Subiectul III

Se consideră funcția $f:(0,+\infty)\to\mathbb{R}, f(x)=x\ln{x}.$

Calculați $f'(x), x\in(0,+\infty).$
Calculați $\lim_{x\to+\infty}\displaystyle\frac{f(x)}{x^2}.$
Demonstrați că funcția este convexă pe intervalul $(0,+\infty).$

Calculăm

$\begin{align*} f'(x) &= \Big(x\ln{x}\Big)'\\\\ &=x'\cdot\ln{x}+x\cdot(\ln{x})'\\\\ &=1\cdot\ln{x}+x\cdot\displaystyle\frac{1}{x}\\\\ &=\ln{x}+1 \end{align*}$

$\begin{align*} \Leftrightarrow f'(x) &=\ln{x}+1. \end{align*}$

Calculăm $\lim_{x\to+\infty}\displaystyle\frac{f(x)}{x^2}.$

$\begin{align*} \lim_{x\to+\infty}\displaystyle\frac{f(x)}{x^2}&= \lim_{x\to+\infty}\displaystyle\frac{x\ln{x}}{x^2}\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{\ln{x}}{x} \end{align*}$

Fracția $\displaystyle\frac{\ln{x}}{x}$ este un caz de $\displaystyle\frac{\infty}{\infty},$ rezultă că vom aplica regula lui L'Hôpital astfel:

$\begin{align*} \lim_{x\to+\infty}\displaystyle\frac{\ln{x}}{x}&= \lim_{x\to+\infty}\Big(\displaystyle\frac{\ln{x}}{x}\Big)'\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{(\ln{x})'}{x'}\\\\ &=\lim_{x\to+\infty}\displaystyle\frac{\displaystyle\frac{1}{x}}{1}\\ \\&=\lim_{x\to+\infty}\displaystyle\frac{1}{x}\\ \\&=0 \end{align*}$

$\Leftrightarrow \lim_{x\to+\infty}\displaystyle\frac{f(x)}{x^2}=0.$

Pentru a demonstra că funcția este convexă pe intervalul $(0,+\infty)$ , vom calcula derivata de ordinul II.

$\begin{align*} f''(x) &= \Big(f'(x)\Big)\\\\ &= (\ln{x}+1)'\\\\ &=(\ln{x})'+1'\\\\ &=\frac{1}{x}+0\\\\ &=\frac{1}{x} \end{align*}$

$\begin{align*}\Leftrightarrow f''(x) &= \frac{1}{x},\forall\ x\in(0,+\infty). \end{align*}$

Avem că $\begin{align*} f''(x) >0 \end{align*}$ , pentru orice $\begin{align*} x\in(0,+\infty) ,\end{align*}$ de unde ne rezultă că funcția este convexă pe intervalul $(0,+\infty).$

Se consideră funcția $f:\mathbb{R}\to\mathbb{R}, f(x)=\displaystylr\frac{1}{x^2+1}.$

Arătați că $\int_{0}^{1} x f(x) dx = \frac{1}{2} \ln{2}.$
Calculați $\int_{0}^{1} x f'(x) dx.$
Determinați volumul corpului obținut prin rotația în jurul axei a graficului funcției $h:[0,1]\to\mathbb{R}, h(x)=\displaystyle\frac{1}{f(x)}.$

Calculăm $\int_{0}^{1} x f(x) dx .$

$\begin{align*} \int_{0}^{1} x f(x) dx &= \int_{0}^{1} x\cdot\displaystyle\frac{1}{x^2+1} \ dx\\\\ &=\int_{0}^{1} \frac{1}{2}\cdot\displaystyle\frac{2x}{x^2+1}\ dx\\\\ &=\frac{1}{2}\int_{0}^{1}\displaystyle\frac{2x}{x^2+1}\ dx\\ \\&=\frac{1}{2}\int_{0}^{1}\displaystyle\frac{(x^2+1)'}{x^2+1}\ dx\\ \\&=\frac{1}{2} \ln{|x^2+1|}\Big|_0^1\\\\ &=\frac{1}{2}\ln{(x^2+1)}\Big|_0^1\\ \\&=\frac{1}{2}\Big(\ln{(1^2+1)}-\ln{(0^2+1)}\Big)\\\\ &=\frac{1}{2}\Big(\ln{(1+1)}-\ln{(0+1)}\Big)\\ \\&=\frac{1}{2} \Big(\ln{2}-\ln{1}\Big)\\\\ &=\frac{1}{2} \Big(\ln{2}-0\Big)\\ \\&=\frac{1}{2}\cdot\ln{2} \end{align*}$

$\Leftrightarrow \int_{0}^{1} x f(x) dx = \frac{1}{2} \ln{2}.$

Calculăm $\int_{0}^{1} x f'(x) dx.$

$\begin{align*} \int_{0}^{1} x f'(x) dx &= xf(x)\Big|_0^1-\int_{0}^{1} f(x) dx\\\\ &=\Big(1\cdot f(1)-0\cdot f(0)\Big)-\int_{0}^{1} f(x) dx\\ \\&=\Big(f(1)-0\Big)-\int_{0}^{1} f(x) dx\\\\ &=f(1)-\int_{0}^{1} f(x) dx\\ \\&=\displaystyle\frac{1}{1^2+1}-\int_{0}^{1} \displaystyle\frac{1}{x^2+1} dx\\\\ &=\frac{1}{1+1}-\int_{0}^{1} \displaystyle\frac{1}{x^2+1^1} dx\\ \\&=\frac{1}{2}-\frac{1}{1}\cdot \arctan{\displaystyle\frac{x}{1}}\Big|_0^1\\\\ &=\frac{1}{2}-\arctan{x}\Big|_0^1\\\\ &=\frac{1}{2}-\Big(\arctan{1}-\arctan{0}\Big)\\\\ &=\frac{1}{2}-\Big(\frac{\pi}{4}-0\Big)\\\\ &=\frac{1}{2}-\frac{\pi}{4} \end{align*}$

$\Leftrightarrow \int_{0}^{1} x f'(x) dx=\frac{1}{2}-\frac{\pi}{4}.$

Calculăm

$\begin{align*} h(x) &= \frac{1}{f(x)}\\\\ &=\displaystyle\frac{1}{\displaystyle\frac{1}{x^2+1}}\\\\ &=1: \displaystyle\frac{1}{x^2+1}\\ \\&=1\cdot \displaystyle\frac{x^2+1}{1}\\ \\&=x^2+1 \end{align*}$

$\begin{align*}\Leftrightarrow h(x) &= x^2+1. \end{align*}$

Atunci:

$\begin{align*} h^2(x) &= (x^2+1)^2\\\\ &=(x^2)^2+2\cdot x^2\cdot1+1^2\\\\ &=x^{2+2}+2x^2+1\\\\ &=x^4+2x^2+1 \end{align*}$

$\begin{align*}\Leftrightarrow h^2(x) &=x^4+2x^2+1 .\end{align*}$

Calculăm volumul corpului obținut prin rotația în jurul axei a graficului funcției $h:[0,1]\to\mathbb{R}, h(x)=\displaystyle\frac{1}{f(x)}.$

$\begin{align*} V &= \int_{0}^{1} \pi \cdot h^2(x)\ dx \\\\ &= \pi \int_{0}^{1} (x^4+2x^2+1)dx \\\\ &=\pi \cdot \Big(\displaystyle\frac{x^5}{5}+2\cdot\frac{x^3}{3}+x\Big)\Big|_0^1\\\\ &=\pi\cdot\Big[\frac{1^5}{5}+2\cdot\frac{1^3}{3}+1-\Big(\frac{0^5}{5}+2\cdot\frac{0^3}{3}+0\Big)\Big]\\ \\&=\pi\cdot\Big(\frac{1}{5}+2\cdot\frac{1}{3}+1-0-0-0\Big)\\\\ &=\pi\cdot \displaystyle\frac{3\cdot1+2\cdot5\cdot1+15\cdot1}{15}\\\\ &=\pi\cdot\displaystyle\frac{3+10+15}{15}\\\\ &=\pi\cdot\frac{28}{15} \end{align*}$

$\begin{align*}\Leftrightarrow V &=\frac{28\pi}{15}. \end{align*}$