The area of the region bounded by the curve y | vedclass

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(B) The required area $A$ is given by the integral of the function $y = \log x$ from $x=1$ to $x=e$.$A = \int_{1}^{e} \log x \, dx$Using integration b

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$1$ Sq. Units

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(B) The required area $A$ is given by the integral of the function $y = \log x$ from $x=1$ to $x=e$.$A = \int_{1}^{e} \log x \, dx$Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = \log x$ and $dv = dx$:$A = [x \log x]_{1}^{e} - \int_{1}^{e} x \cdot \frac{1}{x} \, dx$$A = [x \log x]_{1}^{e} - \int_{1}^{e} 1 \, dx$$A = [x \log x - x]_{1}^{e}$Substituting the limits:$A = (e \log e - e) - (1 \log 1 - 1)$Since $\log e = 1$ and $\log 1 = 0$:$A = (e(1) - e) - (0 - 1)$$A = (e - e) - (-1)$$A = 0 + 1 = 1 	ext{ Sq. Units}$

The area of the region bounded by the curve $y=\log x$,the $x$-axis,and the lines $x=1, x=e$ is

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