f(x)=cosx
a)
[tex]\int_{0}^{\frac{\pi}{2}} \sin x f(x) d x=\frac{1}{2}[/tex]
[tex]\int_{0}^{\frac{\pi}{2}} \sin x\cdot cosx d x=\int_{0}^{\frac{\pi}{2}} \sin x\cdot (sinx)' d x[/tex]
Stim ca (sinx)'=cosx (conform tabel atasat)
(sin²u)'=2sin u·cos u (conform tabel atasat)
Vom adauga un 2 si "il vom da inapoi" (artificiu de calcul)
[tex]\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ 2sin x\cdot (sinx)' d x=\frac{1}{2} sin^2x\ |_0^{\frac{\pi}{2}}=\frac{1}{2} sin^2\frac{\pi}{2} -\frac{1}{2} sin^20= \frac{1}{2}[/tex]
b)
[tex]\lim_{x \to +\infty} \frac{1}{x}\int\limits^x_0 {cos\ t} \, dt[/tex]
Calculam integrala separat:
[tex]\int\limits^x_0 {cos\ t} \, dt =sin\ t_0^x=sinx-sin0=sinx[/tex]
Vom inlocui integrala aflata pentru a calcula limita
[tex]\lim_{x \to +\infty} \frac{1}{x}\times sinx=\lim_{x \to +\infty} \frac{sinx}{x}=0[/tex]
-1≤sinx≤1
x→+∞
[tex]\lim_{x \to +\infty} \frac{1}{x} =\frac{1}{+\infty}=0[/tex]
Un exercitiu similar cu integrale gasesti aici: https://brainly.ro/tema/2004386
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