text{-}e}\frac{dx}{x}\;\,=\;\,3\ln\left|x\right|\,\bigg| ^{\text{-}e^2}_{\text{-}e}\,\;=\,\;3\ln|\text{-}e^2| - 3\ln|\text{-}e|$

. . . . . $= \;\;3\ln(e^2) - 3\ln(e)\;\;= \;\;3(2) - 3(1) \;\;=\;\;\boxed{3}$

Answer: . $(e)\text{ none of these}$

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