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Cum se afle aria subgraficului funcţiei
f:[0,π]→R, f(x)=x+cos x


Răspuns :

 

[tex]\displaystyle\bf\\\textbf{aria subgraficului func\c{t}iei }~~f:[0,\pi]\to R,~~f(x)=x+cos x~~~este:\\\\\int\limits^{\b\pi}_{\b0} {x+cos\,x}~dx=\frac{~x^2}{2}+sin\,x~\Big|_0^\pi=\\\\\\=\left(\frac{~\pi^2}{2}+sin\,\pi\right)-\left(\frac{~0^2}{2}+sin\,0\right)=\\\\\\=\left(\frac{~\pi^2}{2}+0\right)-\left(0+0\right)=\boxed{\bf\frac{~\pi^2}{2}~cm^2}[/tex]