The area enclosed between the curve y = $l o g_{e} (x + e)$ and the coordinate axis is

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Solution

The correct option is A 1
Area = $\int_{1 - e}^{0} l o g (x + e) d x = x l o g (x + e) - \int_{2}^{4} \frac{x}{x + e} d x = ├ x l o g (x + e) - x + e ∙ l o g (x + e)]_{1 - e}^{0} = 1$

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