The area bounded by the curves y = log_e x an | vedclass

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(A) To find the area bounded by the curves $y = \log_e x$ and $y = (\log_e x)^2$,we first determine the points of intersection by setting $\log_e x =

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$3 - e$

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(A) To find the area bounded by the curves $y = \log_e x$ and $y = (\log_e x)^2$,we first determine the points of intersection by setting $\log_e x = (\log_e x)^2$.Let $u = \log_e x$. Then $u = u^2$,which implies $u^2 - u = 0$,so $u(u - 1) = 0$. Thus,$u = 0$ or $u = 1$.For $u = 0$,$\log_e x = 0 \implies x = 1$. For $u = 1$,$\log_e x = 1 \implies x = e$.The curves intersect at $x = 1$ and $x = e$. In the interval $[1, e]$,$\log_e x \ge (\log_e x)^2$.The required area $A$ is given by:$A = \int_1^e [\log_e x - (\log_e x)^2] \, dx$$A = \int_1^e \log_e x \, dx - \int_1^e (\log_e x)^2 \, dx$Using integration by parts:$\int \log_e x \, dx = x \log_e x - x$$\int (\log_e x)^2 \, dx = x(\log_e x)^2 - 2x \log_e x + 2x$Evaluating the definite integrals:$\int_1^e \log_e x \, dx = [x \log_e x - x]_1^e = (e \log_e e - e) - (1 \log_e 1 - 1) = (e - e) - (0 - 1) = 1$$\int_1^e (\log_e x)^2 \, dx = [x(\log_e x)^2 - 2x \log_e x + 2x]_1^e = (e(1)^2 - 2e(1) + 2e) - (1(0)^2 - 2(1)(0) + 2(1)) = (e - 2e + 2e) - (2) = e - 2$Therefore,$A = 1 - (e - 2) = 3 - e$.

The area bounded by the curves $y = \log_e x$ and $y = (\log_e x)^2$ is

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