Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=\cos x$.
$5 p$ a) Arătați că $\int_{0}^{\frac{\pi}{2}} \sin x f(x) d x=\frac{1}{2} .$
$5 p$ b) Calculați $\lim _{x \rightarrow+\infty} \frac{1}{x} \int_{0}^{x} f(t) d t$.

Answer 1

f(x)=cosx

a)

$\int_{0}^{\frac{\pi}{2}} \sin x f(x) d x=\frac{1}{2}$

$\int_{0}^{\frac{\pi}{2}} \sin x\cdot cosx d x=\int_{0}^{\frac{\pi}{2}} \sin x\cdot (sinx)' d x$

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)

$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \ 2sin x\cdot (sinx)' d x=\frac{1}{2} sin^2x\ |_0^{\frac{\pi}{2}}=\frac{1}{2} sin^2\frac{\pi}{2} -\frac{1}{2} sin^20= \frac{1}{2}$

b)

$\lim_{x \to +\infty} \frac{1}{x}\int\limits^x_0 {cos\ t} \, dt$

Calculam integrala separat:

$\int\limits^x_0 {cos\ t} \, dt =sin\ t_0^x=sinx-sin0=sinx$

Vom inlocui integrala aflata pentru a calcula limita

$\lim_{x \to +\infty} \frac{1}{x}\times sinx=\lim_{x \to +\infty} \frac{sinx}{x}=0$

-1≤sinx≤1

x→+∞

$\lim_{x \to +\infty} \frac{1}{x} =\frac{1}{+\infty}=0$

Un exercitiu similar cu integrale gasesti aici: https://brainly.ro/tema/2004386

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