Question

Explicati-mi modul de calul ,ofer 36 p (ofer și coroană)​

Answer 1

la acest exercitiu este atasat doar a)

Answer 2

$\displaystyle 2.~~f:\mathbb{R}\rightarrow \mathbb{R},~f(x)=x^4+x+1\\\\ a)~\int\limits^1_0(f(x)-x-1)dx=\frac{1}{5} \\\\ Calculam~f(x)-x-1=x^4+x+1-x-1=x^4\\\\ Inlocuim~cu~rezultatul~obtinut:\\\\ \int\limits^1_0 (f(x)-x-1)dx=\underbrace{\int\limits^1_0 x^4dx}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Pentru~\int x^4dx~aplicam~formula:\boxed{\int\limits x^ndx=\frac{x^{n+1}}{n+1} }\\\\ \int\limits^1_0 x^4dx=\frac{x^{4+1}}{4+1} \Bigg|_0^1=\frac{x^5}{5} \Bigg|_0^1=\frac{1^5}{5} -\frac{0^5}{5} =\frac{1}{5}$

$\displaystyle b)~\int\limits^e_1 \Big(f(x)-x^4-1\Big)lnxdx=\frac{e^2+1}{4} \\\\ Calculam~ f(x)-x^4-1=x^4+x+1-x^4-1=x\\\\ Inlocuim~cu~rezultatul~obtinut:~\int\limits^e_1\Big(f(x)-x^4-1\Big)lnxdx=\int\limits^e_1xlnxdx\\\\ \int\limits xlnxdx \rightarrow Aplicam~formula~de~integrare~prin~parti:\\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx}$

$\displaystyle f(x)=lnx \Rightarrow f'(x)=\underbrace{(lnx)'}\Rightarrow f'(x)=\frac{1}{x} \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Pentru~(lnx)'~avem~formula:~\boxed{(lnx)'=\frac{1}{x} }\\\\g'(x)=x \Rightarrow g(x)=\underbrace{\int xdx} \Rightarrow g(x)=\frac{x^{1+1}}{1+1}+C\Rightarrow g(x)=\frac{x^2}{2} +C\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~Pentru~\int xdx~aplicam~formula:~\boxed{\int xdx=\frac{x^2}{2} }$

$\displaystyle Inlocuim~in~formula~ceea~ce~am~obtinut:\\\\ \int\limits^e_1 xlnxdx=lnx \cdot \frac{x^2}{2} \Bigg|_1^e-\int\limits^e_1\Bigg(\frac{1}{\not x} \cdot \frac{x^{\not2}}{2}\Bigg) dx=\frac{x^2lnx}{2}\Bigg|_1^e -\int\limits^e_1\frac{x}{2} dx=\\\\=\frac{x^2lnx}{2}\Bigg|_1^e-\frac{1}{2}\underbrace{\int\limits^e_1xdx}\\~~~~~~~~~~~~~~~~~~~Pentru~\int xdx~aplicam~formula:~\boxed{\int xdx=\frac{x^2}{2} }$

$\displaystyle \int\limits^e_1 xlnxdx=\frac{x^2lnx}{2} \Bigg|_1^e-\frac{1}{2} \int\limits^e_1xdx=\frac{x^2lnx}{2}\Bigg|_1^e -\frac{1}{2} \cdot \frac{x^2}{2} \Bigg|_1^e=\\\\=\frac{e^2lne}{2} -\frac{1^2ln1}{2} -\frac{1}{2} \cdot \Bigg(\frac{e^2}{2} -\frac{1^2}{2} \Bigg)=\frac{e^2}{2} -\frac{1}{2}\cdot \frac{e^2-1}{2} =\frac{e^2}{2} -\frac{e^2-1}{4} =\\\\=\frac{2e^2-e^2+1}{4} =\frac{e^2+1}{4}$