The area bounded by the curve y = x(1 - ln x) | vedclass

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(B) The area $A$ is given by the integral of $y$ with respect to $x$ from $x = e^{-1}$ to $x = e$.$A = \int_{e^{-1}}^{e} x(1 - \ln x) \, dx$Let $I = \

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$\frac{e^2 - 5e^{-2}}{4}$

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(B) The area $A$ is given by the integral of $y$ with respect to $x$ from $x = e^{-1}$ to $x = e$.$A = \int_{e^{-1}}^{e} x(1 - \ln x) \, dx$Let $I = \int x(1 - \ln x) \, dx = \int x \, dx - \int x \ln x \, dx$.Using integration by parts for $\int x \ln x \, dx$:Let $u = \ln x$ and $dv = x \, dx$,then $du = \frac{1}{x} \, dx$ and $v = \frac{x^2}{2}$.$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4}$.Thus,$I = \frac{x^2}{2} - (\frac{x^2}{2} \ln x - \frac{x^2}{4}) = \frac{x^2}{2} - \frac{x^2}{2} \ln x + \frac{x^2}{4} = \frac{3x^2}{4} - \frac{x^2}{2} \ln x$.Now,evaluate the definite integral from $e^{-1}$ to $e$:$A = [\frac{3x^2}{4} - \frac{x^2}{2} \ln x]_{e^{-1}}^{e}$$A = (\frac{3e^2}{4} - \frac{e^2}{2} \ln e) - (\frac{3e^{-2}}{4} - \frac{e^{-2}}{2} \ln e^{-1})$$A = (\frac{3e^2}{4} - \frac{e^2}{2}) - (\frac{3e^{-2}}{4} + \frac{e^{-2}}{2})$$A = \frac{e^2}{4} - \frac{5e^{-2}}{4} = \frac{e^2 - 5e^{-2}}{4}$.

The area bounded by the curve $y = x(1 - \ln x)$,the line $x = e^{-1}$,and the positive $X$-axis between $x = e^{-1}$ and $x = e$ is:

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