Question

Se consideră funcţia $f: \mathbb{R} \rightarrow \mathbb{R}, f(x)=x \sqrt{x^{2}+1}$.

$5 p$ a) Arătați că $\int_{1}^{e} \frac{\sqrt{x^{2}+1}}{f(x)} d x=1$.

$5 \mathbf{p}$ b) Calculați $\int_{1}^{2} f^{2}(x) d x$.

5p c) Demonstrați că $\int_{0}^{2020} f(x) d x \leq \int_{0}^{a} f(x) d x$, pentru orice $a \in[2020,+\infty)$.

Answer 1

$f(x)=x \sqrt{x^{2}+1}$

a)

$\int\limits^e_1{\frac{\sqrt{x^2+1} }{x\sqrt{x^2+1} } } \, dx =\int\limits^e_1 {\frac{1}{x} } \, dx =lnx|_1^e=lne-ln1=1-0=1$

(vezi tabel integrale in atasament)

b)

$\int\limits^2_1 {x^2(x^2+1)} \, dx =\int\limits^2_1 x^4+x^2\ dx=\int\limits^2_1x^4\ dx+\int\limits^2_1x^2\ dx=\frac{x^5}{5}|_1^2+\frac{x^3}{3}|_1^2=\\=\frac{32}{5}-\frac{1}{5}+\frac{8}{3}-\frac{1}{3} =\frac{31}{5}+\frac{7}{3}=\frac{128}{15}$

c)

a≥2020

Pentru x≥0, f(x)≥0

Deci $\int\limits^a_{2020} {f(x)} \, dx \geq 0$

$\int\limits^a_0 {f(x)} \, dx -\int\limits^{2020}_0 {f(x)} \, dx =\int\limits^{2020}_0 {f(x)} \, dx+\int\limits^a_{2020} {f(x)} \, dx -\int\limits^{2020}_0 {f(x)} \, dx=\int\limits^a_{2020} {f(x)} \ dx$

Am demonstrat mai sus ca integrala aceea este ≥0

Deci $\int\limits^a_{2020} {f(x)} \geq 0\\\\\int\limits^{2020}_0 {f(x)} \leq \int\limits^a_0{f(x)}$

Un alt exercitiu similar de bac il gasesti aici: https://brainly.ro/tema/4786722

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