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Integration question, help please! (1 Viewer)

Thread starter tk8
Start date Mar 25, 2022

tk8

Member

#1

Hey guys,
Could you help me with the attached question, to find 'a'? The answer is a=-1/e by the way

Screen Shot 2022-03-25 at 4.05.20 pm.png

5

5uckerberg

Well-Known Member

#2

$\left[log_{e}x\right]_{-e}^{a}=log_{e}\left(\frac{a}{-e}\right)=-2$
$\frac{a}{-e}=e^{-2}$
$-ae^{-1}=e^{-2}$
$-a=e^{-1}$
$a=-e^{-1}$
$a=-\frac{1}{e}$

U

Undefined

New Member

#3

Isn't it $\pm\frac{1}{e}$ because $\int\frac{f'(x)}{f(x)}dx$ is $ln|f(x)| + C$ not $ln(f(x)+C$ ?

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