The area bounded by the curve y = log x betwe | vedclass

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(B) We have the curve $y = \log x$. To find the limits of integration,we find the intersection with the $x$-axis by setting $y = 0$: $0 = \log x \impl

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$1$

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(B) We have the curve $y = \log x$. To find the limits of integration,we find the intersection with the $x$-axis by setting $y = 0$: $0 = \log x \implies x = 1$. The area $A$ bounded by the curve between $x = 1$ and $x = e$ is given by the integral: $A = \int_{1}^{e} \log x \, dx$. Using integration by parts,$\int \log x \, dx = x \log x - x$. Evaluating the definite integral: $A = [x \log x - x]_{1}^{e} = (e \log e - e) - (1 \log 1 - 1)$. Since $\log e = 1$ and $\log 1 = 0$: $A = (e(1) - e) - (0 - 1) = (e - e) + 1 = 1$. Thus,the area is $1$ square unit.

The area bounded by the curve $y = \log x$ between the $x$-axis and the ordinate $x = e$ is

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