Question

Buna! Ma ajuta si pe mine cineva la subiectul 1 ex 1 si subiectul 3 ex 2 b,c?? Va rog :)

Answer 1

$\displaystyle \mathtt{Subiectul~1}\\ \\\mathtt{ \frac{2}{ \sqrt{3}-1 }- \sqrt{3}=1 } \\ \\ \mathtt{1.~~~ \frac{2}{\sqrt{3}-1}- \sqrt{3}= \frac{2\left( \sqrt{3}+1\right) }{3-1}- \sqrt{3}= \frac{\not2\left(\sqrt{3} +1\right)} {\not2} } - \sqrt{3}= }\\ \\ \mathtt{= \sqrt{3}+1-\sqrt{3}=1}$

$\displaystyle \mathtt{Subiectul~3}\\ \\ \mathtt{2.~~~b)~F(x)=\int\limits \left(x^2+2\right)dx=\int\limits x^2dx+\int\limits 2dx= \frac{x^3}{3} +2x+C}\\ \\ \mathtt{F(3)= \frac{3^3}{3}+2 \cdot 3+C= \frac{27}{3}+6+C=9+6+C=15+C}\\ \\ \mathtt{F(3)=5\Rightarrow 15+C=5\Rightarrow C=5-15\Rightarrow C=-10}\\ \\ \mathtt{F(x)= \frac{x^3}{3}+2x-10}$

$\displaystyle \mathtt{c)~g:\mathbb{R}\rightarrow\mathbb{R},~g(x)=e^x\cdot f(x)~~~~~~~~~~~~~~~~~~~~~x=0~~~~~~~~~~~~x=1}~\\ \\ \mathtt{\mathcal{A}=3e-4}\\ \\ \mathtt{g(x)=e^x \cdot f(x)=e^x\left(x^2+2\right)=\left(x^2+2\right)e^x}\\ \\ \mathtt{\int\limits_0^1\left(x^2+2\right)e^xdx}$

$\displaystyle \mathtt{\int\limits f(x)g'(x)dx=f(x)g(x)-\int\limits f'(x)g(x)dx}\\ \\ \mathtt{f(x)=x^2+2\Rightarrow f'(x)=\left(x^2+2\right)'=\left(x^2\right)'+2'=2x+0=2x}\\ \\ \mathtt{g'(x)=e^x \Rightarrow g(x)=\int\limits e^xdx=e^x} \\ \\ \mathtt{\int\limits \left(x^2+2\right)e^xdx=\left(x^2+2\right)e^x-\int\limits 2xe^xdx=\left(x^2+2\right)e^x-2\int\limits xe^xdx}$

$\displaystyle \mathtt{\int\limits xe^xdx}\\ \\ \mathtt{f(x)=x\Rightarrow f'(x)=x'=1}\\ \\ \mathtt{g'(x)=e^x\Rightarrow g(x)=\int\limits e^xdx=e^x}\\ \\ \mathtt{\int\limits xe^xdx=xe^x-\int\limits e^xdx=xe^x-e^x+C=e^x(x-1)+C}$

$\displaystyle \mathtt{\int\limits \left(x^2+2\right)e^xdx=\left(x^2+2\right)e^x-\int\limits 2xe^xdx=\left(x^2+2\right)e^x-2\int\limits xe^xdx=}\\ \\ \mathtt{=\left(x^2+2\right)e^x-2\left(e^x(x-1)\right)+C=x^2e^x+2e^x-2\left(xe^x-e^x\right)+C=}\\ \\ \mathtt{=x^2e^x+2e^x-2xe^x+2e^x+C=x^2e^x-2xe^x+4e^x+C=}\\ \\ \mathtt{=e^x\left(x^2-2x+4\right)+C}$

$\displaystyle \mathtt{\int\limits_0^1\left(x^2+2\right)e^xdx=e^x\left(x^2-2x+4\right)\Bigg|_0^1=}\\ \\ \mathtt{=e^1\left(1^2-2 \cdot 1+4\right)-e^0\left(0^2-2\cdot0+4\right)=e(1-2+4)-1(0-0+4)=}\\ \\ \mathtt{=e \cdot 3-1 \cdot 4=3e-4\Rightarrow \mathcal{A}=3e-4}$