קיץ 2023 · מועד א

$f'(x) = \dfrac{1}{x} - \dfrac{1}{x^2} = \dfrac{x-1}{x^2}$

$\begin{array}{c|c|c|c} x & 0<x<1 & 1 & 1<x \\ \hline f'(x) & - & 0 & + \\ f(x) & \searrow & & \nearrow \end{array}$

$g(x) = (x+1)(1 - \ln x) = 0$

$1 - \ln x = 0 \Rightarrow \ln x = 1$

$x = e$

$g'(x) = (1)(1-\ln x) + (x+1)\left(-\dfrac{1}{x}\right)$

$= 1 - \ln x - \dfrac{x+1}{x} = 1 - \ln x - 1 - \dfrac1x = -\ln x - \dfrac1x$

$g'(x) = -\left(\ln x + \dfrac1x\right) = -f(x)$

$\begin{array}{c|c} x & x>0 \\ \hline g'(x) & - \\ g(x) & \searrow \end{array}$

$g''(x) = -f'(x) = -\dfrac{x-1}{x^2} = \dfrac{1-x}{x^2}$

$g''(x) = 0 \Rightarrow 1 - x = 0$

$x = 1$

$\begin{array}{c|c|c|c} x & 0<x<1 & 1 & 1<x \\ \hline g''(x) & + & 0 & - \\ g(x) & \cup & & \cap \end{array}$

$h(x) = \dfrac1x\,g'(x) = \dfrac1x\cdot\bigl(-f(x)\bigr) = -\dfrac1x\left(\ln x + \dfrac1x\right)$

$S = \int_1^e \bigl(0 - h(x)\bigr)dx = \int_1^e \dfrac1x f(x)\,dx = \int_1^e \left(\dfrac{\ln x}{x} + \dfrac{1}{x^2}\right)dx$

$= \left[\dfrac{(\ln x)^2}{2} - \dfrac1x\right]_1^e$

$= \left(\dfrac{1}{2} - \dfrac1e\right) - \left(0 - 1\right)$

$= \dfrac{3}{2} - \dfrac1e$