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(A) The curve $y = \log_e x$ intersects the $x$-axis at $(1, 0)$.To find the area bounded by the coordinate axes and the curve,we consider the region

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$1$

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(A) The curve $y = \log_e x$ intersects the $x$-axis at $(1, 0)$.To find the area bounded by the coordinate axes and the curve,we consider the region between $x = 0$ and $x = 1$.The area $A$ is given by the integral of the absolute value of the function from $x = 0$ to $x = 1$:$A = \int_0^1 |\log_e x| \, dx = \int_0^1 -\log_e x \, dx$Using integration by parts,$\int \log_e x \, dx = x \log_e x - x$.Thus,$A = -[x \log_e x - x]_0^1 = -[(1 \cdot \log_e 1 - 1) - \lim_{x 	o 0^+} (x \log_e x - x)]$.Since $\log_e 1 = 0$ and $\lim_{x 	o 0^+} x \log_e x = 0$,we have:$A = -[0 - 1 - 0] = 1$.Therefore,the area is $1$ square unit.

The measurement of the area bounded by the coordinate axes and the curve $y = \log_e x$ is

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