Question

Se consideră funcția :

$f:\mathbb{R}\setminus\left\{1\right\} \: \: \: ,f(x)= \frac{ {x}^{2} + x + 2}{x - 1}$

a)Să se determine aria subgraficului funcției cuprins între axa Ox şi dreptele de ecuații x=-3 şi x=-2.

b)Să se calculeze :

$\lim_{x\rightarrow\infty}\frac{1}{x^{2}}\int_{2}^{x}f(t)dt$
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Answer 1

$f:\mathbb{R}\backslash \{1\}\to \mathbb{R},\quad f(x) = \dfrac{x^2+x+2}{x-1} \\ \\ \Delta = 1-8 = -7 \Rightarrow x^2+x+2 > 0\\ \\ \Rightarrow \text{Pentru }x < 1,~~f(x) <0 \\ \\ a)\quad A = \displaystyle \int_{-3}^{-2}\Big[0-f(x)\Big]\, dx = \int_{-2}^{-3}f(x)\, dx =\int_{-2}^{-3}\dfrac{x^2+x+2}{x-1}\,dx= \\ \\ =\int_{-2}^{-3}\dfrac{x^2-1+x-1+4}{x-1}\, dx=$

$\displaystyle = \int_{-2}^{-3}\dfrac{(x-1)(x+1)+(x-1)+4}{x-1}\, dx = \\ \\ = \int_{-2}^{-3}(x+1)\,dx +\int_{-2}^{-3}1 \, dx + 4\int_{-2}^{-3}\dfrac{1}{x-1}\, dx= \\ \\ = \Big(\dfrac{(x+1)^2}{2}+x+4\ln|x-1|\Big)\Big|_{-2}^{-3} = \\ \\ = (2-3+4\ln4) - \Big(\dfrac{1}{2}-2+4\ln3\Big) = \\ \\ =4\ln \dfrac{4}{3}+\dfrac{1}{2}$

$\displaystyle b)\quad \lim\limits_{x\to \infty}\dfrac{1}{x^2}\int_{2}^xf(t)\, dt = \\ \\ =\lim\limits_{x\to \infty}\dfrac{1}{x^2}\int_{2}^x\dfrac{t^2+t+2}{t-1}\, dt = \lim\limits_{x\to \infty}\dfrac{1}{x^2}\int_{2}^x\dfrac{(t-1)(t+2)+4}{t-1}\, dt = \\ \\ = \lim\limits_{x\to \infty}\dfrac{1}{x^2}\Big(\dfrac{(t+2)^2}{2}+4\ln(t-1)\Big)\Big|_{2}^x =$

$= \lim\limits_{x\to \infty}\dfrac{1}{x^2}\Big(\dfrac{(x+2)^2}{2}+4\ln(x-1)-8\Big) = \\ \\ = \lim\limits_{x\to \infty}\dfrac{(x+2)^2}{2x^2}+\lim\limits_{x\to \infty}\dfrac{4\ln(x-1)-8}{x^2} = \\ \\ = \dfrac{1}{2}+\lim\limits_{x\to \infty}\dfrac{\dfrac{4}{x-1}}{2x} =\dfrac{1}{2}+0= \dfrac{1}{2}$