Question

ma poate ajuta si pe mine cineva la exercitiul 13, va rog frumos?

Answer 1

$\displaystyle \mathtt{\mathcal{A}=\int\limits_1^e \big|f(x)\big|dx=\int\limits_1^e\big(x-ln~x\big)dx=\int\limits_1^e xdx-\int\limits_1^e ln~xdx}\\\\\mathtt{\int\limits ln~xdx}\\\\\mathtt{\int\limits f(x)g'(x)dx=f(x)g(x)-\int\limits f'(x)g(x)dx}\\\\\mathtt{f(x)=\big(ln~x\big)'= \frac{1}{x};~~g'(x)=1\Rightarrow g(x)=\int\limits1dx=x}$
$\displaystyle \mathtt{\int\limits ln~xdx=ln~x \cdot x-\int\limits \left(\frac{1}{x}\cdot x\right)dx=xlnx-\int\limits1dx=xln~x-x+C=}\\ \\ \mathtt{=x(ln~x-1)+C}\\\\\mathtt{\int\limits ln~xdx=x(ln~x-1)+C}$
$\displaystyle \mathtt{\mathcal{A}=\int\limits_1^e\big(x-ln~x\big)dx=\int\limits_1^e xdx-\int\limits_1^e ln~xdx= \frac{x^2}{2}\Bigg|_1^e-x\big(ln~x -1\big)\Bigg|_1^e=}\\ \\ \mathtt{= \frac{e^2}{2}- \frac{1^2}{2}-\big[e\big(ln~e-1\big)-1\big(ln~1-1\big)\big]=}\\ \\ \mathtt{= \frac{e^2-1}{2}-\big[e\big(1-1\big)-1\big(0-1\big)\big]= \frac{e^2-1}{2}-\big[e \cdot 0-1 \cdot \big(-1\big)\big]=}\\ \\ \mathtt{= \frac{e^2-1}{2}-\big(0+1\big)= \frac{e^2-1}{2}-1= \frac{e^2-1-2}{2}= \frac{e^2-3}{2} }$

$\displaystyle \mathtt{\mathcal{A}=\frac{e^2-3}{2} }}$